题面里讲述的题意很明白
一开始wa掉到的点:1、有的点会在没扩展前就被覆盖掉;
2、x,y都要取min(zz错误)
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
using namespace std;
const int MAXN = 200 + 50;
struct zt{
double x,y;
}node[MAXN << 1];
int n;
double len[MAXN][MAXN];
double R[MAXN];
double x,y,X,Y,xx,yy;
double tot;
double ans,now,r;
int temp[MAXN];
void solve(){
now = 0;
for(int i = 1;i <= n;i ++){
xx = node[temp[i]].x;
yy = node[temp[i]].y;
r = 1061109567.0;
for(int j = 1;j < i;j ++){
if(R[temp[j]] >= len[temp[i]][temp[j]]){
r = 0;
continue;
}
r = min(r,len[temp[i]][temp[j]] - R[temp[j]]);
}
r = min(r,abs(x - xx));
r = min(r,abs(X - xx));
r = min(r,abs(Y - yy));
r = min(r,abs(yy - y));
R[temp[i]] = r;
now += r*r*M_PI;
}
ans = max(ans,now);
}
int main(){
scanf("%d",&n);
scanf("%lf%lf%lf%lf",&x,&y,&X,&Y);
tot = abs(x - X)*abs(y - Y);
for(int i = 1;i <= n;i ++){
scanf("%lf%lf",&node[i].x,&node[i].y);
}
for(int i = 1;i <= n;i ++){
for(int j = 1;j < i;j ++){
len[i][j] = sqrt((node[i].x - node[j].x)*(node[i].x - node[j].x) +
(node[i].y - node[j].y)*(node[i].y - node[j].y));
len[j][i] = len[i][j];
}
}
for(int i = 1;i <= n;i ++)temp[i] = i;
do{
solve();
}while(next_permutation(temp + 1,temp + n + 1));
printf("%.lf",tot - ans);
}

本文介绍了一种解决圆覆盖问题的算法,通过排列组合的方法找到最优解,确保每个点都被覆盖的同时最小化总面积。
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