HOJ 2739 The Chinese Postman Problem

题目链接

带权有向图上的中国邮路问题：一名邮递员需要经过每条有向边至少一次，最后回到出发点，一条边多次经过权值要累加，问最小总权值是多少。

首先，每条边进过一次的话我们就先统计所有边sum值，然后就是判断哪些边重复走了。

统计所有点的入度与出度，设i点入度与出度的差为di；

如果di>0,addedge(s,i,di,0),即要走多di次去覆盖

如果di<0,addedge(i,t,|di|,0),同理

然后就是对于原来的边addedge(i,j,INF,cij)

这样跑一次最小费用最大流，然后费用加上sum   就是 所求；

注意的是，要图要连通，入度和出度>0

#include <iostream>
#include <queue>
#include <cstdio>
#include <cstring>
using namespace std;
const int MAXN = 10000;
const int MAXM = 10000;
const int INF = 0x3f3f3f3f;
struct Edge{
    int to,next,cap,flow,cost;
}edge[MAXM];
int head[MAXN],tol;
int pre[MAXN],dis[MAXN];
bool vis[MAXN];
int N;
void init(int n){
    N = n;
    tol = 0;
    memset(head,-1,sizeof(head));
}
void addedge(int u, int v, int cap, int cost){
    edge[tol].to = v;
    edge[tol].cap = cap;
    edge[tol].cost = cost;
    edge[tol].flow = 0;
    edge[tol].next = head[u];
    head[u] = tol++;
    edge[tol].to = u;
    edge[tol].cap = 0;
    edge[tol].cost = - cost;
    edge[tol].flow = 0;
    edge[tol].next = head[v];
    head[v] = tol++;
}
bool spfa(int s,int t){
    queue<int> q;
    for(int i = 0; i < N; i++){
        dis[i] = INF;
        vis[i] = false;
        pre[i] = -1;
    }
    dis[s] = 0;
    vis[s] = true;
    q.push(s);
    while(!q.empty()){

        int u = q.front();
        q.pop();
        vis[u] = false;
        for(int i = head[u]; ~i; i = edge[i].next){
            int v = edge[i].to;
            if(edge[i].cap > edge[i].flow && dis[v] > dis[u]+edge[i].cost){
                dis[v] = dis[u] + edge[i].cost;
                pre[v] = i;
                if(!vis[v]){
                    vis[v] = true;
                    q.push(v);
                }
            }
        }
    }
    if(pre[t] == -1)return false;
    return true;
}
int mincost(int s, int t, int &cost){
    int flow = 0;
    cost = 0;
    while(spfa(s,t)){
        int Min = INF;
        for(int i = pre[t]; i != -1; i = pre[edge[i^1].to]){
            if(Min > edge[i].cap - edge[i].flow)
                Min = edge[i].cap - edge[i].flow;
        }
        for(int i = pre[t]; i != -1; i = pre[edge[i^1].to]){
            edge[i].flow += Min;
            edge[i^1].flow -= Min;
            cost += edge[i].cost*Min;
        }
        flow += Min;
    }
    return flow;
}
int in[MAXN],out[MAXN],viss[MAXN];
void dfs(int u){
	viss[u] = 1;
	for(int i = head[u]; ~i; i = edge[i].next){
		if(!viss[edge[i].to]){
			dfs(edge[i].to);
		}
	}
}
int main(){
    //freopen("in.txt","r",stdin);
    int st, en, sum, u, v, c, n, m;
    int t;
    scanf("%d",&t);
    while(t--){
		scanf("%d%d",&n,&m);

        init(n+3);
        memset(in,0,sizeof(in));
        memset(out,0,sizeof(out));
        memset(viss,0,sizeof(vis));
        st = 0;
        en = n+1;
        sum = 0;
        for(int i = 1; i <= m; i++){
            scanf("%d%d%d",&u,&v,&c);
            u++,v++;
			addedge(u,v,INF,c);
			sum+=c;
			in[v]++,out[u]++;
        }
        dfs(1);
        bool flag = false;
        for(int i = 1; i <= n; i++){
			if(in[i] == 0 || out[i] == 0 || !viss[i]){
				flag = true;
				break;
			}
			if(in[i] > out[i]){
				addedge(st,i,in[i]-out[i],0);
			}
			else if(in[i] < out[i]){
				addedge(i,en,out[i]-in[i],0);
			}
        }
        if(flag){
			puts("-1");
			continue;
        }
        mincost(st,en,c);
        printf("%d\n",c+sum);
    }
    return 0;
}