hdu4462--曼哈顿距离

题目大意：有N*N个点的田野,然后有k个点是用来放稻草人的，每个稻草人对周围满足曼哈顿距离的庄稼有保护作用

问最小的稻草人的个数能够保护所有庄稼,如果不能保护则输出-1

注意的地方：

1.放稻草人的点不需要计算，因为不是庄稼

2.可能存在0的解，也就是k=N*N时

思路：二进制枚举所有情况,找到最小解

代码如下:

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#include<cstdio> #include<cstring> #include<cmath> #include<iostream> using namespace std; const int maxs = 51; int n,k; bool vis[maxs][maxs]; struct Point {    int x,y;    int dis; }point[11];   bool judge(Point p[],int t,Point goal) {     for(int i=0;i<t;i++)         //满足曼哈顿距离的条件         if(fabs(p[i].x-goal.x)+fabs(p[i].y-goal.y)<=p[i].dis)             return true;     return false; }   bool solve(Point p[],int t) {     memset(vis,false,sizeof(vis));     for(int i=0;i<k;i++)         vis[point[i].x][point[i].y]=true;     Point goal;     for(int x=1;x<=n;x++)         for(int y=1;y<=n;y++)             if(!vis[x][y])             {                 goal.x=x; goal.y=y;                 if(!judge(p,t,goal))                     return false;             }     return true; } int main() {     freopen("in.txt","r",stdin);     while(scanf("%d",&n)!=EOF&&n)     {         int ans = 20;//因为最大为10         scanf("%d",&k);         for(int i=0;i<k;i++)             scanf("%d%d",&point[i].x,&point[i].y);         for(int i=0;i<k;i++)             scanf("%d",&point[i].dis);         if(k==n*n)         {             printf("0\n");             continue;         }         int total = 1<<k;         //二进制枚举         for(int i=1;i<total;i++)         {             int temp;             Point p[11];             memset(p,0,sizeof(p));             int cnt=0;             for(int j=0,s=1;j<k;j++)             {                 if(i&s)                     p[cnt++]=point[j];                 s=s<<1;             }             if(solve(p,cnt))             {                 temp=cnt;                 if(temp<ans)                     ans=temp;             }         }         if(ans==20)             printf("-1\n");         else             printf("%d\n",ans);     } }