*【CodeForces - 799C】Fountains （线段树或树状数组，类似二元组问题）

题干：

Arkady plays Gardenscapes a lot. Arkady wants to build two new fountains. There are n available fountains, for each fountain its beauty and cost are known. There are two types of money in the game: coins and diamonds, so each fountain cost can be either in coins or diamonds. No money changes between the types are allowed.

Help Arkady to find two fountains with maximum total beauty so that he can buy both at the same time.

Input

The first line contains three integers n, c and d (2 ≤ n ≤ 100 000, 0 ≤ c, d ≤ 100 000) — the number of fountains, the number of coins and diamonds Arkady has.

The next n lines describe fountains. Each of these lines contain two integers bi and pi (1 ≤ bi, pi ≤ 100 000) — the beauty and the cost of the i-th fountain, and then a letter "C" or "D", describing in which type of money is the cost of fountain i: in coins or in diamonds, respectively.

Output

Print the maximum total beauty of exactly two fountains Arkady can build. If he can't build two fountains, print 0.

Examples

Input

Output

Input

2 4 5
2 5 C
2 1 D

Output

Input

Output

Note

In the first example Arkady should build the second fountain with beauty 4, which costs 3 coins. The first fountain he can't build because he don't have enough coins. Also Arkady should build the third fountain with beauty 5 which costs 6 diamonds. Thus the total beauty of built fountains is 9.

In the second example there are two fountains, but Arkady can't build both of them, because he needs 5 coins for the first fountain, and Arkady has only 4 coins.

题目大意：

解题报告：

比赛的时候让这个线段树搞垮我了，，很多特殊情况要考虑，并且，最后卡住我的不是线段树那一块，而是之前的C和D两种货币各选一个的那种情况，我是直接，找到就break，但是其实应该是维护一个最大值。。。因为你直接break的话只能保证cost是可符合的最大的，而保证不了val的最大性！我们需要的是val！（%%%WJH大佬）

AC代码1：

#include<bits/stdc++.h>
using namespace std;
const int MAX = 100000 +5;
int n,c,d,totc,totd;
int cc[MAX];
int dd[MAX];
struct Node{int val,cost;Node(){}Node(int val,int cost):val(val),cost(cost){} 
} C[MAX],D[MAX];
struct TREE {int l,r;int val;
} tree[MAX*4]; 
bool cmp(const Node &a,const Node &b) {return a.cost < b.cost;
}
void pushup(int cur) {tree[cur].val = max(tree[cur*2].val,tree[cur*2+1].val);
}
void buildc(int l,int r,int cur) {tree[cur].l=l;tree[cur].r=r;if(l == r) {tree[cur].val = C[l].val;return ;}int m = (l +r)/2;buildc(l,m,cur*2);buildc(m+1,r,cur*2+1);pushup(cur);
}
void buildd(int l,int r,int cur) {tree[cur].l=l;tree[cur].r=r;if(l == r) {tree[cur].val = D[l].val;return ;}int m = (l +r)/2;buildd(l,m,cur*2);buildd(m+1,r,cur*2+1);pushup(cur);
}
int query(int pl,int pr,int cur) {if(pl <= tree[cur].l && pr >= tree[cur].r) return tree[cur].val;int res = 0;if(pl <= tree[cur*2].r) res = query(pl,pr,cur*2);if(pr >= tree[cur*2+1].l) res = max(res,query(pl,pr,cur*2+1));return res;
}
void update(int tar,int val,int cur) {if(tree[cur].l == tree[cur].r) {tree[cur].val = val;return ;}int m = (tree[cur].l + tree[cur].r)/2;if(tar <= m) update(tar,val,cur*2);else update(tar,val,cur*2+1);pushup(cur);
}
signed main()
{char op[10];int val,cost,flag1=0,flag2=0;cin>>n>>c>>d;for(int i = 1; i<=n; i++) {scanf("%d%d%s",&val,&cost,op);if(op[0] == 'C') {C[++totc] = Node(val,cost);}else {D[++totd] = Node(val,cost);}}sort(C+1,C+totc+1,cmp);	sort(D+1,D+totd+1,cmp);int maxx=0,tmp1 = 0,tmp2 = 0,pos;for(int i = totc; i>=1; i--) {if(C[i].cost <= c) {flag1=1;tmp1 = max(tmp1,C[i].val);}}for(int i = totd; i>=1; i--) {if(D[i].cost <= d) {flag2=1;tmp2 = max(tmp2,D[i].val);}}if(flag1==0 && flag2 == 0) {printf("0\n");return 0;} if(flag1 == 1 && flag2 == 1) {maxx = tmp1 + tmp2;}if(totc == 1 && totd == 1) {if(C[1].cost > c || D[1].cost > d) {printf("0\n");return 0;}}for(int i = 1; i<=totc; i++) cc[i] = C[i].cost;for(int i = 1; i<=totd; i++) dd[i] = D[i].cost;	
//	printf("maxx = %d\n",maxx);//处理Cif(totc != 0 ) {buildc(1,totc,1);for(int i =1; i<=totc; i++) {if(c-cc[i] < 0) break; update(i,0,1);pos = upper_bound(cc+1,cc+totc + 1,c-cc[i]) - cc-1;if(pos == 0) continue;maxx = max(maxx,C[i].val + query(1,pos,1));update(i,C[i].val,1);}}for(int i = 1; i<=MAX; i++) update(i,0,1);
//	printf("%d  %d\n",query(1,1,1),query(1,2,1));if(totd !=0) {buildd(1,totd,1);for(int i =1; i<=totd; i++) {if(d-dd[i] < 0) break;update(i,0,1);pos = upper_bound(dd+1,dd+totd + 1,d-dd[i])- dd -1;if(pos == 0) continue;//		if(dd[pos] > d-dd[i]) pos--;maxx = max(maxx,D[i].val + query(1,pos,1));update(i,D[i].val,1);}}printf("%d\n",maxx);return 0 ;
}
//6 4 9
//6 6 D
//1 4 D
//6 7 C
//7 6 D
//5 7 D
//2 5 D//6 68 40
//1 18 D
//6 16 D
//11 16 D
//7 23 D
//16 30 D
//2 20 D//3 9 5
//10 7 C
//8 6 C
//5 3 C

AC代码2：（o(n^2)）

#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int MAX = 1e5 + 5 ;
int n,c,d,ans;
struct Node {int cost,val;char op[5];
} node[MAX];
bool cmp(const Node & a,const Node & b) {if(a.val != b.val) return a.val > b.val;return a.cost < b.cost;
}
int main()
{cin>>n>>c>>d;for(int i = 1; i<=n; i++) {scanf("%d%d%s",&node[i].val,&node[i].cost,node[i].op);}sort(node+1,node+n+1,cmp);int cc,dd,flag,tmp;for(int i = 1; i<=n; i++) {cc = c;dd = d;flag = 0 ;if(node[i].op[0] == 'C' && node[i].cost <= c) {flag=1;cc-=node[i].cost;tmp=node[i].val;}else if(node[i].op[0] == 'D' && node[i].cost <= d) {flag=1;dd-=node[i].cost;tmp = node[i].val;}if(flag == 0) continue;for(int j = i+1; j<=n; j++) {if(node[j].op[0] == 'C' && node[j].cost <= cc) {ans = max(ans,tmp+node[j].val);break;}else if(node[j].op[0] == 'D' && node[j].cost <= dd) {ans = max(ans,tmp+node[j].val);break;}}}printf("%d\n",ans);return 0 ;
}

AC代码3：（树状数组）

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>using namespace std;const int maxn = 100000;int C[maxn+10],D[maxn+10];void add(int *tree,int k,int num)
{while(k<=maxn){tree[k] = max(tree[k],num);k+=k&-k;}
}int read(int *tree,int k)
{int res=0;while(k){res = max(res,tree[k]);k-=k&-k;}return res;
}int main(void)
{int n,c,d,i,j;while(scanf("%d%d%d",&n,&c,&d)==3){memset(C,0,sizeof(C));memset(D,0,sizeof(D));int ans = 0;for(i=1;i<=n;i++){int b,p;char t[5];scanf("%d%d%s",&b,&p,t);int maxn;if(t[0] == 'C'){maxn = read(D,d);if(p > c)continue;maxn = max(maxn,read(C,c-p));add(C,p,b);}else{maxn = read(C,c);if(p > d)continue;maxn = max(maxn,read(D,d-p));add(D,p,b);}if(maxn)ans = max(ans,maxn + b);}cout << ans << endl;}return 0;
}

AC代码4：（tourist）

#include <bits/stdc++.h>using namespace std;const int N = 1234567;int pref[N], a[N], b[N];
char c[N];int solve(vector < pair <int, int> > z, int b) {sort(z.begin(), z.end());int cnt = z.size();if (cnt < 2) {return 0;}pref[0] = 0;for (int i = 0; i < cnt; i++) {pref[i + 1] = max(pref[i], z[i].second);}int i = 0;int res = 0;for (int j = cnt - 1; j >= 0; j--) {while (i < j && z[i].first + z[j].first <= b) {i++;}i = min(i, j);if (i > 0) {res = max(res, pref[i] + z[j].second);}}return res;
}int main() {int n, C, D;scanf("%d %d %d", &n, &C, &D);for (int i = 0; i < n; i++) {scanf("%d %d", a + i, b + i);c[i] = getchar();while (c[i] != 'C' && c[i] != 'D') {c[i] = getchar();}}int ans = 0;{int x = 0, y = 0;for (int i = 0; i < n; i++) {if (c[i] == 'C' && b[i] <= C) {x = max(x, a[i]);}}for (int i = 0; i < n; i++) {if (c[i] == 'D' && b[i] <= D) {y = max(y, a[i]);}}if (x > 0 && y > 0) {ans = max(ans, x + y);}}{vector < pair <int, int> > z;for (int i = 0; i < n; i++) {if (c[i] == 'C') {z.push_back(make_pair(b[i], a[i]));}}ans = max(ans, solve(z, C));}{vector < pair <int, int> > z;for (int i = 0; i < n; i++) {if (c[i] == 'D') {z.push_back(make_pair(b[i], a[i]));}}ans = max(ans, solve(z, D));}printf("%d\n", ans);return 0;
}

AC代码5：（在线搞，以cost建树（而不是排好序后用下标建树），维护最大值）

#include<cstdio>
#include<algorithm>
#include<cstring>
#define ll long long
using namespace std;
const int maxn = 100010;
int a[maxn],acost[maxn],b[maxn],bcost[maxn];
struct node
{int Max;int l,r;
}t[maxn*4];
void build(int root,int l,int r)
{t[root].l=l;t[root].r=r;//t[root].Max=-1;if(l==r)return;int mid=(l+r)>>1;build(2*root,l,mid);build(2*root+1,mid+1,r);
}
void update(int root,int index,int val)
{int l=t[root].l,r=t[root].r;if(l==r){t[root].Max=max(t[root].Max,val);return;//}int mid=(l+r)>>1;if(index<=mid)update(2*root,index,val);elseupdate(2*root+1,index,val);t[root].Max=max(t[root*2].Max,t[root*2+1].Max);
}
int query(int root,int ql,int qr)
{int l=t[root].l;int r=t[root].r;if(l==ql&&r==qr){return t[root].Max;}int mid=(l+r)>>1;if(qr<=mid)return query(2*root,ql,qr);else if(ql>mid)return query(2*root+1,ql,qr);elsereturn max(query(2*root,ql,mid),query(2*root+1,mid+1,qr));//2*root+1
}
int main()
{int N,C,D;while(~scanf("%d%d%d",&N,&C,&D)){int acnt=0,bcnt=0;char ch;int c,w;int Maxc=-1;for(int i=0;i<N;i++){scanf("%d%d %c",&c,&w,&ch);if(ch=='C'){a[acnt]=c;acost[acnt++]=w;}else if(ch=='D'){b[bcnt]=c;bcost[bcnt++]=w;}Maxc=max(w,Maxc);}int ans=-1;int res1=-1,res2=-1;for(int i=0;i<acnt;i++){if(C>=acost[i])res1=max(res1,a[i]);}for(int i=0;i<bcnt;i++){if(D>=bcost[i])res2=max(res2,b[i]);}if(res1!=-1&&res2!=-1)ans=res1+res2;Maxc=max(Maxc,max(C,D));build(1,1,Maxc+10);for(int i=0;i<acnt;i++){int res=-1;if(C>acost[i])res=query(1,1,C-acost[i]);update(1,acost[i],a[i]);if(res!=-1)ans=max(ans,a[i]+res);}build(1,1,Maxc);for(int i=0;i<bcnt;i++){int res=-1;if(D>bcost[i])res=query(1,1,D-bcost[i]);update(1,bcost[i],b[i]);if(res!=-1)ans=max(ans,b[i]+res);}if(ans!=-1)printf("%d\n",ans);elseprintf("0\n");}
}

AC代码6：（维护最大价值和次大价值 + 前缀和）（感觉这个题解有问题？）

#include<bits/stdc++.h>
using namespace std;
#pragma comment(linker, "/STACK:102400000,102400000")
#define rep(i,a,b) for (int i=a; i<=b; ++i)
#define per(i,b,a) for (int i=b; i>=a; --i)
#define mes(a,b)  memset(a,b,sizeof(a))
#define INF 0x3f3f3f3f
#define MP make_pair
#define PB push_back
#define fi  first
#define se  second
#define PII  pair<ll , ll >
typedef long long ll;
const int N = 100005;int n;
ll  C, D;
PII  c[N], d[N];
ll ans1[N][2], ans2[N][2];
int  cnt1=0, cnt2=0;
//init函数中均可加等号
void Init()
{sort(c+1, c+1+cnt1);sort(d+1, d+1+cnt2);ll mx1=0, mx2=0;rep(i,1,cnt1){if(mx1 < c[i].se){mx2=mx1,  mx1=c[i].se;ans1[c[i].fi][0]=mx1;ans1[c[i].fi][1]=mx2;}else if(mx2 < c[i].se){mx2=c[i].se;ans1[c[i].fi][0]=mx1;//可注释ans1[c[i].fi][1]=mx2;}}mx1=0, mx2=0;rep(i,1,cnt2){if(mx1 < d[i].se){mx2=mx1,  mx1=d[i].se;ans2[d[i].fi][0]=mx1;ans2[d[i].fi][1]=mx2;}else if(mx2 < d[i].se){mx2=d[i].se;ans2[d[i].fi][0]=mx1;//可注释ans2[d[i].fi][1]=mx2;}}rep(i,1,max(C,D)) rep(j,0,1){ans1[i][j]=max(ans1[i][j], ans1[i-1][j]);ans2[i][j]=max(ans2[i][j], ans2[i-1][j]);}
}
int main()
{scanf("%d%lld%lld", &n, &C, &D);ll bi, pi;    char ch;rep(i,1,n){scanf("%lld%lld%*c%c", &bi, &pi, &ch);if(ch=='C') c[++cnt1]=MP(pi, bi);else d[++cnt2]=MP(pi, bi);}Init();ll sum1=0;if(ans1[C][0] && ans2[D][0])sum1 = max(sum1, ans1[C][0]+ans2[D][0]);ll sum2=0, sum3=0;rep(i,1,cnt1){ll  ret=C-c[i].fi;if(ret>0){if(ans1[ret][0]==c[i].se){if(ans1[ret][1])sum2 = max(sum2, c[i].se+ans1[ret][1]);}else{if(ans1[ret][0])sum2 = max(sum2, c[i].se+ans1[ret][0]);}}}rep(i,1,cnt2){ll  ret=D-d[i].fi;if(ret>0){if(ans2[ret][0]==d[i].se){if(ans2[ret][1])sum3 = max(sum3, d[i].se+ans2[ret][1]);}else{if(ans2[ret][0])sum3 = max(sum3, d[i].se+ans2[ret][0]);}}}printf("%lld\n", max(sum1, max(sum2, sum3)));return 0;
}

AC代码7：（还是自己写的那个线段树，只是将upperbound部分进行了优化，快了将近50ms）

#include<bits/stdc++.h>
using namespace std;
const int MAX = 100000 +5;
int n,c,d,totc,totd;
int cc[MAX];
int dd[MAX];
struct Node{int val,cost;Node(){}Node(int val,int cost):val(val),cost(cost){} 
} C[MAX],D[MAX];
struct TREE {int l,r;int val;
} tree[MAX*4]; 
bool cmp(const Node &a,const Node &b) {if(a.cost != b.cost)return a.cost < b.cost;return a.val > b.val;
}
void pushup(int cur) {tree[cur].val = max(tree[cur*2].val,tree[cur*2+1].val);
}
void buildc(int l,int r,int cur) {tree[cur].l=l;tree[cur].r=r;if(l == r) {tree[cur].val = C[l].val;return ;}int m = (l +r)/2;buildc(l,m,cur*2);buildc(m+1,r,cur*2+1);pushup(cur);
}
void buildd(int l,int r,int cur) {tree[cur].l=l;tree[cur].r=r;if(l == r) {tree[cur].val = D[l].val;return ;}int m = (l +r)/2;buildd(l,m,cur*2);buildd(m+1,r,cur*2+1);pushup(cur);
}
int query(int pl,int pr,int cur) {if(pl <= tree[cur].l && pr >= tree[cur].r) return tree[cur].val;int res = 0;if(pl <= tree[cur*2].r) res = query(pl,pr,cur*2);if(pr >= tree[cur*2+1].l) res = max(res,query(pl,pr,cur*2+1));return res;
}
void update(int tar,int val,int cur) {if(tree[cur].l == tree[cur].r) {tree[cur].val = val;return ;}int m = (tree[cur].l + tree[cur].r)/2;if(tar <= m) update(tar,val,cur*2);else update(tar,val,cur*2+1);pushup(cur);
}
int main()
{char op[10];int val,cost,flag1=0,flag2=0;cin>>n>>c>>d;for(int i = 1; i<=n; i++) {scanf("%d%d%s",&val,&cost,op);if(op[0] == 'C') {C[++totc] = Node(val,cost);}else {D[++totd] = Node(val,cost);}}sort(C+1,C+totc+1,cmp);	sort(D+1,D+totd+1,cmp);int maxx=0,tmp1 = 0,tmp2 = 0,pos;for(int i = totc; i>=1; i--) {if(C[i].cost <= c) {flag1=1;tmp1 = max(tmp1,C[i].val);}}for(int i = totd; i>=1; i--) {if(D[i].cost <= d) {flag2=1;tmp2 = max(tmp2,D[i].val);}}if(flag1==0 && flag2 == 0) {printf("0\n");return 0;} if(flag1 == 1 && flag2 == 1) {maxx = tmp1 + tmp2;}if(totc == 1 && totd == 1) {if(C[1].cost > c || D[1].cost > d) {printf("0\n");return 0;}}for(int i = 1; i<=totc; i++) cc[i] = C[i].cost;for(int i = 1; i<=totd; i++) dd[i] = D[i].cost;
//	printf("maxx = %d\n",maxx);//处理Cif(totc != 0 ) {buildc(1,totc,1);for(int i =1; i<totc; i++) {if(c-cc[i] < 0) break; 
//			update(i,0,1);pos = upper_bound(cc+i+1,cc+totc + 1,c-cc[i]) - (cc/*+i*/)-1;if(pos == i) continue;maxx = max(maxx,C[i].val + query(i+1,pos,1));
//			update(i,C[i].val,1);}}
//	printf("%d  %d\n",query(1,1,1),query(1,2,1));if(totd !=0) {buildd(1,totd,1);for(int i =1; i<totd; i++) {if(d-dd[i] < 0) break;
//			update(i,0,1);pos = upper_bound(dd+i+1,dd+totd + 1,d-dd[i])- (dd/*+i*/) -1;if(pos == i) continue;//		if(dd[pos] > d-dd[i]) pos--;maxx = max(maxx,D[i].val + query(i+1,pos,1));
//			update(i,D[i].val,1);}}printf("%d\n",maxx);return 0 ;
}
//6 4 9
//6 6 D
//1 4 D
//6 7 C
//7 6 D
//5 7 D
//2 5 D//6 68 40
//1 18 D
//6 16 D
//11 16 D
//7 23 D
//16 30 D
//2 20 D//3 9 5
//10 7 C
//8 6 C
//5 3 C

总结：

对于这一道经典的题目，我想，可以总结的东西还是很多的。

对于任何数据结构（包括线段树等，包括栈队列等），首先需要看的就是能否进入，或者说，这项功能能否执行的问题。比如栈，进行pop和top操作的时候要想，是否需要先判断是否栈为空呢？因为为空了就没法进行操作了呀。同理，对于线段树，我们也要想，我们能否建树呢？建树后，能否进行查询呢？即，这个传参是合法的吗？（这两条思考，分别对应了AC代码1中的，if(totc!=0),,,,if(c-cc[i] < 0) break; 这两句）

其次是这一块的处理，也经常容易忽略：（一个比较可行的解决办法就是自己造数据的时候这些边界情况都考虑一下）（好像说了一堆废话）

	if(flag1==0 && flag2 == 0) {printf("0\n");return 0;} if(flag1 == 1 && flag2 == 1) {maxx = tmp1 + tmp2;}if(totc == 1 && totd == 1) {if(C[1].cost > c || D[1].cost > d) {printf("0\n");return 0;}}

将ac代码1修改成AC代码7：

改1：for(int i =1; i<totc; i++) {

改2：if(pos == i) continue;

改3：maxx = max(maxx,C[i].val + query(i+1,pos,1));//这里query中是从i+1开始的

改4：每个for中都可以注释掉两个update函数。（因为就没用了）

还有一些可总结的：

lowerbound upperbound的时候，如果不是对整个数列进行，比如是对[ l , r ]进行，那么upper的可能值就是[ l , r+1 ]这个区间内的值。（即upper出来的pos，肯定大于等于l，小于等于r+1），同理 lowerbound出来的值略有不同，区间是[ l , r ]（我感觉是这样？）

还有就是，对[ l , r ]进行的时候，但是还是要 -a 而不是-(a+i-1) 也就是int pos = lower_bound(a+i,a+i+1,val) - a;这样就可以，因为你最后需要的还是在原数组中的下标呀。

总之做题还是要在仔细一些吧。

另一个代码:https://blog.csdn.net/SSimpLe_Y/article/details/71702915