/*
題意:無源無彙,并且每條邊的容量有上下界限的網絡流問題!既然無源無彙,那麼素有的節點都應該滿足“入流==出流”!
輸出每一條邊的流量,使得滿足上面的條件。(如果u->v有流量,那麼v->u就不會有流量)
思路:如果增加了源點s和彙點t,對于u->v(下限為l, 上限為f) 将這一條邊拆成3條,s->v(容量為l), u->v(容量為f-l)
u->t(容量為l)這樣就變成了每一個點的流入或者流出的流量至少是b!然後從s->t走一遍最大流,如果所有的附件邊都已經
滿載,則就是所有s->v的邊和u->t的邊(或者隻判斷其中一者就可以),那麼就存在答案!
*/
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#include<queue>
#define INF 0x3f3f3f3f
#define N 205
#define M 500000
using namespace std;
struct EDGE{
int v, cap, tot, nt, b;
EDGE(){};
EDGE(int v, int cap, int nt, int b) : v(v), cap(cap), nt(nt), b(b), tot(cap){}
};
EDGE edge[M];
int n, m;
int first[N];
int pre[N], d[N];
int sz;
int s, t;
int full, fout;
void addEdge(int u, int v, int b, int cap){
edge[sz] = (EDGE(v, cap, first[u],b));
first[u] = sz++;
edge[sz] = (EDGE(u, 0, first[v], 0));
first[v] = sz++;
edge[sz] = (EDGE(v, b, first[s], 0));
first[s] = sz++;
edge[sz] = (EDGE(s, 0, first[v], 0));
edge[sz] = (EDGE(t, b, first[u], 0));
full += b;
edge[sz] = (EDGE(u, 0, first[t], 0));
first[t] = sz++;
}
bool bfs(){
queue<int>q;
memset(d, 0, sizeof(d));
d[s] = 1;
q.push(s);
while(!q.empty()){
int u = q.front(); q.pop();
for(int i = first[u]; ~i; i = edge[i].nt){
int v = edge[i].v;
if(!d[v] && edge[i].cap >0){
d[v] = d[u] + 1;
q.push(v);
}
}
}
if(d[t] == 0) return false;
return true;
int dfs(int u, int totf){
int ff;
if( u == t) return totf;
int flow = 0;
for(int i = first[u]; ~i && totf > flow; i = edge[i].nt){
int v = edge[i].v;
int cap = edge[i].cap;
//流入u節點的目前總的流量為totf,可以得到 u->v1, u->v2, u->v3....這些路徑上的最大流的和為flow+=f(u->vi)
//f(u->vi)表示u節點沿着vi節點方向的路徑上的最大流;如果u->vi+1的容量為wi+1,那麼u->vi+1所允許流過的最大
//的流量就是 min(totf - cost, wi+1)了!
if(d[v] == d[u] + 1 && cap > 0 ){
ff = dfs(v, min(totf - flow, cap));
if(ff){
edge[i].cap -= ff;
edge[i^1].cap += ff;
flow += ff;
else
d[v] = -1;//表示v這個點無法在繼續增廣下去了
return flow;//傳回從u節點向外流出的最大流量!
bool Dinic(){
while(bfs())
fout += dfs(0, INF);//這一塊沒想到寫成while(dfs())會逾時....
if( fout != full) return false;
int main(){
scanf("%d%d", &n, &m);
memset(first, -1, sizeof(first));
sz = 0;
fout = full = 0;
s = 0; t = n+1;
int u, v, l, f;
for(int i = 1; i <= m; ++i){
scanf("%d%d%d%d", &u, &v, &l, &f);
addEdge(u, v, l, f-l);
if(!Dinic()){
printf("NO\n");
return 0;
printf("YES\n");
int j = (i-1)*6;
printf("%d\n", edge[j].tot - edge[j].cap + edge[j].b);//輸出這條邊實際流過的流量+下限
return 0;
本文轉自 小眼兒 部落格園部落格,原文連結:http://www.cnblogs.com/hujunzheng/p/4006435.html,如需轉載請自行聯系原作者