NOIP 模拟赛总结
NOIP 模拟赛 6 & 多校 1
挂分啦
T1 汉谟拉比(crazy)
不简单签到题，但是放过了 \(O(n m ^ 2)\)。
正解是发现要进行 \(n\) 次卷积。
所以考虑广义快速幂即可。
点击查看代码
#include <bits/stdc++.h>
#define con putchar_unlocked(' ')
#define ent putchar_unlocked('\n')
#define Blue_Archive return 0
using namespace std;
constexpr int N = 5e3 + 3;
constexpr int INF = 1e9;int n;
int m;
int mx;
int ans;
int a[N];
int w[N];
int dp[N][N];inline int read() // 不读负数的普通快读
{int k = 0,f = 1;int c = getchar_unlocked();while(c < '0' || c > '9') c = getchar_unlocked();while(c >= '0' && c <= '9') k = (k << 3) + (k << 1) + (c ^ 48),c = getchar_unlocked();return k * f;
}inline void write(int x) // 普通快写
{if(x < 0) putchar_unlocked('-'),x = -x;if(x > 9) write(x / 10);putchar_unlocked(x % 10 + '0');
}inline int max(int a,int b){return a > b ? a : b;}
inline int min(int a,int b){return a < b ? a : b;}signed main()
{freopen("crazy.in","r",stdin);freopen("crazy.out","w",stdout);// freopen("data.in","r",stdin);freopen("ans.out","w",stdout);n = read();m = read();for(int i = 1;i <= n;i ++){a[i] = read();w[a[i]] ++;mx = max(mx,a[i]);} mx = min(mx,m);for(int i = 1;i <= mx;i ++) w[i] += w[i - 1];if(1ll * n * mx * m <= 1e9){for(int i = 1;i <= n;i ++){for(int k = 0;k <= mx;k ++){for(int j = k;j <= m;j ++){if(dp[i][j] < dp[i - 1][j - k] + w[k]) dp[i][j] = dp[i - 1][j - k] + w[k];}}}}else {for(int i = 1;i <= n;i ++){for(int k = 0;k <= min(800,mx);k ++){for(int j = k;j <= m;j ++){if(dp[i][j] < dp[i - 1][j - k] + w[k]) dp[i][j] = dp[i - 1][j - k] + w[k];}}for(int k = max(mx - 800,0);k <= mx;k ++){for(int j = k;j <= m;j ++){if(dp[i][j] < dp[i - 1][j - k] + w[k]) dp[i][j] = dp[i - 1][j - k] + w[k];}}}}write(n * n - dp[n][m]),ent;Blue_Archive;
}
T2 虫洞折跃(flip)
判无解死亡导致又挂 20 pts。
简单 dp + 超级多判无解。
点击查看代码
#include <bits/stdc++.h>
#define int long long
#define con putchar_unlocked(' ')
#define ent putchar_unlocked('\n')
#define Blue_Archive return 0
using namespace std;
constexpr int N = 1e3 + 3;
constexpr int M = 1e6 + 3;
constexpr int INF = 1e9;int T;
int n;
int m;
int ans;
int dis[N][N];bool a[N][N];inline int read() // 不读负数的普通快读
{int k = 0,f = 1;int c = getchar_unlocked();while(c < '0' || c > '9') c = getchar_unlocked();while(c >= '0' && c <= '9') k = (k << 3) + (k << 1) + (c ^ 48),c = getchar_unlocked();return k * f;
}inline void write(int x) // 普通快写
{if(x < 0) putchar_unlocked('-'),x = -x;if(x > 9) write(x / 10);putchar_unlocked(x % 10 + '0');
}inline int max(int a,int b){return a > b ? a : b;}
inline int min(int a,int b){return a < b ? a : b;}inline void work()
{n = read();m = read();for(int i = 1;i <= n;i ++){for(int j = 1;j <= m;j ++){a[i][j] = read();}}if(n == 1 && m == 1 && a[1][1]) return void(puts("Impossible"));if(n == 1 && m == 2 && a[1][1] + a[1][2] == 1) return void(puts("Impossible"));if(n == 2 && m == 1 && a[1][1] + a[2][1] == 1) return void(puts("Impossible"));if(n == 1 && m == 4 && a[1][4] && !a[1][1] && !a[1][2] && !a[1][3]) return void(puts("2"));memset(dis,0x3f,sizeof(dis));dis[1][1] = a[1][1];for(int i = 1;i <= n;i ++){for(int j = 1;j <= m;j ++){if(a[i][j]) {if(i < n) dis[i + 1][j] = min(dis[i + 1][j],dis[i][j]);if(j < m) dis[i][j + 1] = min(dis[i][j + 1],dis[i][j]);}else {if(i < n && a[i + 1][j]) dis[i + 1][j] = min(dis[i + 1][j],dis[i][j] + 1);if(i < n && !a[i + 1][j]) dis[i + 1][j] = min(dis[i + 1][j],dis[i][j]);if(j < m && a[i][j + 1]) dis[i][j + 1] = min(dis[i][j + 1],dis[i][j] + 1);if(j < m && !a[i][j + 1]) dis[i][j + 1] = min(dis[i][j + 1],dis[i][j]);}}}if(dis[n][m] >= INF) puts("Impossible");else write(dis[n][m]),ent;
}signed main()
{freopen("flip.in","r",stdin);freopen("flip.out","w",stdout);// freopen("data.in","r",stdin);freopen("data.out","w",stdout);T = read();while(T --) work();Blue_Archive;
}
T3：深巢温泉(dnspring)
不会，咕咕。
T4：逃离冰场(skate)
神秘图论建模题。
发现每到相邻的一格需要 2 步，
每到最近的冰块的前一格需要一步。
然后直接连边跑最短路即可。
点击查看代码
#include <iostream>
#include <string.h>
#include <queue>
#define id(u,v) ((u - 1) * m + v)
#define add(u,v,c) to[++ tot] = v,w[tot] = c,nxt[tot ] = h[u],h[u] = tot
#define con putchar_unlocked(' ')
#define ent putchar_unlocked('\n')
#define Blue_Archive return 0
using namespace std;
constexpr int N = 1e3 + 3;
constexpr int L = 1e6 + 3;
constexpr int M = 1e7 + 3;
constexpr int INF = 1e9;int n;
int m;
int tot;
int ans;
int sx,sy;
int ex,ey;
int h[L];
int w[M];
int to[M];
int nxt[M];
int dis[L];bool vis[M];
bool a[N][N];inline int max(int a,int b){return a > b ? a : b;}
inline int min(int a,int b){return a < b ? a : b;}priority_queue<pair<int,int>,vector<pair<int,int>>,greater<pair<int,int>>> q;inline void dijk(int s)
{memset(dis,0x3f,sizeof(dis));dis[s] = 0;q.push({0,s});while(!q.empty()){int u = q.top().second;q.pop();if(vis[u]) continue;vis[u] = 1;for(int i = h[u];i;i = nxt[i]){if(dis[to[i]] > dis[u] + w[i]){dis[to[i]] = dis[u] + w[i];q.push({dis[to[i]],to[i]});}}}
}signed main()
{freopen("skate.in","r",stdin);freopen("skate.out","w",stdout);// freopen("data.in","r",stdin);freopen("data.out","w",stdout);ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);char op;cin >> n >> m;for(int i = 1;i <= n;i ++){for(int j = 1;j <= m;j ++){cin >> op;if(op == '#') a[i][j] = 1;}}for(int i = 1;i <= n;i ++){for(int j = 1;j <= m;j ++){if(a[i][j]) continue;if(!a[i - 1][j]) add(id(i - 1,j),id(i,j),2);if(!a[i + 1][j]) add(id(i + 1,j),id(i,j),2);if(!a[i][j - 1]) add(id(i,j - 1),id(i,j),2);if(!a[i][j + 1]) add(id(i,j + 1),id(i,j),2);}}for(int i = 1,las = -1;i <= n;i ++){for(int j = 2;j <= m;j ++){if(a[i][j]) continue;if(a[i][j - 1]) las = j;else add(id(i,j),id(i,las),1);}for(int j = m - 1;j >= 1;j --){if(a[i][j]) continue;if(a[i][j + 1]) las = j;else add(id(i,j),id(i,las),1);}}for(int i = 1,las = -1;i <= m;i ++){for(int j = 2;j <= n;j ++){if(a[j][i]) continue;if(a[j - 1][i]) las = j;else add(id(j,i),id(las,i),1);}for(int j = n - 1;j >= 1;j --){if(a[j][i]) continue;if(a[j + 1][i]) las = j;else add(id(j,i),id(las,i),1);}}cin >> sx >> sy;cin >> ex >> ey;dijk(id(sx,sy));ans = (dis[id(ex,ey)] >= INF ? -1 : dis[id(ex,ey)]);cout << ans << '\n';Blue_Archive;
}
