# Dinic重边处理模板

Dinic重边处理模板

https://www.luogu.com.cn/problem/P2936

本题相比普通最大流题目只是多了一个重边的处理，意义不大，但还是想记录一下，反正也花不了多少时间。

这里的处理方式是使用二维数组预处理边集，将重边合并，再将该二维数组作为Dinic的输入，暂时没有想到更好的处理方式，望大佬指教。

#include <bits/stdc++.h>
using namespace std;
#define fre freopen("data.in","r",stdin);
#define ms(a) memset((a),0,sizeof(a))
#define go(i,a,b) for(register int (i)=(a);(i)<(b);++(i))
#define rep(i,a,b) for(register int (i)=(a);(i)<=(b);++(i))
#define sf(x) scanf("%d",&(x))
#define reg register
typedef long long LL;
const int inf=(0x7f7f7f7f);
const int maxn=1e2;
const int maxm=1e3;
int n;
struct node{int to,flow,next;}e[maxm<<1];
int head[maxn],cur[maxn],deep[maxn];
int cnt;
int m[100][100];
queue<int> q;
inline void add(int x,int y,int w){
    e[cnt].to=y,e[cnt].flow=w,e[cnt].next=head[x];
    head[x]=cnt++;
}
inline void in(){
    sf(n);
    memset(head,-1,sizeof(head));
    char x,y;int w;
    go(i,0,n){
        scanf(" %c %c %d",&x,&y,&w);
      //  cout<<x<<y<<w;
        m[x-'A'][y-'A']+=w;  //将重边合并
    }

    go(i,0,100)go(j,0,100)
        if(m[i][j]){
            add(i,j,m[i][j]);
            add(j,i,0);
        }
}
inline bool bfs(){
    ms(deep);
    deep[0]=1;q.push(0);
    int u,v;
    while(!q.empty()){
        u=q.front();q.pop();
        for(int i=head[u];~i;i=e[i].next){
            v=e[i].to;
            if(!deep[v]&&e[i].flow){
                deep[v]=deep[u]+1;
                q.push(v);
            }
        }
    }
    return deep[25];
}
int dfs(int now,int nowFlow){
    if(now==25)return nowFlow;
    int totFlow=0;
    int v;
    for(int i=cur[now];~i;i=e[i].next){
        cur[now]=i;v=e[i].to;
        if(deep[v]==deep[now]+1&&e[i].flow){
            int canFlow=0;
            canFlow=dfs(v,min(nowFlow,e[i].flow));
            if(!canFlow)continue;
            nowFlow-=canFlow,totFlow+=canFlow;
            e[i].flow-=canFlow,e[i^1].flow+=canFlow;
            if(nowFlow<=0)break;
        }
    }
    if(totFlow<=0)deep[now]=-2;
    return totFlow;
}
inline void Dinic(){
    int maxFlow=0;
    while(bfs()){
        memcpy(cur,head, sizeof(head));
        maxFlow+=dfs(0,inf);
    }
    printf("%d\n",maxFlow);
}
int main(){
    in();
    Dinic();
    return 0;
}