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>
    
#define MIN(x,y) ((x)<(y)?(x):(y))
int V,E;
int et[120];
int map[120][120];

int route(int s,int ed,int c){
  int i,r,res;
  res=0;
  if(c==0)return 0;
  if(s==ed)return c;
  et[s]=1;
  for(i=V-1;i>=0;i--){
    if(et[i]==0&&map[s][i]>0){
      r=route(i,ed,c<map[s][i]?c:map[s][i]);
      if(r>0){
        map[s][i]-=r;
        map[i][s]+=r;
        res+=r;
        c-=r;
        if(c==0){
          et[s]=0;
          return res;
        }
      }
    }
  }
  et[s]=0;
  return res;
}
    
int main(){
  int i,j;
  int res,cmax;
  scanf("%d %d",&V,&E);
  for(i=0;i<V;i++){
    for(j=0;j<V;j++){
      map[i][j]=0;
    }
  }
  for(i=0;i<E;i++){
    int a,b,c;
    scanf("%d %d %d",&a,&b,&c);
    map[a][b]+=c;
  }

  cmax=0;
  for(i=0;i<V;i++){
    cmax+=map[i][V-1];
  }
  //printf("cmax:%d\n",cmax);
  //return 0;
  res=0;
  for(i=0;i<120;i++){
    et[i]=0;
  }
  while(1){
  //for(i=0;i<3;i++){
    int x=0;
    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;
}