标签：continue nim namespace bre iss als should time tle
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;
}
标签：continue nim namespace bre iss als should time tle
原文地址：http://www.cnblogs.com/zhangliu/p/7058051.html
