hdu3549--Flow Problem(初识最大流)

Flow Problem

Time Limit: 5000/5000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 7870    Accepted Submission(s): 3664


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)  
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 第一次接触最大流，水题练手，使用了EK的方法，这个题不需要考虑建图，直接使用模板 

#include 
#include 
#include 
#include 
using namespace std;
#define maxn 2100
#define INF 0x3f3f3f3f
struct edge{
int v , w ;
int next ;
} p[maxn];
int head[maxn] , cnt , vis[maxn] , pre[maxn];
queue  q ;
void add(int u,int v,int w)
{
p[cnt].v = v ; p[cnt].w = w ;
p[cnt].next = head[u] ; head[u] = cnt++ ;
p[cnt].v = u ; p[cnt].w = 0 ;
p[cnt].next = head[v] ; head[v] = cnt++ ;
}
int bfs(int s,int t)
{
int u , v , i , min1 = INF ;
while( !q.empty() )
q.pop() ;
memset(vis,0,sizeof(vis));
vis[s] = 1 ;
q.push(s) ;
while(!q.empty())
{
u = q.front() ;
q.pop();
for(i = head[u] ; i != -1 ; i = p[i].next )
{
v = p[i].v ;
if( !vis[v] && p[i].w )
{
vis[v] = 1 ;
min1 = min(min1,p[i].w);
pre[v] = i ;
q.push(v) ;
}
}
}
if(vis[t])
return min1 ;
return -1 ;
}
int main()
{
int tt , t , n , m , i , u , v , w , max_flow;
scanf("%d", &t);
for(tt = 1 ; tt <= t ; tt++)
{
cnt = 0 ; max_flow = 0 ;
memset(head,-1,sizeof(head));
scanf("%d %d", &n, &m);
while(m--)
{
scanf("%d %d %d", &u, &v, &w);
add(u,v,w);
}
memset(pre,-1,sizeof(pre));
while(1)
{
int k = bfs(1,n);
if(k == -1)
break;
max_flow += k ;
for(i = pre[n] ; i != -1 ; i = pre[ p[ i^1 ].v ] )
{
p[i].w -= k ;
p[i^1].w += k ;
}
}
printf("Case %d: %d\n", tt, max_flow);
}
return 0;
}