Tour

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 2925    Accepted Submission(s): 1407

Problem Description

In
the kingdom of Henryy, there are N (2 <= N <= 200) cities, with M
(M <= 30000) one-way roads connecting them. You are lucky enough to
have a chance to have a tour in the kingdom. The route should be
designed as: The route should contain one or more loops. (A loop is a
route like: A->B->……->P->A.)
Every city should be just in one route.
A
loop should have at least two cities. In one route, each city should be
visited just once. (The only exception is that the first and the last
city should be the same and this city is visited twice.)
The total distance the N roads you have chosen should be minimized.

Input

An integer T in the first line indicates the number of the test cases.
In
each test case, the first line contains two integers N and M,
indicating the number of the cities and the one-way roads. Then M lines
followed, each line has three integers U, V and W (0 < W <=
10000), indicating that there is a road from U to V, with the distance
of W.
It is guaranteed that at least one valid arrangement of the tour is existed.
A blank line is followed after each test case.

Output

For each test case, output a line with exactly one integer, which is the minimum total distance.

Sample Input

1
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

Sample Output

42

题意:和hdu 1853题意和解法几乎一样,但是这题我看英文硬是没看懂。。。题意就是n个城市,每个城市都必须在一个环里面并且也只能出现在一个环里面?问最小的花费是多少?

题解:解法一:最小费用最大流:要去重 不然TLE。每个点只能出现一次,那么一个点容量限制为1，然后拆点跑最小费用最大流即可.

#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
using namespace std;
const int INF = 999999999;
const int N = 405;
const int M = 200005;
struct Edge{
    int u,v,cap,cost,next;
}edge[M];
int head[N],tot,low[N],pre[N];
int total ;
bool vis[N];
int flag[N][N];
void addEdge(int u,int v,int cap,int cost,int &k){
    edge[k].u=u,edge[k].v=v,edge[k].cap = cap,edge[k].cost = cost,edge[k].next = head[u],head[u] = k++;
    edge[k].u=v,edge[k].v=u,edge[k].cap = 0,edge[k].cost = -cost,edge[k].next = head[v],head[v] = k++;
}
void init(){
    memset(head,-1,sizeof(head));
    tot = 0;
}
bool spfa(int s,int t,int n){
    memset(vis,false,sizeof(vis));
    for(int i=0;i<=n;i++){
        low[i] = (i==s)?0:INF;
        pre[i] = -1;
    }
    queue<int> q;
    q.push(s);
    while(!q.empty()){
        int u = q.front();
        q.pop();
        vis[u] = false;
        for(int k=head[u];k!=-1;k=edge[k].next){
            int v = edge[k].v;
            if(edge[k].cap>0&&low[v]>low[u]+edge[k].cost){
                low[v] = low[u] + edge[k].cost;
                pre[v] = k; ///v为终点对应的边
                if(!vis[v]){
                    vis[v] = true;
                    q.push(v);
                }
            }
        }
    }
    if(pre[t]==-1) return false;
    return true;
}
int MCMF(int s,int t,int n){
    int mincost = 0,minflow,flow=0;
     while(spfa(s,t,n))
    {
        minflow=INF+1;
        for(int i=pre[t];i!=-1;i=pre[edge[i].u])
            minflow=min(minflow,edge[i].cap);
        flow+=minflow;
        for(int i=pre[t];i!=-1;i=pre[edge[i].u])
        {
            edge[i].cap-=minflow;
            edge[i^1].cap+=minflow;
        }
        mincost+=low[t]*minflow;
    }
    total=flow;
    return mincost;
}
int n,m;
int main(){
     int tcase;
     scanf("%d",&tcase);
     while(tcase--){
        init();
        scanf("%d%d",&n,&m);
        int src = 0,des = 2*n+1;
        for(int i=1;i<=n;i++){
            addEdge(src,i,1,0,tot);
            addEdge(i+n,des,1,0,tot);
        }
        memset(flag,-1,sizeof(flag));
        for(int i=1;i<=m;i++){  ///去重
            int u,v,w;
            scanf("%d%d%d",&u,&v,&w);
            if(flag[u][v]==-1||w<flag[u][v]){
                flag[u][v] = w;
            }
        }
        for(int i=1;i<=n;i++){
            for(int j=1;j<=n;j++){
                if(flag[i][j]!=-1){
                    addEdge(i,j+n,1,flag[i][j],tot);
                }
            }
        }
        int mincost = MCMF(src,des,2*n+2);
        if(total!=n) printf("-1\n");
        else printf("%d\n",mincost);
     }
}

题解二:KM算法,也是将一个点看成两个点,算最优匹配即可.

#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
using namespace std;
const int INF = 999999999;
const int N = 405;
int graph[N][N];
int lx[N],ly[N];
int linker[N];
bool x[N],y[N];
int n,m;
void init(){
    memset(lx,0,sizeof(lx));
    memset(ly,0,sizeof(ly));
    memset(linker,-1,sizeof(linker));
    for(int i=1;i<=n;i++){
        for(int j=1;j<=n;j++){
            if(lx[i]<graph[i][j]) lx[i] = graph[i][j];
        }
    }
}
bool dfs(int u){
    x[u] = true;
    for(int i=1;i<=n;i++){
        if(!y[i]&&graph[u][i]==lx[u]+ly[i]){
            y[i] = true;
            if(linker[i]==-1||dfs(linker[i])){
                linker[i] = u;
                return true;
            }
        }
    }
    return false;
}
int KM(){
    int sum = 0;
    init();
    for(int i=1;i<=n;i++){
        while(1){
            memset(x,false,sizeof(x));
            memset(y,false,sizeof(y));
            if(dfs(i)) break;
            int d = INF;
            for(int j=1;j<=n;j++){
                if(x[j]){
                    for(int k=1;k<=n;k++){
                        if(!y[k]) d = min(d,lx[j]+ly[k]-graph[j][k]);
                    }
                }
            }
            if(d==INF) break;
            for(int j=1;j<=n;j++){
                if(x[j]) lx[j]-=d;
                if(y[j]) ly[j]+=d;
            }
        }
    }
    for(int i=1;i<=n;i++){
        sum+=graph[linker[i]][i];
    }
    return sum;
}
int main()
{
    int tcase;
    scanf("%d",&tcase);
    while(tcase--){
        scanf("%d%d",&n,&m);
        for(int i=1;i<=n;i++){
            for(int j=1;j<=n;j++){
                graph[i][j] = -INF;
            }
        }
        for(int i=1;i<=m;i++){
            int u,v,w;
            scanf("%d%d%d",&u,&v,&w);
            graph[u][v] = max(graph[u][v],-w);
        }
        int ans = KM();
        printf("%d\n",-ans);
    }
    return 0;
}