题意:给定一个矩形网格,再给你k个点和这几个点的覆盖范围(曼哈顿距离),让你求选择最少的点覆盖整个网格
解题思路:爆搜!(数据量大小考虑不周到,WA了一次)
解题代码:
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
// File Name: j.c // Author: darkdream // Created Time: 2013年06月01日 星期六 09时24分43秒 #include<stdio.h> #include<string.h> #include<stdlib.h> #include<time.h> #include<math.h> struct point { int x, y; }; struct yuan { int x, y; int k ; struct point points[3000]; }yuans[12]; int Visit[100][100]; int n , last,ans, ok,t; int dis(int i ,int x, int y ) { return abs(yuans[i].x-x ) + abs(yuans[i].y-y); } void expend(int k ,int temp ) { yuans[k].k =0 ; for(int i = 1;i <= n;i++) { for(int j = 1; j <= n; j ++) if(dis(k,i,j) <= temp ) { yuans[k].points[yuans[k].k].x = i ; yuans[k].points[yuans[k].k].y = j ; yuans[k].k++; } } } void dfs(int k , int step ,int sum) { if(last == sum) { if(step < ans) ans = step; ok = 1; return ; } if( step >= ans) return; for(int s = k+1; s <= t; s++) { int temp =0 ; struct point points[3000] = {0}; for(int i = 0 ;i < yuans[s].k ; i ++) { if(!Visit[yuans[s].points[i].x][yuans[s].points[i].y]) { points[temp].x = yuans[s].points[i].x; points[temp].y = yuans[s].points[i].y; Visit[points[temp].x][points[temp].y] = 1; temp++; } } dfs(s,step+1,sum+temp); for(int i = 0 ;i < temp; i++) Visit[points[i].x][points[i].y] = 0; } return ; } int main(){ //freopen("/home/plac/problem/input.txt","r",stdin); //freopen("/home/plac/problem/output.txt","w",stdout); while(scanf("%d",&n) != EOF) { if(n == 0 ) break; memset(yuans,0,sizeof(yuans)); memset(Visit,0,sizeof(Visit)); scanf("%d",&t); last = n*n - t; ans = t; ok = 0 ; for(int i = 1;i <= t;i ++) { scanf("%d %d",&yuans[i].x,&yuans[i].y); Visit[yuans[i].x][yuans[i].y] = 1; } int temp; for(int i =1 ;i <= t; i ++) { scanf("%d",&temp); expend(i,temp); } dfs(0,0,0); if(1 == ok) printf("%d\n",ans); else printf("-1\n"); } return 0 ; }
没有梦想,何谈远方