题目
题目链接:http://poj.org/problem?id=2536
题目来源:http://www.cnblogs.com/vongang/archive/2012/02/21/2361882.html
题解
最小化死掉的人,也就是最大化存活的人。
人和洞建二分图,根据题中信息算出某人对于某个洞是否能够到达,能到达就连上边。
图上的最大匹配的规模就是最多的存活的人, n−|M| 就是答案。
代码
#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <cstring>
#include <stack>
#include <queue>
#include <string>
#include <vector>
#include <set>
#include <map>
#define fi first
#define se second
using namespace std;
typedef long long LL;
typedef pair<int,int> PII;
typedef pair<double, double> pdd;
// head
const int N = 105;
const int M = N;
const int E = N * M;
struct Gragh {
struct Edge {
int to, nxt;
Edge(int to, int nxt) : to(to), nxt(nxt) {}
Edge() {}
};
Edge e[E];
int head[N], match[M], ec;
bool vis[N];
void addEdge(int from, int to) {
e[ec] = Edge(to, head[from]);
head[from] = ec++;
}
void init(int n, int m) {
ec = 0;
for (int i = 0; i <= n; i++) head[i] = -1;
for (int i = 0; i <= m; i++) match[i] = -1;
}
bool dfs(int x) {
vis[x] = true;
for (int i = head[x]; ~i; i = e[i].nxt) {
int to = e[i].to, w = match[to];
if (w == -1 || (!vis[w] && dfs(w))) {
match[to] = x;
return true;
}
}
return false;
}
int bipatiteMatch(int n) {
int ans = 0;
for (int i = 0; i < n; i++) {
memset(vis, 0, sizeof vis);
if (dfs(i)) ans++;
}
return ans;
}
};
inline double sq(double x) {
return x * x;
}
double dis(pdd &a, pdd &b) {
return sqrt(sq(a.fi - b.fi) + sq(a.se - b.se));
}
Gragh g;
pdd a[N], b[N];
int main() {
int n, m, s, v;
while (scanf("%d%d%d%d", &n, &m, &s, &v) == 4) {
s *= v;
g.init(n, m);
for (int i = 0; i < n; i++) {
scanf("%lf%lf", &a[i].fi, &a[i].se);
}
for (int i = 0; i < m; i++) {
scanf("%lf%lf", &b[i].fi, &b[i].se);
}
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
if (dis(a[i], b[j]) > s) continue;
g.addEdge(i, j);
}
}
printf("%d\n", n - g.bipatiteMatch(n));
}
return 0;
}