题意：你的任务是在n*n的棋盘上放 n 小于5000 个车 使得任意两个车不互相攻击 且第i个车在一个给定的矩形ri之内 给出该矩形左上角坐标和右下角坐标四个点 必须满足放车的位置在矩形内 边上也行 如果无解输出IMPSSIBLE
行与列是独立的 所以可以分割成两个一模一样的子问题 贪心
要以右边界升序排序 我一开始按照左边界排序错了 举个例子 1-1 1-3 2-2 这样的话就会错 1-1 2-2 1-3才对
还有就是注意细节 sort 从一开始的话 都要加一，，，
#include<bits/stdc++.h> using namespace std; #define N 5010 int n,k; int vis[N]; struct node {int id;int x,y;int x1,x2,y1,y2;}chess[N];bool cmp1(node a,node b){return a.x2<b.x2||(a.x2==b.x2&&a.x1<b.x1);} bool cmp2(node a,node b){return a.y2<b.y2||(a.y2==b.y2&&a.y1<b.y1);} bool cmp3(node a,node b){return a.id<b.id;} int main() {while(cin>>n,n){int flag=1;for(int i=0;i<n;i++){chess[i].id=i;scanf("%d%d%d%d",&chess[i].x1,&chess[i].y1,&chess[i].x2,&chess[i].y2);}sort(chess,chess+n,cmp1);memset(vis,0,sizeof vis);for(int i=0;i<n;i++){int ok=0;for(int j=chess[i].x1;j<=chess[i].x2;j++){if(!vis[j]){ok=1;vis[j]=1;chess[i].x=j;break; }//把chess里面的i写成了j 强行将自己dubug了半个小时。。。 }if(!ok){flag=0;}}sort(chess,chess+n,cmp2);memset(vis,0,sizeof vis);for(int i=0;i<n;i++){int ok=0;for(int j=chess[i].y1;j<=chess[i].y2;j++){if(!vis[j]){ok=1;vis[j]=1;chess[i].y=j;break; }}if(!ok){flag=0;}}sort(chess,chess+n,cmp3);if(flag)for(int i=0;i<n;i++)printf("%d %d\n",chess[i].x,chess[i].y);else printf("IMPOSSIBLE\n");} }
LRJ的代码 更慢
#include<cstdio> #include<cstring> #include <algorithm> using namespace std;// solve 1-D problem: find c so that a[i] <= c[i] <= b[i] (0 <= i < n) bool solve(int *a, int *b, int *c, int n) {fill(c, c+n, -1);for(int col = 1; col <= n; col++) {// find a rook with smalleset b that is not yet assignedint rook = -1, minb = n+1;for(int i = 0; i < n; i++)if(c[i] < 0 && b[i] < minb && col >= a[i]) { rook = i; minb = b[i]; }if(rook < 0 || col > minb) return false;c[rook] = col;}return true; }const int maxn = 5000 + 5; int n, x1[maxn], y1[maxn], x2[maxn], y2[maxn], x[maxn], y[maxn];int main() {while(scanf("%d", &n) == 1 && n) {for (int i = 0; i < n; i++)scanf("%d%d%d%d", &x1[i], &y1[i], &x2[i], &y2[i]);if(solve(x1, x2, x, n) && solve(y1, y2, y, n))for (int i = 0; i < n; i++) printf("%d %d\n", x[i], y[i]);elseprintf("IMPOSSIBLE\n");}return 0; }
