网络流24题11

https://loj.ac/problem/6122

来回相当于同时在起点开始不相交

跑一个最大费用最大流

最后输出答案时沿着反向边有容量dfs就行了

#include <iostream>
#include <map>
#include <cstring>
#include <queue>
using namespace std;
map<string,int> mapp;
const int maxn = 110;
string str[maxn];
const int inf = 0x3f3f3f3f;
int head[maxn*2];
int tot = 0;
struct edge{
    int v,nex,w,c;
}e[maxn*maxn*2];
void addedge(int u,int v,int w,int c){
    e[tot] = (edge){v,head[u],w,c,};
    head[u] = tot++;
    e[tot] = (edge){u,head[v],0,-c,};
    head[v] = tot++;
}
int dis[maxn*2];
int vis[maxn*2];
int pre[maxn*2];
bool spfa(int S,int T){
    memset(dis,0x3f,sizeof(dis));
    memset(vis,0,sizeof(vis));
    queue<int> q;
    q.push(S);
    dis[S] = 0;
    vis[S] = 1;
    while(!q.empty()){
        int now  = q.front();
        vis[now] = 0;
        q.pop();
        for(int i=head[now];i!=-1;i=e[i].nex){
            int v = e[i].v;
            int w = e[i].w;
            int c = e[i].c;
            if(w>0 && dis[v]>dis[now]+c){
                dis[v] = dis[now] +c;
                pre[v] = i;
                if(vis[v]==0){
                    vis[v] = 1;
                    q.push(v);
                }
            }
        }
    }
    if(dis[T]==0x3f3f3f3f) return false;
    return true;
}
int ed(int S,int T,int &flow){
    int p;
    int sum = 0;
    int maxflow = 0x3f3f3f3f;
    for(int i=T;i!=S;i=e[p^1].v){
        p = pre[i];
        maxflow = min(e[p].w,maxflow);
    }
    flow+=maxflow;
    for(int i=T;i!=S;i=e[p^1].v){
        p = pre[i];
        sum+=e[p].c*maxflow;
        e[p].w-=maxflow;
        e[p^1].w+=maxflow;
    }
    return sum;
}
int ans1[maxn];
int ans2[maxn];
int dfs(int now,int* p,int deep){
    p[deep] = now;
    for(int i=head[now];i!=-1;i=e[i].nex){
        int w = e[i^1].w;
        int v = e[i].v;
        if(w>0){
            e[i^1].w -=1;
            return dfs(e[i].v,p,deep+1);
        }
    }
    return deep+1;
}
    int n,v;
void solve(int S,int T){
    int flow = 0;
    int ans = 0;
    while(spfa(S,T)){
        ans+=ed(S,T,flow);
    }
    if(flow!=2) {
        cout<<"No Solution!"<<endl;
    }else{
        cout<<(-ans-2)<<endl;
        int k1,k2;
        k1 = dfs(1,ans1,0);
        k2 = dfs(1,ans2,0);
        for(int i=0;i<k1;i+=2){
            if(ans1[i]==T) continue;
            cout<<str[ans1[i]]<<endl;
        }
        for(int i=k2-1;i>=0;i-=2){
            if(ans2[i]==T) continue;
            if(ans2[i]==n) continue;
            cout<<str[ans2[i]]<<endl;
        }
    }
}
int main()
{
    memset(head,-1,sizeof(head));

    cin>>n>>v;
    int S = 0,T = 2*n+1;
    for(int i=1;i<=n;i++){
        cin>>str[i];
        mapp[str[i]] = i;
    }
    for(int i=1;i<=v;i++){
        string u,v;
        cin>>u>>v;
        addedge(mapp[u]+n,mapp[v],inf,0);
    }
    for(int i=2;i<n;i++){
        addedge(i,i+n,1,-1);
    }
    addedge(1,1+n,2,-1);
    addedge(n,n*2,2,-1);
    addedge(S,1,2,0);
    addedge(n*2,T,2,0);
    solve(S,T);
    return 0;
}

原文地址：https://www.cnblogs.com/tjucxz/p/8637642.html