Hdu 6598 Harmonious Army 最小割

对于每一个u,v,a,b,c

建(S,u,a) (u,v,a+c-2*b) (v,T,c) (S,v,a) (v,u,a+c-2*b) (u,T,c)

最后答案为(2*sum(a+c)-最大流)/2

思路可以参考洛谷P1361

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int MAXN = 505;
const int MAXM = 100000;
int Head[MAXN], cur[MAXN], lev[MAXN], to[MAXM << 1], nxt[MAXM << 1], f[MAXM << 1], ed = 1, S, T;
inline void addedge(int u, int v, int cap) {
        to[++ed] = v;
        nxt[ed] = Head[u];
        Head[u] = ed;
        f[ed] = cap;
        to[++ed] = u;
        nxt[ed] = Head[v];
        Head[v] = ed;
        f[ed] = 0;
        return;
}
inline bool BFS(int x) {
        int u;
        for (int i = 0; i <= x; i++)
                lev[i] = -1;
        queue<int>q;
        lev[S] = 0;
        q.push(S);
        while (q.size()) {
                u = q.front();
                q.pop();
                for (int i = Head[u]; i; i = nxt[i])
                        if (f[i] && lev[to[i]] == -1) {
                                lev[to[i]] = lev[u] + 1;
                                q.push(to[i]);
                        }
        }
        for (int i = 0; i <= x; i++)
                cur[i] = Head[i];
        return lev[T] != -1;
}
inline int DFS(int u, int maxf) {
        if (u == T || !maxf) {
                return maxf;
        }
        int cnt = 0;
        for (int &i = cur[u], tem; i; i = nxt[i])
                if (f[i] && lev[to[i]] == lev[u] + 1) {
                        tem = DFS(to[i], min(maxf, f[i]));
                        maxf -= tem;
                        f[i] -= tem;
                        f[i ^ 1] += tem;
                        cnt += tem;
                        if (!maxf) {
                                break;
                        }
                }
        if (!cnt) {
                lev[u] = -1;
        }
        return cnt;
}
ll Dinic(int x) {
        ll ans = 0;
        while (BFS(x)) {
                ans += DFS(S, 2147483647);
        }
        return ans;
}
void init(int SS, int TT) {
        for (int i = 0; i <= TT; i++)
                Head[i] = 0;
        ed = 1;
        S = SS;
        T = TT;
        return;
}
ll answer = 0;
int main() {
        int n, m;
        int u, v;
        ll a, b, c;
        while (scanf("%d %d", &n, &m) == 2) {
                answer = 0;
                init(0, n + 2);
                for (int i = 1; i <= m; i++) {
                        scanf("%d %d %lld %lld %lld", &u, &v, &a, &b, &c);
                        addedge(S, u, a), addedge(u, v, a + c - 2 * b), addedge(v, T, c);
                        addedge(S, v, a), addedge(v, u, a + c - 2 * b), addedge(u, T, c);
                        answer += 2 * a + 2 * c;
                }
                printf("%lld\n", (answer - Dinic(n + 2)) / 2);
        }
        return 0;
}