Problem:
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Maximum Flow

A flow network is a directed graph which has a $source$ and a $sink$. In a flow network, each edge $(u, v)$ has a capacity $c(u, v)$. Each edge receives a flow, but the amount of flow on the edge can not exceed the corresponding capacity. Find the maximum flow from the $source$ to the $sink$.

Input

A flow network is given in the following format.

$|V|\;|E|$

$u_0\;v_0\;c_0$

$u_1\;v_1\;c_1$

:

$u_{|E|-1}\;v_{|E|-1}\;c_{|E|-1}$

$|V|$, $|E|$ is the number of vertices and edges of the flow network respectively. The vertices in $G$ are named with the numbers 0, 1,..., $|V|-1$. The source is 0 and the sink is $|V|-1$.

$u_i$, $v_i$, $c_i$ represent $i$-th edge of the flow network.
A pair of $u_i$ and $v_i$ denotes that there is an edge from $u_i$ to $v_i$ and $c_i$ is the capacity of $i$-th edge.

Output

Print the maximum flow.

Constraints

$2 \leq |V| \leq 100$

$1 \leq |E| \leq 1000$

$0 \leq c_i \leq 10000$

Sample Input 1

4 5
0 1 2
0 2 1
1 2 1
1 3 1
2 3 2

Sample Output 1

3

Solution in C (judge verdict: Time Limit Exceeded)

#include<stdio.h>
#include<stdlib.h>
    
typedef struct{
  int fmax;
  int flow;
  int n[2];
}TEdge;
    
typedef struct{
  int num;
  int e[1200];
}TVertex;
    
TEdge e[1200];
TVertex v[120];
  
#define MIN(x,y) ((x)<(y)?(x):(y))
int V,E;
int et[120];
  
int route(int s, int ed,int c){
  int res=0,r=-1,m=0,i=0;
  //printf("{%d,%d",s,c);
  if(c==0)return 0;
  r=-1;
  et[s]=1;
  for(i=0;i<v[s].num;i++){
    if(e[v[s].e[i]].n[0]==s){
      if(et[e[v[s].e[i]].n[1]]==0){
        if(e[v[s].e[i]].fmax-e[v[s].e[i]].flow>0){
          m=MIN(c,e[v[s].e[i]].fmax-e[v[s].e[i]].flow);
          if(e[v[s].e[i]].n[1]==ed){
            r=m;
            e[v[s].e[i]].flow+=r;
            //printf(":%d}",r);
            res+=r;
            if(res>=c){
              et[s]=0;
              //printf("}");
              return res;
            }
            c-=r;
          }else{
            r=route(e[v[s].e[i]].n[1],ed,m);
            if(r>0){
              e[v[s].e[i]].flow+=r;
              res+=r;
              if(res>=c){
                et[s]=0;
                return res;
              }
              c-=r;
            }
          }
        }
      }
    }else{
      if(et[e[v[s].e[i]].n[0]]==0){
        if(e[v[s].e[i]].flow>0){
          m=MIN(c,e[v[s].e[i]].flow);
          r=route(e[v[s].e[i]].n[0],ed,m);
          if(r>0){
            e[v[s].e[i]].flow-=r;
            res+=r;
            if(res>=c){
              et[s]=0;
              return res;
            }
            c-=r;
          }
        }
      }
    }//*/
  }
  et[s]=0;
  //printf(":-1}");
  return res;
}
    
int main(){
  int i;
  int res,cmax;
  scanf("%d %d",&V,&E);
  for(i=0;i<V;i++){
    v[i].num=0;
  }
  cmax=0;
  for(i=0;i<E;i++){
    int a,b,c;
    scanf("%d %d %d",&a,&b,&c);
    e[i].n[0]=a;
    e[i].n[1]=b;
    v[a].e[v[a].num++]=i;
    v[b].e[v[b].num++]=i;
    e[i].fmax=c;
    e[i].flow=0;
    if(b==V-1){
      v[a].e[v[a].num-1]=v[a].e[0];
      v[a].e[0]=i;
      cmax+=c;
    }
  }
  //return 0;
  res=0;
  for(i=0;i<100;i++){
    et[i]=0;
  }
  while(1){
  //for(i=0;i<3;i++){
    int x;
    x=route(0,V-1,cmax);
    //printf("\n");
    if(x<=0){
      printf("%d\n",res);
      return 0;
    }
    res+=x;
    cmax-=x;
  }
  printf("er\n");
  return 0;
}