Drainage Ditches

思路：裸的最大流，有一点是用链式前向星写的网络流可以很好的解决重边问题

#include<bits/stdc++.h>
#define PI acos(-1.0)
#define pb push_back
#define F first
#define S second
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int N=1e5+5;
const int M=1e5+5;
const int MOD=1e9+7;
const int inf=1<<30;
template <class T>
bool sf(T &ret){ //Faster Input

    char c; int sgn; T bit=0.1;
    if(c=getchar(),c==EOF) return 0;
    while(c!='-'&&c!='.'&&(c<'0'||c>'9')) c=getchar();
    sgn=(c=='-')?-1:1;
    ret=(c=='-')?0:(c-'0');
    while(c=getchar(),c>='0'&&c<='9') ret=ret*10+(c-'0');
    if(c==' '||c=='\n'){ ret*=sgn; return 1; }
    while(c=getchar(),c>='0'&&c<='9') ret+=(c-'0')*bit,bit/=10;
    ret*=sgn;
    return 1;
}
int ver[M],edge[M],Next[M],head[N],d[N];
int s,t,ks;
int tot;
int maxflow;
void add(int x,int y,int z){
    ver[++tot]=y,edge[tot]=z,Next[tot]=head[x],head[x]=tot;
    ver[++tot]=x,edge[tot]=0,Next[tot]=head[y],head[y]=tot;
}

void init(){
    tot=1;
    maxflow=0;
//    memset(ver,0,sizeof ver);
//    memset(edge,0,sizeof edge);
//    memset(Next,0,sizeof Next);
    memset(head,-1,sizeof head);
}

bool bfs(){
    memset(d,0,sizeof d);
    queue<int> q;
    q.push(s);
    d[s]=1;
    while(!q.empty()){
        int x=q.front();q.pop();
        for(int i=head[x];i!=-1;i=Next[i]){
            if(edge[i] && !d[ver[i]]){
                q.push(ver[i]);
                d[ver[i]]=d[x]+1;
                if(ver[i]==t)   return true;
            }
        }
    }
    return false;
}

int dinic(int x,int flow){
    if(x==t)    return flow;
    int res=0;
    for(int i=head[x];i!=-1;i=Next[i]){
        if(edge[i]&&d[ver[i]]==d[x]+1){
            int k=dinic(ver[i],min(flow-res,edge[i]));
            if(!k)  d[ver[i]]=0;
            edge[i]-=k;
            edge[i^1]+=k;
//            cout <<i <<"  and  "<<(1^i)<<endl;
            res+=k;
        }
    }
    return res;
}
int n,m;
void solve(){
    swap(n,m);
    init();
    s=1,t=n;
    for(int i=1;i<=m;i++){
        int u,v,w;
        sf(u);sf(v);sf(w);
        add(u,v,w);
    }
    int flow;
    while(bfs())
        while(flow=dinic(s,inf))    maxflow+=flow;
    printf("%d\n",maxflow);
}

int main(void){
//    int T=1;
//    sf(T);
    while((scanf("%d%d",&n,&m))==2){
        solve();
    }

    return 0;
}