HDU 1853 Cyclic Tour（最小费用最大流）

Posted 2020-09-27 mfmdaoyou

Cyclic Tour

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/65535 K (Java/Others)
Total Submission(s): 1879    Accepted Submission(s): 938

Problem Description

There are N cities in our country, and M one-way roads connecting them. Now Little Tom wants to make several cyclic tours, which satisfy that, each cycle contain at least two cities, and each city belongs to one cycle exactly. Tom wants the total length of all the tours minimum, but he is too lazy to calculate. Can you help him?

 

Input

There are several test cases in the input. You should process to the end of file (EOF).
The first line of each test case contains two integers N (N ≤ 100) and M, indicating the number of cities and the number of roads. The M lines followed, each of them contains three numbers A, B, and C, indicating that there is a road from city A to city B, whose length is C. (1 ≤ A,B ≤ N, A ≠ B, 1 ≤ C ≤ 1000).

 

Output

Output one number for each test case, indicating the minimum length of all the tours. If there are no such tours, output -1.

 

Sample Input

6 9
1 2 5
2 3 5
3 1 10
3 4 12
4 1 8
4 6 11
5 4 7
5 6 9
6 5 4
6 5
1 2 1
2 3 1
3 4 1
4 5 1
5 6 1

 

Sample Output

42
-1

Hint
 In the first sample, there are two cycles, (1->2->3->1) and (6->5->4->6) whose length is 20 + 22 = 42. 
 

 

Author

[email protected]

 

Source

HDU 2007 Programming Contest - Final 

题意：给一个有向图，问每一个点都仅仅在一个简单环上。能够有多个环，边权总和最小为多少，假设不满足条件就输出-1。

#include<stdio.h>
#include<string.h>
#include<queue>
using namespace std;
const int MAXN = 10010;
const int MAXM = 100100;
const int INF = 1<<30;
struct EDG{
    int to,next,cap,flow;
    int cost;  //每条边的单位价格
}edg[MAXM];
int head[MAXN],eid;
int pre[MAXN], cost[MAXN]  ; //点0~(n-1)

void init(){
    eid=0;
    memset(head,-1,sizeof(head));
}
void addEdg(int u,int v,int cap,int cst){
    edg[eid].to=v; edg[eid].next=head[u]; edg[eid].cost = cst;
    edg[eid].cap=cap; edg[eid].flow=0; head[u]=eid++;

    edg[eid].to=u; edg[eid].next=head[v]; edg[eid].cost = -cst;
    edg[eid].cap=0; edg[eid].flow=0; head[v]=eid++;
}

bool inq[MAXN];
bool spfa(int sNode,int eNode,int n){
    queue<int>q;
    for(int i=0; i<n; i++){
        inq[i]=false; cost[i]= INF;
    }
    cost[sNode]=0; inq[sNode]=1; pre[sNode]=-1;
    q.push(sNode);
    while(!q.empty()){
        int u=q.front(); q.pop();
        inq[u]=0;
        for(int i=head[u]; i!=-1; i=edg[i].next){
            int v=edg[i].to;
            if(edg[i].cap-edg[i].flow>0 && cost[v]>cost[u]+edg[i].cost){ //在满足可增流的情况下，最小花费
                cost[v] = cost[u]+edg[i].cost;
                pre[v]=i;   //记录路径上的边
                if(!inq[v])
                    q.push(v),inq[v]=1;
            }
        }
    }
    return cost[eNode]!=INF;    //推断有没有增广路
}
//反回的是最大流，最小花费为minCost
int minCost_maxFlow(int sNode,int eNode ,int& minCost,int n){
    int ans=0;
    while(spfa(sNode,eNode,n)){
        ans++;
        for(int i=pre[eNode]; i!=-1; i=pre[edg[i^1].to]){
            edg[i].flow+=1; edg[i^1].flow-=1;
            minCost+=edg[i].cost;
        }
    }
    return ans;
}
void scanf(int &ans){
    char ch;
    while(ch=getchar()){
        if(ch>='0'&&ch<='9')
            break;
    }
    ans=ch-'0';
    while(ch=getchar()){
        if(ch<'0'||ch>'9')
            break;
        ans=ans*10+ch-'0';
    }
}
int  mapt[1005][1005];
int main(){
    int n,m , u, v, d ;
    while(scanf("%d%d",&n,&m)>0){
        init();
        int s=0, t=2*n+1;

        for(int i=1; i<=n; i++){
            addEdg(s , i , 1 , 0);
            addEdg(i+n , t , 1 , 0);
            for(int j=1; j<=n; j++)
                mapt[i][j]=INF;
        }
        while(m--){
            scanf(u); scanf(v); scanf(d);
            if(mapt[u][v]>d)
                mapt[u][v]=d;
        }
        for( u=1; u<=n; u++)
            for(v=1; v<=n; v++)
            if(mapt[u][v]!=INF)
                addEdg(u,v+n,1,mapt[u][v]);

        int mincost=0;
       n-= minCost_maxFlow(s , t , mincost , t+1);
        if(n==0)
            printf("%d\n",mincost);
        else
            printf("-1\n");
    }
}