題目來源:http://codeforces.com/gym/101666/attachments
由於是求最小距離的最大值,因此很容易想到二分答案。
在驗證每一個值的過程中,我們可以將距離<mid的邊保留,>mid的邊刪除,因此變成了求二分圖的最大獨立集的問題。
相關概念及定理:https://blog.csdn.net/moon_sky1999/article/details/81331795
由定理知:|最大獨立集|+|最小頂點覆蓋|=|V|
對於二分圖:|最大匹配|=|最小頂點覆蓋|。
因此可以對處理後的圖進行二分圖匹配,求得最大匹配。即可驗證。
代碼:
#include <bits/stdc++.h>
using namespace std;
const double eps = 1e-9;
int n, b, r, from[510];
double xb[510], yb[510], xr[510], yr[510], g[510][510], e[510][510];
bool vis[510];
bool match(int x) {
for (int i = 1; i <= n; ++i) {
if (e[x][i] > eps && vis[i] == 0) {
vis[i] = 1;
if (from[i] == -1 || match(from[i])) {
from[i] = x;
return 1;
}
}
}
return 0;
}
int hungry() {
int tot = 0;
memset(from, -1, sizeof(from));
for (int i = 1; i <= b; ++i) {
memset(vis, 0, sizeof(vis));
if (match(i))
++tot;
}
return tot;
}
bool can(double x) {
memset(e, 0, sizeof(e));
for (int i = 1; i <= b; ++i) {
for (int j = 1; j <= r; ++j)
if (g[i][j] < x)
e[i][j] = g[i][j];
}
int tot = hungry();
if (b + r - tot >= n)return 1;
return 0;
}
int main() {
scanf("%d%d%d", &n, &b, &r);
for (int i = 1; i <= b; ++i)
scanf("%lf%lf", &xb[i], &yb[i]);
for (int i = 1; i <= r; ++i)
scanf("%lf%lf", &xr[i], &yr[i]);
for (int i = 1; i <= b; ++i)
for (int j = 1; j <= r; ++j)
g[i][j] = sqrt((xb[i] - xr[j]) * (xb[i] - xr[j]) + (yb[i] - yr[j]) * (yb[i] - yr[j]));
double le = 0.0, ri = 10000.0;
while (ri - le > eps) {
double mid = (le + ri) / 2;
if (can(mid))le = mid;
else ri = mid;
}
printf("%.9f\n", ri);
return 0;
}