sgu194：Reactor Colling(无源无汇上下界最大流)

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题意：

求一个无源无汇的有上下界的最大流。

分析：

把每条边的边容量重设为上界-下界，记录每个点的最小进入in和最小输出流out，重设一个s、t，将s、t连向每一个点，边容量分别为每个点的in和out，从s到t跑最大流即

可。如果s的每条边满流，则有解，否则无解。

#include <cstdio>#include <algorithm>#include <queue>using namespace std;const int MAXN = 209, INF = 1e9;int n, m;struct Edge{int v, ne;int f, c, l;}edge[MAXN*MAXN];int edgehead[MAXN], p = 1;int sum, in[MAXN], out[MAXN];int dis[MAXN];void add(int u, int v, int f, int l){edge[p].v = v;edge[p].c = edge[p].f = f;edge[p].l = l;edge[p].ne = edgehead[u];edgehead[u] = p++;}int adv(int k) {return ((k-1)^1)+1;}bool bfs(int s, int t){for(int i = 1; i <= n; ++i) dis[i] = INF;dis[t] = 0;queue<int> q;q.push(t);while(!q.empty()){int u = q.front();q.pop();for(int i = edgehead[u]; i; i = edge[i].ne){int v = edge[i].v, f = edge[adv(i)].f; if(f && dis[v] > dis[u]+1) { dis[v] = dis[u]+1;q.push(v); }}}return dis[s] != INF;}int dfs(int now, int t, int flow){if(now == t || flow == 0) return flow;int re = 0, tmp;for(int i = edgehead[now]; i; i = edge[i].ne){int to = edge[i].v;if(dis[to] == dis[now]-1 && edge[i].f && (tmp = dfs(to, t, min(flow, edge[i].f)))){re += tmp;flow -= tmp;edge[i].f -= tmp;edge[adv(i)].f += tmp;if(flow == 0) break;}}dis[now] = INF;return re;}int dinic(int s, int t, int maxf){int re = 0;while(bfs(s, t))re += dfs(s, t, maxf);return re;}int main(){scanf("%d%d", &n, &m);for(int i = 1; i <= m; ++i){int a, b, c, d;scanf("%d%d%d%d", &a, &b, &c, &d);in[b] += c;out[a] += c;sum += c;add(a, b, d-c, c);add(b, a, 0, 0);}for(int i = 1; i <= n; ++i){if(in[i]) add(n+1, i, in[i], 0), add(i, n+1, 0, 0);if(out[i]) add(i, n+2, out[i], 0), add(n+2, i, 0, 0);}n += 2;if(dinic(n-1, n, INF) != sum){puts("NO");goto end;}puts("YES");for(int i = 1; i <= m; ++i)printf("%d\n", edge[2*i-1].c-edge[2*i-1].f+edge[2*i-1].l);end:;return 0;}

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