Cyclic Tour

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/65535 K (Java/Others)

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. 

和HDU 3488同类型的题，具体的解析请看 ：HDU3488

水一遍熟悉熟悉模板。

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#define INF 0x3f3f3f3f
#define maxn 500
#define maxm 77000
using namespace std;
int n, m;
int outset;
int inset;
struct node {
    int u, v, cap, flow, cost, next;
};

node edge[maxm];
int head[maxn], cnt;
int per[maxn];//记录增广路径上 到达点i的边的编号
int dist[maxn], vis[maxn];

void init(){
    cnt = 0;
    memset(head, -1, sizeof(head));
}

void add(int u, int v, int w, int c){
    int i;
    for(i = head[u]; i != -1; i = edge[i].next){
        node E = edge[i];
        if(v == E.v)
            break;
    }
    if(i != -1){
        if(edge[i].cost > c)
            edge[i].cost = c, edge[i ^ 1].cost = -c;
        return ;
    }
    //edge[cnt] = {u, v, w, 0, c, head[u]}这样写就直接超时了。
    node E1 = {u, v, w, 0, c, head[u]};
    edge[cnt] = E1;
    head[u] = cnt++;
    node E2 = {v, u, 0, 0, -c, head[v]};
    edge[cnt] = E2;
    head[v] = cnt++;
}

void getmap(){
    outset = 0;
    inset = n * 2 + 1;
    for(int i = 1; i <= n; ++i){
        add(outset, i, 1, 0);
        add(i + n, inset, 1, 0);
    }
    while(m--){
        int a, b, c;
        scanf("%d%d%d", &a, &b, &c);
        add(a, b + n, 1, c);
    }
}

bool SPFA(int st, int ed){
    queue<int>q;
    for(int i = 0; i <= inset; ++i){
        dist[i] = INF;
        vis[i] = 0;
        per[i] = -1;
    }
    dist[st] = 0;
    vis[st] = 1;
    q.push(st);
    while(!q.empty()){
        int u = q.front();
        q.pop();
        vis[u] = 0;
        for(int i = head[u]; i != -1; i = edge[i].next){
            node E = edge[i];
            if(dist[E.v] > dist[u] + E.cost && E.cap > E.flow){//可以松弛 且 没有满流
                dist[E.v] = dist[u] + E.cost;
                per[E.v] = i;//记录到达这个点的边的编号
                if(!vis[E.v]){
                    vis[E.v] = 1;
                    q.push(E.v);
                }
            }
        }
    }
    return per[ed] != -1;
}

void MCMF(int st, int ed, int &cost, int &flow){
    flow = 0;//总流量
    cost = 0;//总费用
    while(SPFA(st, ed)){//每次寻找花销最小的路径
        int mins = INF;
        for(int i = per[ed]; i != -1; i = per[edge[i ^ 1].v]){
            mins = min(mins, edge[i].cap - edge[i].flow);
        }
         //增广
        for(int i = per[ed]; i != -1; i = per[edge[i ^ 1].v]){
            edge[i].flow += mins;
            edge[i ^ 1].flow -= mins;
            cost += edge[i].cost * mins;
        }
        flow += mins;
    }
}
int main (){
    while(scanf("%d%d", &n, &m) != EOF){
        init();
        getmap();
        int cost, flow;
        MCMF(outset, inset, cost, flow);
        if(flow == n)
            printf("%d\n", cost);
        else
            printf("-1\n");
    }
    return 0;
}

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