题目来源:http://poj.org/problem?id=3469
由于maxm开小了,RE了好多发。另外,G++交的TLE了,用C++过的。
最小割问题,由定理知,最小割=最大流,因此可用最大流Dinic算法解决。
代码:
#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <queue>
using namespace std;
const int maxn=2e4+10;
const int maxm=2e6+10;
struct edge{
int to,next,cap,rev;
}e[maxm];
int n,m,cnt=0,head[maxn],dis[maxn],iter[maxn];
queue<int> q;
void ins(int x,int y,int z) {
e[++cnt].to = y;
e[cnt].next = head[x];
head[x] = cnt;
e[cnt].cap = z;
e[cnt].rev = cnt + 1;
e[++cnt].to = x;
e[cnt].next = head[y];
head[y] = cnt;
e[cnt].cap = 0;
e[cnt].rev = cnt - 1;
}
void bfs(int s) {
memset(dis, -1, sizeof(dis));
while (!q.empty())q.pop();
dis[s] = 0;
q.push(s);
while (!q.empty()) {
int u = q.front();
q.pop();
for (int i = head[u]; i; i = e[i].next) {
if (e[i].cap && dis[e[i].to] < 0) {
q.push(e[i].to);
dis[e[i].to] = dis[u] + 1;
}
}
}
}
int dfs(int v,int t,int f) {
if (v == t)return f;
for (int i = (iter[v] ? iter[v] : head[v]); i; i = e[i].next, iter[v] = i) {
if (e[i].cap && dis[v] < dis[e[i].to]) {
int d = dfs(e[i].to, t, min(f, e[i].cap));
if (d > 0) {
e[i].cap -= d;
e[e[i].rev].cap += d;
return d;
}
}
}
return 0;
}
int max_flow(int s,int t) {
int flow = 0;
for (;;) {
bfs(s);
if (dis[t] < 0)return flow;
memset(iter, 0, sizeof(iter));
int f;
while ((f = dfs(s, t, 1e9)) > 0)
flow += f;
}
}
int main() {
while (~scanf("%d%d", &n, &m)) {
memset(head, 0, sizeof(head));
cnt = 0;
int s = n + 1, t = n + 2;
for (int i = 1; i <= n; ++i) {
int x, y;
scanf("%d%d", &x, &y);
ins(s, i, x);
ins(i, t, y);
}
for (int i = 1; i <= m; ++i) {
int x, y, z;
scanf("%d%d%d", &x, &y, &z);
ins(x, y, z);
ins(y, x, z);
}
printf("%d\n", max_flow(s, t));
}
return 0;
}