水题二道

版权声明： https://blog.csdn.net/qq_38234381/article/details/82817410

https://www.luogu.org/problemnew/show/P1034

这道题的数据是真的水，唯一的难点就是确定好爆搜的状态，我们爆搜每个矩形包含那几个点，存在数组里，再加一个最优化剪枝和可行性剪枝，判断答案是否更优或者矩形是否相交即可。

code:

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;

const int N=60, K=5, INF=0x3f3f3f3f;
struct point {
    int x, y;
} p[N];
int n, k, ans=INF, cnt[K], a[K][N];

bool check(int pos) {
    int i, j, S=0, top[K], bottom[K], left[K], right[K];
    memset(top, -1, sizeof(top));
    memset(bottom, INF, sizeof(bottom));
    memset(left, INF, sizeof(left));
    memset(right, -1, sizeof(right));
    for (i=1; i<=k; i++) {
        for (j=1; j<=cnt[i]; j++) {
            top[i]=max(p[a[i][j]].y, top[i]);
            bottom[i]=min(p[a[i][j]].y, bottom[i]);
            right[i]=max(p[a[i][j]].x, right[i]);
            left[i]=min(p[a[i][j]].x, left[i]);
        }
        //cout << top[i] << " " << bottom[i] << " " << right[i] << " " << left[i] << endl;  
    }
    for (i=1; i<=k; i++) {
        for (j=1; j<=k; j++) {
            if (cnt[i]==0 || cnt[j]==0) continue;
            if (i==j) continue;
            if (right[i]<left[j] || left[i]>right[j] || bottom[i]>top[j] || top[i]<bottom[j]) continue;
            return false;
        }
    }
    for (i=1; i<=k; i++) {
        if (cnt[i]==0) continue;
        S+=(top[i]-bottom[i])*(right[i]-left[i]);
        if (S>=ans) return false;
    }
    if (pos==n+1) ans=S;
    return true;
}

void dfs(int pos) {
    if (!check(pos)) return ;
    int i;
    for (i=1; i<=k; i++) {
        a[i][++cnt[i]]=pos;
        dfs(pos+1);
        a[i][cnt[i]--]=0;
    }
    return ;
}

int main() {
    int i;
    cin >> n >> k;
    for (i=1; i<=n; i++)
        cin >> p[i].x >> p[i].y;
    dfs(1);
    cout << ans;
    return 0;
}

https://www.luogu.org/problemnew/show/P1072

非常水的蓝题，是不是标签错了啊。

洛谷博客：https://www.luogu.org/blog/user16215/solution-p1072