题意:
P≤500个人要分配到S≤20个地方,每个地方的容量都是C≤100,给定m条边(u,v)表示u可以分配到v,问最终可以分配多少人?
分析:
点(地方)上有容量限制,裸的拆点最大流
代码:
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>
using namespace std;
#define pr(x) cout << #x << " = " << x << " "
#define prln(x) cout << #x << " = " << x << endl
const int N = 1e3 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;
const int M = 1e6 + 10;
int head[N], pnt[M], cap[M], nxt[M], cnt;
void add_edge(int u, int v, int w) {
pnt[cnt] = v;
cap[cnt] = w;
nxt[cnt] = head[u];
head[u] = cnt++;
}
void add_double(int u, int v, int w1, int w2 = 0) {
add_edge(u, v, w1);
add_edge(v, u, w2);
}
int lev[N], cur[N];
bool bfs(int s, int t) {
queue<int> q;
memset(lev, 0, sizeof lev);
q.push(s); lev[s] = 1;
while(q.size() && !lev[t]) {
int u = q.front(); q.pop();
for(int i = head[u]; ~i; i = nxt[i]) {
int v = pnt[i];
if(cap[i] > 0 && !lev[v]) {
lev[v] = lev[u] + 1;
q.push(v);
}
}
}
return lev[t];
}
int dfs(int u, int t, int delta) {
if(u == t || !delta) return delta;
int ret = 0;
for(int i = cur[u]; ~i; i = nxt[i]) {
int v = pnt[i];
if(cap[i] > 0 && lev[v] == lev[u] + 1) {
int d = dfs(v, t, min(delta, cap[i]));
cur[u] = i;
ret += d; delta -= d;
cap[i] -= d;
cap[i ^ 1] += d;
if(delta == 0) return ret;
}
}
lev[u] = 0;
return ret;
}
int dinic(int s, int t) {
int ret = 0;
while(bfs(s, t)) {
for(int i = s; i <= t; ++i) cur[i] = head[i];
ret += dfs(s, t, INF);
}
return ret;
}
int n, m, S, C;
int main() {
#ifdef LOCAL
freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);
// freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);
#endif
ios_base::sync_with_stdio(0);
int t; scanf("%d", &t);
while(t--) {
scanf("%d%d%d%d", &n, &S, &C, &m);
cnt = 0; memset(head, -1, sizeof head);
int s = 0, t = n + S * 2 + 1;
for(int i = 1; i <= n; ++i) add_double(s, i, 1);
while(m--) {
int u, v; scanf("%d%d", &u, &v);
add_double(u, n + v, INF);
}
for(int i = 1; i <= S; ++i) {
add_double(n + i, n + i + S, C);
add_double(n + i + S, t, INF);
}
printf("%d\n", dinic(s, t));
}
return 0;
}