Hdu 1853 Cyclic Tour

Posted 2020-09-23 寂地沉

Cyclic Tour

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/65535 K (Java/Others)
Total Submission(s): 2674    Accepted Submission(s): 1371

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

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<climits>
using namespace std;
#define N 505
#define MAXN 1<<28int map[N][N];
int lx[N], ly[N];
int slack[N];
int match[N];
bool visitx[N], visity[N];
int n;

bool Hungary(int u)
{
    visitx[u] = true;
    for(int i = 1; i <= n; ++i)
    {
        if(visity[i])
            continue;
        else
        {
            if(lx[u] + ly[i] == map[u][i])
            {
                visity[i] = true;
                if(match[i] == -1 || Hungary(match[i]))
                {
                    match[i] = u;
                    return true;
                }
            }
            else
                slack[i] = min(slack[i], lx[u] + ly[i] - map[u][i]);
        }
    }
    return false;
}

void KM_perfect_match()
{
    int temp;
    for(int i = 1; i <= n; ++i)
        lx[i] = -MAXN;
    memset(ly, 0, sizeof(ly));
    for(int i = 1; i <= n; ++i)
        for(int j = 1; j <= n; ++j)
            lx[i] = max(lx[i], map[i][j]);
    for(int i = 1; i <= n; ++i)
    {
        for(int j = 1; j <= n; ++j)
            slack[j] = MAXN;
        while(1)
        {
            memset(visitx, false, sizeof(visitx));
            memset(visity, false, sizeof(visity));
            if(Hungary(i))
                break;
            else
            {
                temp = MAXN;
                for(int j = 1; j <= n; ++j)
                    if(!visity[j])
                        temp = min(temp, slack[j]);
                for(int j = 1; j <= n; ++j)
                {
                    if(visitx[j])
                        lx[j] -= temp;
                    if(visity[j])
                        ly[j] += temp;
                    else
                        slack[j] -= temp;
                }
            }
        }
    }
}

int main()
{
    int m;
    int a, b, cost;
    int ans;
    bool flag;
    while(scanf("%d%d", &n, &m) != EOF)
    {
        ans = 0;
        flag = true;
        memset(match, -1, sizeof(match));
        for(int i = 1; i <= n; ++i)
            for(int j = 1; j <= n; ++j)
                map[i][j] = -MAXN;
        for(int i = 1; i <= m; ++i)
        {
            scanf("%d%d%d", &a, &b, &cost); //防止有重边。取较小值（负数实现）            if(-cost > map[a][b])
                map[a][b] = -cost;
        }
        KM_perfect_match();
        for(int i = 1; i <= n; ++i) //是否有完美匹配        {
            if(match[i] == -1 || map[ match[i] ][i] == -MAXN)
            {
                flag = false;
                break;
            }
            ans += map[match[i]][i];
        }
        if(flag)
            printf("%d\n", -ans);
        else
            printf("-1\n");
    }
    return 0;
}