Flow Problem
Time Limit: 5000/5000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 11733 Accepted Submission(s): 5565
Total Submission(s): 11733 Accepted Submission(s): 5565
Problem Description
Network flow is a well-known difficult problem for ACMers. Given a graph, your task is to 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)
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
Sample Output
Case 1: 1 Case 2: 2
Author
HyperHexagon
Source
Recommend
数组大小也是一门学问啊
#include<stdio.h>
#include<string.h>
#include<queue>
#include<vector>
#include<algorithm>
using namespace std;
#define MAX 200+10
#define INF 10000000+10
struct node
{
int u,v,cap,flow,next;
}edge[2100];
int head[MAX],cur[MAX];
int vis[MAX],dis[MAX],m,n,top;
void init()
{
top=0;
memset(head,-1,sizeof(head));
}
void add(int a,int b,int c)
{
node E1={a,b,c,0,head[a]};
edge[top]=E1;
head[a]=top++;
node E2={b,a,0,0,head[b]};
edge[top]=E2;
head[b]=top++;
}
bool bfs(int s,int e)
{
memset(vis,0,sizeof(vis));
memset(dis,-1,sizeof(dis));
queue<int>q;
while(!q.empty()) q.pop();
q.push(s);
vis[s]=1;
dis[s]=0;
while(!q.empty())
{
int u=q.front();
q.pop();
for(int i=head[u];i!=-1;i=edge[i].next)
{
node E=edge[i];
if(!vis[E.v]&&E.cap>E.flow)
{
vis[E.v]=1;
dis[E.v]=dis[E.u]+1;
if(E.v==e)
return true;
q.push(E.v);
}
}
}
return false;
}
int dfs(int x,int a,int e)
{
if(x==e||a==0) return a;
int flow=0,f;
for(int i=cur[x];i!=-1;i=edge[i].next)
{
node& E=edge[i];
if(dis[x]+1==dis[E.v]&&(f=dfs(E.v,min(a,E.cap-E.flow),e))>0)
{
E.flow+=f;
flow+=f;
edge[i^1].flow-=f;
a-=f;
if(a==0) break;
}
}
return flow;
}
int MAXflow(int s,int e)
{
int flow=0;
while(bfs(s,e))
{
memcpy(cur,head,sizeof(head));
flow+=dfs(s,INF,e);
}
return flow;
}
void getmap()
{
int a,b,c;
while(m--)
{
scanf("%d%d%d",&a,&b,&c);
add(a,b,c);
}
}
int main()
{
int t;
int k=1;
scanf("%d",&t);
while(t--)
{
scanf("%d%d",&n,&m);
init();
getmap();
printf("Case %d: %d
",k++,MAXflow(1,n));
}
return 0;
}