hdu 4462 - Scaring the Birds（状态压缩＋bfs）

原创于 2017-09-23 20:06:14 发布 · 391 阅读

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本文介绍了一个有趣的算法问题：如何利用有限数量的稻草人保护农田免受鸟类侵扰。通过状态压缩和暴力枚举的方法，确定放置最少稻草人的位置以确保所有作物得到保护。

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

Sample Output

-1
1

题意：有一个n*n的地图，有k个空地可以放稻草人，给出每个空地可以放的稻草人属性，属性中有个R代表这个位置可以守卫的范围，问最少需要放多少个稻草人才可以守卫这个地图。

基本的状态压缩，然后暴力枚举一遍就可以了

int n,k;
int vis[51][51];
int x[11],y[11],r[11];
int  main()
{
    ios::sync_with_stdio(false);
    while(cin>>n&&n)
    {
        cin>>k;
        for(int i=0;i<k;i++)cin>>x[i]>>y[i];
        for(int i=0;i<k;i++)cin>>r[i];
        int ans=11;
        for(int i=0;i<(1<<k);i++)
        {
            memset(vis,0,sizeof(vis));
            int tmp=0;
            for(int j=0;j<k;j++)
            {
                vis[x[j]][y[j]]=1;
                if(i&(1<<j))
                {
                    tmp++;
                    for(int ii=1;ii<=n;ii++)
                        for(int jj=1;jj<=n;jj++)
                            if(abs(ii-x[j])+abs(jj-y[j])<=r[j])
                                vis[ii][jj]=1;
                }
            }
            int flag=1;
            for(int ii=1;ii<=n;ii++)
                for(int jj=1;jj<=n;jj++)
                    if(!vis[ii][jj])
                        flag=0;
            if(flag) ans=min(ans,tmp);
        }
        if(ans>k)cout<<-1<<endl;
        else cout<<ans<<endl;
    }
    return 0;
}