SGU 194. Reactor Cooling(无源汇有上下界的网络流)

时间限制：0.5s

空间限制：6M

题意：

显然就是求一个无源汇有上下界的网络流的可行流的问题

Solution：

没什么好说的，直接判定可行流，输出就好了

code

/*
       无汇源有上下界的网络流
*/
#include <iostream>
#include <cstring>
#define ms(a,b) memset(a,b,sizeof a)
using namespace std;
const int MAXN = ;

struct node {
    int u, v, c, ne;
} edge[MAXN * MAXN << ];
int pHead[MAXN*MAXN], SS, ST, T, ncnt, ans;
int Gup[MAXN][MAXN], Glow[MAXN][MAXN], st[MAXN], ed[MAXN], cap[MAXN][MAXN], tflow;

void addEdge (int u, int v, int c) {
    edge[++ncnt].v = v, edge[ncnt].c = c, edge[ncnt].u = u;
    edge[ncnt].ne = pHead[u], pHead[u] = ncnt;
    edge[++ncnt].v = u, edge[ncnt].c = , edge[ncnt].u = v;
    edge[ncnt].ne = pHead[v], pHead[v] = ncnt;
}

int SAP (int pStart, int pEnd, int N) {
    int numh[MAXN], h[MAXN], curEdge[MAXN], pre[MAXN];
    int cur_flow, flow_ans = , u, neck, i, tmp;
    ms (h, ); ms (numh, ); ms (pre, -);
    for (i = ; i <= N; i++) curEdge[i] = pHead[i];
    numh[] = N;
    u = pStart;
    while (h[pStart] <= N) {
        if (u == pEnd) {
            cur_flow = 1e9;
            for (i = pStart; i != pEnd; i = edge[curEdge[i]].v)
                if (cur_flow > edge[curEdge[i]].c) neck = i, cur_flow = edge[curEdge[i]].c;
            for (i = pStart; i != pEnd; i = edge[curEdge[i]].v) {
                tmp = curEdge[i];
                edge[tmp].c -= cur_flow, edge[tmp ^ ].c += cur_flow;
            }
            flow_ans += cur_flow;
            u = neck;
        }
        for ( i = curEdge[u]; i != ; i = edge[i].ne) {
            if (edge[i].v > N) continue; //重要!!!
            if (edge[i].c && h[u] == h[edge[i].v] + )     break;
        }
        if (i != ) {
            curEdge[u] = i, pre[edge[i].v] = u;
            u = edge[i].v;
        }
        else {
            if ( == --numh[h[u]]) continue;
            curEdge[u] = pHead[u];
            for (tmp = N, i = pHead[u]; i != ; i = edge[i].ne) {
                if (edge[i].v > N) continue; //重要!!!
                if (edge[i].c)  tmp = min (tmp, h[edge[i].v]);
            }
            h[u] = tmp + ;
            ++numh[h[u]];
            if (u != pStart) u = pre[u];
        }
    }
    return flow_ans;
}
int solve (int n) {
    SS = n + , ST = n + ;
    for (int i = ; i <= n; i++) {
        if (ed[i]) addEdge (SS, i, ed[i]);
        if (st[i]) addEdge (i, ST, st[i]);
    }
    int tem = SAP (SS, ST, ST);
    if (tem != tflow) return ;
    else
              return ;
}
int n, m;
int main() {
    ios::sync_with_stdio ();
    ncnt = ;
    cin >> n >> m;
    for (int i = , u, v, x, y; i <= m; i++) {
        cin >> u >> v >> x >> y;
        Gup[u][v] = y, Glow[u][v] = x;
        st[u] += x, ed[v] += x;
        tflow += x;
        addEdge (u, v, y - x);
    }
    if (solve (n) ) {
        cout << "YES\n";
        for (int i = , x, y; i <= m * ; i += ) {
            x = edge[i].u, y = edge[i].v;
            cout << Gup[x][y]-edge[i].c << '\n';
        }
    }
    else
        cout << "NO\n";
    return ;
}