CodeForces1082G Petya and Graph 最小割

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网络流裸题

\(s\)向点连边\((s, i, a[i])\)

给每个边建一个点

边\((u, v, w)\)抽象成\((u, E, inf)\)和\((v, E, inf)\)以及边\((E, t, w)\)

最小割建模...

然后就没了....复习一下板子吧

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#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;

#define ll long long
#define ri register int
#define rep(io, st, ed) for(ri io = st; io <= ed; io ++)
#define drep(io, ed, st) for(ri io = ed; io >= st; io --)

#define gc getchar
inline int read() {
    int p = 0, w = 1; char c = gc();
    while(c > '9' || c < '0') { if(c == '-') w = -1; c = gc(); }
    while(c >= '0' && c <= '9') p = p * 10 + c - '0', c = gc();
    return p * w;
}

const int sid = 5e4 + 5;
const ll inf = 1e17;

int n, m, s, t, ip, cnp = 1;
int a[sid], q[sid], go[sid], num[sid], L[sid];
int cap[sid], nxt[sid], node[sid];
ll res[sid];

inline void addedge(int u, int v, ll w) {
    nxt[++ cnp] = cap[u]; cap[u] = cnp; node[cnp] = v; res[cnp] = w;
    nxt[++ cnp] = cap[v]; cap[v] = cnp; node[cnp] = u; res[cnp] = 0;
}

#define cur node[i]
inline void bfs() {
    int fr = 1, to = 0;
    L[t] = 1; num[1] ++; q[++ to] = t;
    while(fr <= to) {
        int o = q[fr ++];
        for(int i = cap[o]; i; i = nxt[i])
            if(res[i ^ 1] && !L[cur]) {
                q[++ to] = cur;
                L[cur] = L[o] + 1; num[L[cur]] ++;
            }
    }
    for(int i = 1; i <= t; i ++) go[i] = cap[i];
}

inline ll dfs(int o, ll flow) {
    ll tt = 0, tmp;
    if(o == t) return flow;
    for(int &i = go[o]; i; i = nxt[i]) {
        if(L[cur] + 1 != L[o]) continue;
        tmp = dfs(cur, min(flow, res[i]));
        res[i] -= tmp; res[i ^ 1] += tmp;
        flow -= tmp; tt += tmp;
        if(!flow) return tt;
    }
    -- num[L[o]]; if(!num[L[o]]) L[s] = t + 1;
    L[o] ++; num[L[o]] ++;
    go[o] = cap[o];
    return tt;
}

inline ll isap(int s, int t) {
    bfs();
    ll ret = dfs(s, inf);
    while(L[s] <= t) ret += dfs(s, inf);
    return ret;
}

int main() {
    ll ans = 0;
    n = read(); m = read();
    rep(i, 1, n) a[i] = read();
    s = n + m + 1; t = s + 1;
    rep(i, 1, n) addedge(s, i, a[i]);
    rep(i, 1, m) {
        int u = read(), v = read(), w = read();
        addedge(u, n + i, inf); addedge(v, n + i, inf);
        addedge(n + i, t, w); ans += w;
    }
    ans -= isap(s, t);
    printf("%lld\n", ans);
    return 0;
}