可行流会求了，这次求最大流
1 #include <bits/stdc++.h> 2 using namespace std; 3 const int inf = 0x3f3f3f3f; 4 const int maxn = 210; 5 const int maxm = 50010; 6 struct G 7 { 8 int v, cap, next, num; 9 G() {} 10 G(int v, int cap, int next, int num) : v(v), cap(cap), next(next), num(num) {} 11 } E[maxm]; 12 int p[maxn], T; 13 int d[maxn], temp_p[maxn], qw[maxn]; //d顶点到源点的距离标号,temp_p当前狐优化，qw队列 14 int n,m,s,t; 15 void init() 16 { 17 memset(p, -1, sizeof(p)); 18 T = 0; 19 } 20 void add(int u, int v, int cap, int num) 21 { 22 E[T] = G(v, cap, p[u], num); 23 p[u] = T++; 24 E[T] = G(u, 0, p[v], num); 25 p[v] = T++; 26 } 27 bool bfs(int st, int en, int n) 28 { 29 int i, u, v, head, tail; 30 for(i = 0; i <= n; i++) d[i] = -1; 31 head = tail = 0; 32 d[st] = 0; 33 qw[tail] = st; 34 while(head <= tail) 35 { 36 u = qw[head++]; 37 for(i = p[u]; i + 1; i = E[i].next) 38 { 39 v = E[i].v; 40 if(d[v] == -1 && E[i].cap > 0) 41 { 42 d[v] = d[u] + 1; 43 qw[++tail] = v; 44 } 45 } 46 } 47 return (d[en] != -1); 48 } 49 int dfs(int u, int en, int f) 50 { 51 if(u == en || f == 0) return f; 52 int flow = 0, temp; 53 for(; temp_p[u] + 1; temp_p[u] = E[temp_p[u]].next) 54 { 55 G& e = E[temp_p[u]]; 56 if(d[u] + 1 == d[e.v]) 57 { 58 temp = dfs(e.v, en, min(f, e.cap)); 59 if(temp > 0) 60 { 61 e.cap -= temp; 62 E[temp_p[u] ^ 1].cap += temp; 63 flow += temp; 64 f -= temp; 65 if(f == 0) break; 66 } 67 } 68 } 69 return flow; 70 } 71 int dinic(int st, int en, int n) 72 { 73 int i, ans = 0; 74 while(bfs(st, en, n)) 75 { 76 for(i = 0; i <= n; i++) temp_p[i] = p[i]; 77 ans += dfs(st, en, inf); 78 } 79 return ans; 80 } 81 int du[maxn],sum=0; 82 int main() 83 { 84 scanf("%d%d%d%d",&n,&m,&s,&t); 85 init(); 86 memset(du,0,sizeof(du)); 87 for(int i=1; i<=m; i++){ 88 int u,v,l,r; 89 scanf("%d %d %d %d", &u,&v,&l,&r); 90 add(u,v,r-l, i); 91 du[u]-=l; 92 du[v]+=l; 93 } 94 int ss = 0, tt = n+1; 95 for(int i=1; i<=n; i++){ 96 if(du[i]>0) sum+=du[i],add(ss,i,du[i], 0); 97 if(du[i]<0) add(i,tt,-du[i], 0); 98 } 99 add(t, s, inf, 0); 100 if(sum == dinic(ss,tt,n+2)){ 101 sum = E[p[t]^1].cap; 102 for(int i=0; i<T; i++){ 103 if(!E[i].num) E[i].v=0; 104 } 105 p[ss]=p[tt]=-1; 106 ss=s; 107 tt=t; 108 printf("%d ", sum+dinic(ss,tt,n+2)); 109 } 110 else{ 111 puts("please go home to sleep"); 112 } 113 return 0; 114 }