HDU 4462Scaring the Birds(枚举所有状态)

Scaring the Birds

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1257    Accepted Submission(s): 420

Problem Description

It’s harvest season now! 

Farmer John plants a lot of corn. There are many birds living around his corn field. These birds keep stealing his corn all the time. John can't stand with that any more. He decides to put some scarecrows in the field to drive the birds away. 

John's field can be considered as an N×N grid which has N×N intersections. John plants his corn on every intersection at first. But as time goes by, some corn were destroyed by rats or birds so some vacant intersections were left. Now John wants to put scarecrows on those vacant intersections and he can put at most one scarecrow on one intersection. Because of the landform and the different height of corn, every vacant intersections has a scaring range R meaning that if John put a scarecrow on it, the scarecrow can only scare the birds inside the range of manhattan distance R from the intersection.

The figure above shows a 7×7 field. Assuming that the scaring range of vacant intersection (4,2) is 2, then the corn on the marked intersections can be protected by a scarecrow put on intersection (4,2). 

Now John wants to figure out at least how many scarecrows he must buy to protect all his corn.

 

Input

There are several test cases. 

For each test case: 

The first line is an integer N ( 2 <= N <= 50 ) meaning that John's field is an N×N grid. 

The second line is an integer K ( 0<= K <= 10) meaning that there are K vacant intersections on which John can put a scarecrow.

The third line describes the position of K vacant intersections, in the format of r
₁,c
₁,r
₂,c
₂ …. r
_K,c
_k . (r
_i,c
_i) is the position of the i-th intersection and 1 <= r
₁,c
₁,r
₂,c
₂…. r
_K,c
_k <= N. 

The forth line gives the scaring range of all vacant intersections, in the format of R
₁,R
₂…R
_K and 0 <= R
₁,R
₂…R
_K <= 2 × N. 

The input ends with N = 0.

 

Output

For each test case, print the minimum number of scarecrows farmer John must buy in a line. If John has no way to protect all the corn, print -1 instead.

 

Sample Input

4
2
2 2 3 3
1 3
4
2
2 2 3 3
1 4
0

 

Sample Output

-1
1

 

Source

2012 Asia Hangzhou Regional Contest

 

题目大意：给你一张地图，然后有n个地点是空的用来放稻草人，其它的都是田地，每个稻草人位置有自己的横纵坐标以及可以保护田地的"半径"fabs(x-a)+fabs(y-b)<=r。问你找最少的稻草人使得所有田地被保护。

  解题思路：根据这个题目，以后可以把一个集合的所有子集全部摸出来了。开始想用DFS写，后来觉得求最小的，应该用BFS，最后思路全乱了，还是回到最初的枚举所有的状态数目。由于状态数目是1<<p，即为2^p，然后把0~1<<p转化为p位二进制存储，刚好唯一，就是题目中的VIS数组。

  题目地址：Scaring the Birds

AC代码：

#include<iostream>
#include<cstring>
#include<string>
#include<cmath>
#include<cstdio>
using namespace std;
int n,p,res;   //p指的是有多少个空地可以放稻草人
int visi[55][55];
int vis[12];  //点访问的情况
int num;  //记录用了多少点
struct mq
{
    int x;
    int y;
    int r;
};
mq node[12];

void init()
{
    memset(visi,0,sizeof(visi));
    memset(vis,0,sizeof(vis));
    for(int i=0;i<p;i++)
        visi[node[i].x][node[i].y]=1;
}

void fun()
{
    int j,k,ra,rb,ca,cb;
    num=0;
    for(int i=0;i<p;i++)
    {
        if(vis[i])
        {
            num++;
            ra=node[i].x-node[i].r;
            rb=node[i].x+node[i].r;
            ca=node[i].y-node[i].r;
            cb=node[i].y+node[i].r;
            if(ra<1) ra=1;
            if(rb>n) rb=n;
            if(ca<1) ca=1;
            if(cb>n) cb=n;
            for(j=ra;j<=rb;j++)
                for(k=ca;k<=cb;k++)
                   if(abs(j-node[i].x)+abs(k-node[i].y)<=node[i].r)  //范围之类
                        visi[j][k]=1;
        }
    }
}

int over()   //是否全部覆盖
{
    int i,j;
    for(i=1; i<=n; i++)
    {
        for(j=1; j<=n; j++)
            if(!visi[i][j])
            {
                return 0;
            }
    }
    return 1;
}

void solve()
{
    int i,j;
    res=100;
    for(i=0;i<(1<<p);i++)   //枚举所有的状态
    {
        init();
        int tmp=i;
        for(j=0;j<p;j++)
        {
            vis[j]=tmp&1;  //刚好二进制是这样存储，唯一! 所有枚举子集
            tmp>>=1;
            //cout<<vis[j]<<" ";
        }
        //cout<<endl;
        fun();
        if(over())  //可以覆盖了
            res=min(res,num);
    }
}
int main()
{
    int i;
    while(scanf("%d",&n)&&n)
    {
        scanf("%d",&p);
        for(i=0; i<p; i++)
            scanf("%d%d",&node[i].x,&node[i].y);
        for(i=0; i<p; i++)
            scanf("%d",&node[i].r);
        init();
        if(over())   //说明不需要稻草人。。。
        {
            puts("0");
            continue;
        }
        solve();
        if(res==100) puts("-1");  //说明覆盖不了
        else printf("%d\n",res);
    }
    return 0;
}

//31MS