sgu194 Reactor Cooling【无源汇有上下界可行流】

　　这是模板题了吧，先建立附加源汇，然后保留每个点的in-out，如果这个值是正的，那么就从附加源先这个点连一个边权为in-out的边，否则从这个点向附加汇连一条相反数的边，剩下题目中的边就用上界-下界连就好了。

 #include <bits/stdc++.h>
 #define rep(i, a, b) for (int i = a; i <= b; i++)
 #define drep(i, a, b) for (int i = a; i >= b; i--)
 #define REP(i, a, b) for (int i = a; i < b; i++)
 #define mp make_pair
 #define pb push_back
 #define clr(x) memset(x, 0, sizeof(x))
 #define xx first
 #define yy second
 using namespace std;
 typedef long long i64;
 typedef pair<int, int> pii;
 const int inf = ~0U >> ;
 const i64 INF = ~0ULL >> ;
 //***********************************

 const int maxn = , maxm = ;

 struct Ed {
     int u, v, nx, c; Ed() {}
     Ed(int _u, int _v, int _nx, int _c) :
         u(_u), v(_v), nx(_nx), c(_c) {}
 } E[maxm << ];
 int G[maxn], edtot = ;
 void addedge(int u, int v, int c) {
     E[++edtot] = Ed(u, v, G[u], c);
     G[u] = edtot;
     E[++edtot] = Ed(v, u, G[v], );
     G[v] = edtot;
 }

 int level[maxn], s, t;
 bool bfs() {
     static int que[maxn]; int qh(), qt();
     clr(level);
     level[que[++qt] = s] = ;
     while (qh != qt) {
         int x = que[++qh]; if (qh == maxn - ) qh = ;
         for (int i = G[x]; i; i = E[i].nx) if (E[i].c && !level[E[i].v]) {
             level[que[++qt] = E[i].v] = level[x] + ;
             if (qt == maxn - ) qt = ;
         }
     }
     return !!level[t];
 }
 int dfs(int u, int rm) {
     if (u == t) return rm;
     int rm1 = rm;
     for (int i = G[u]; i; i = E[i].nx) {
         if (E[i].c && level[E[i].v] == level[u] + ) {
             int flow = dfs(E[i].v, min(E[i].c, rm));
             E[i].c -= flow, E[i ^ ].c += flow;
             if ((rm -= flow) == ) break;
         }
     }
     if (rm1 == rm) level[u] = ;
     return rm1 - rm;
 }

 int l[maxm], in[maxn], out[maxn];
 int cnt;
 int main() {
     clr(G), edtot = , clr(in), clr(out);
     int n, m;
     scanf("%d%d", &n, &m);
     s = n + , t = n + ;
     rep(i, , m) {
         int x, y, a, b; scanf("%d%d%d%d", &x, &y, &a, &b); l[i] = a;
         addedge(x, y, b - a);
         in[y] += a, out[x] += a;
     }
     int sum();
     rep(i, , n) {
         if (in[i] > out[i]) addedge(s, i, in[i] - out[i]), sum += in[i] - out[i];
         else addedge(i, t, out[i] - in[i]);
     }
     int ans();
     while (bfs()) ans += dfs(s, 0x3f3f3f3f);
     if (ans != sum) { puts("NO"); return ; }
     puts("YES");
     for (int i = ; i <= m; i++) printf("%d\n", E[(i << ) ^ ].c + l[i]);
     return ;
 }