hdu 4462 Scaring the Birds 解题报告

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：

// 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 ;
}