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Flow Problem
Time Limit: 5000/5000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 13315 Accepted Submission(s): 6372
Problem Description
Network flow is a well-known difficult problem for ACMers. Given a graph, your task is to //代码效果参考：https://v.youku.com/v_show/id_XNjQwNjg1MjE3Ng==.html
find out the maximum flow for the weighted directed graph.Input
The first line of input contains an integer T, denoting the number of test cases.
For each test case, the first line contains two integers N and M, denoting the number of vertexes and edges in the graph. (2 <= N <= 15, 0 <= M <= 1000)
Next M lines, each line contains three integers X, Y and C, there is an edge from X to Y and the capacity of it is C. (1 <= X, Y <= N, 1 <= C <= 1000)
Output
For each test cases, you should output the maximum flow from source 1 to sink N.
Sample Input
2
3 2
1 2 1
2 3 1
3 3
1 2 1
2 3 1
1 3 1
#include
#include
#include
#include
#define maxn 16
#define inf 1[29
using namespace std;
struct edge
{
int to,cap,rev;
};
int n,m,vis【maxn】;
vector fuck【maxn】;
void add_edge(int from,int to,int cap)//构图过程
{
fuck【from】.push_back((edge){to,cap,fuck【to】.size()});//fuck【to】.size() 表示反边在反向邻接表的位置--即fuck【to】中的位置
fuck【to】.push_back((edge){from,0,fuck【from】.size()-1});//同上
}
void init()
{
for(int i=1;i<=n;i++) fuck【maxn】.clear();
}
int dfs(int s,int e,int f)
{
if(s==e) return f;
vis【s】=1;
for(int i=0;i0)
{
int d=dfs(temp.to,e,min(f,temp.cap));
if(d>0)
{
temp.cap-=d;
fuck【temp.to】【temp.rev】.cap+=d;// 回溯处理容量问题 以及对反向边的处理
//一直回溯到流量最小的那个点上去 然后去走其他路
return d;//回溯过程肯定是要更新的
}
}
}
return 0;
}
int max_flow(int s,int e)
{
int flow=0;
while(1)//不断的从s走到e这个点 当所有可能的通道cap为0的时候 贪心完成（因为每次过程都会cap进行一次更新）
{
memset(vis,0,sizeof(vis));
int temp=dfs(s,e,inf);
if(temp==0) return flow;
else flow+=temp;
}
}
int //代码效果参考：https://v.youku.com/v_show/id_XNjQwMDM5NzIyOA==.html
main(){
cin.sync_with_stdio(false);
int t,Case=0;
cin]t;
while(t--)
{
cin]n]m;
init();
while(m--)
{
int a,b,c;
cin]a]b]c;
add_edge(a,b,c);//构图
}
cout[""Case ""[++Case["": ""[max_flow(1,n)[endl;
}
return 0;
}
"
