https://vjudge.net/problem/UVA-820
有一个计算机网络,输入节点数n,输入网络流源点和汇点src,des,再输入双向边数m。给出m条边的负载,求最大流。
#include<vector>
#include<algorithm>
#include<string>
#include<iostream>
#include<queue>
#include<stdio.h>
#include<cstring>
using namespace std;
const int maxn = 10000 + 5;
const int INF = 1e8;
struct Edge{
int from,to,cap,flow;
Edge(int u,int v,int c,int f):from(u),to(v),cap(c),flow(f) {}
};
struct EdmondsKarp{
int n,m;
vector<Edge> edges; //边数的两倍
vector<int> G[maxn];
int a[maxn]; //当起点到i的可改进量
int p[maxn]; //最短路上p的入弧编号
void init(int n){
for(int i=0;i<n;i++) G[i].clear();
edges.clear();
}
void AddEdge(int from,int to,int cap){
edges.push_back(Edge(from,to,cap,0));
edges.push_back(Edge(to,from,cap,0));//反向弧
m = edges.size();
G[from].push_back(m-2);
G[to].push_back(m-1);//G[i][j]表示节点i的第j条边在e数组的序号
}
int Maxflow(int s,int t){
int flow=0;
for(;;){
memset(a,0,sizeof(a));
queue<int> Q;
Q.push(s);
a[s]=INF;
while(!Q.empty()){
int x=Q.front(); Q.pop();
for(int i=0;i<G[x].size();i++){
Edge& e=edges[ G[x][i] ];
//以下松弛操作
if(!a[e.to]&&e.cap>e.flow){
p[e.to] = G[x][i];//p数组相当于记录了 e.to的入弧,
a[e.to] = min(a[x],e.cap-e.flow);//找到这一条路上残量最小值
Q.push(e.to);
}
}
if(a[t]) break;
}
if(!a[t]) break;
for(int u=t;u!=s;u=edges[p[u] ].from){
edges[p[u] ].flow += a[t];
edges[p[u]^1 ].flow -= a[t];
}
flow += a[t];
}
return flow;
}
};
EdmondsKarp solver;
int main()
{
int n,m,icount=1;
int s,t;
cin>>n;
while(n)
{
solver.init(4*n+10);
cin>>s>>t>>m;
for(int i=0;i<m;i++)
{
int s1,s2,icost;
cin>>s1>>s2>>icost;
solver.AddEdge(s1,s2,icost);
}
cout<<"Network "<<icount++<<endl;
printf("The bandwidth is %d.\n\n", solver.Maxflow(s, t));
cin>>n;
}
}