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。