2013
12-23

# Cyclic Tour

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?

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 one number for each test case, indicating the minimum length of all the tours. If there are no such tours, output -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
6 5
1 2 1
2 3 1
3 4 1
4 5 1
5 6 1

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.  

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<climits>
using namespace std;
#define N 505
#define MAXN 1<<28
int 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;
}