You are part of a group of n teenage girls armed with cellphones. You have some news you want
to tell everyone in the group. The problem is that no two of you are in the same room, and you must
communicate using only cellphones. What’s worse is that due to excessive usage, your parents have
refused to pay your cellphone bills, so you must distribute the news by calling each other in the cheapest
possible way. You will call several of your friends, they will call some of their friends, and so on until
everyone in the group hears the news.
Each of you is using a different phone service provider, and you know the price of girl A calling girl B for all possible A and B. Not all of your friends like each other, and some of them will never
call people they don’t like. Your job is to find the cheapest possible sequence of calls so that the news
spreads from you to all n-1 other members of the group.
The first line of input gives the number of cases, N (N < 150). N test cases follow. Each one starts
with two lines containing n (0 ≤ n ≤ 1000) and m (0 ≤ m ≤ 40, 000). Girls are numbered from 0 to
n-1, and you are girl 0. The next m lines will each contain 3 integers, u, v and w, meaning that a call
from girl u to girl v costs w cents (0 ≤ w ≤ 1000). No other calls are possible because of grudges,
rivalries and because they are, like, lame. The input file size is around 1200 KB.
For each test case, output one line containing ‘Case #x:’ followed by the cost of the cheapest method
of distributing the news. If there is no solution, print ‘Possums!’ instead.
4
2
1
0 1 10
2
1
1 0 10
4
4
0 1 10
0 2 10
1 3 20
2 3 30
4
4
0 1 10
1 2 20
2 0 30
2 3 100
Sample Output
Case #1: 10
Case #2: Possums!
Case #3: 40
Case #4: 130
#include <bits/stdc++.h>
using namespace std;
const int inf = 0x3f3f3f3f;
const int mn = 1010, mm = 40010;
int n, m;
int s[mn][mn];
bool flag[mn];
bool vis[mn];
int pre[mn];
int zhuliu(int r)
{
int ans = 0;
memset(flag, 0, sizeof flag);
while(1)
{
for (int i = 0; i < n; i++)
{
if (i == r || flag[i])
continue;
pre[i] = i;
s[i][i] = inf;
for (int j = 0; j < n; j++)
if (!flag[j] && s[j][i] < s[pre[i]][i])
pre[i] = j;
if (pre[i] == i)
return -1;
}
int i;
for (i = 0; i < n; i++)
{
if (i == r || flag[i])
continue;
int j = i, cnt = 0;
while(j != r && pre[j] != i &&cnt <= n)
j = pre[j], cnt ++;
if (j == r || cnt > n)
continue;
break;
}
if (i == n)
{
for (int j = 0; j < n; j++)
if (j != r && !flag[j])
ans += s[pre[j]][j];
return ans;
}
memset(vis, 0, sizeof vis);
int j = i;
do
{
ans += s[pre[j]][j];
j = pre[j];
vis[j] = 1;
flag[j] = 1;
} while(j != i);
flag[i] = 0;
for (j = 0; j < n; j++)
{
if (!vis[j])
continue;
for (int k = 0; k < n; k++)
{
if (vis[k])
continue;
if (s[i][k] > s[j][k])
s[i][k] = s[j][k];
if (s[k][j] < inf && (s[k][j] - s[pre[j]][j]) < s[k][i])
s[k][i] = s[k][j] - s[pre[j]][j];
}
}
}
}
int main()
{
int T;
scanf("%d", &T);
for (int u = 1; u <= T; u++)
{
scanf("%d %d", &n, &m);
for (int i = 0; i < n; i++)
for (int j = 0; j < n; j++)
s[i][j] = inf;
for (int i = 0; i < m; i++)
{
int a, b, c;
scanf("%d %d %d", &a, &b, &c);
s[a][b] = c;
}
int ans = zhuliu(0);
printf("Case #%d: ", u);
if (ans == -1)
printf("Possums!\n");
else
printf("%d\n", ans);
}
return 0;
}