Crawling in process... Crawling failed Time Limit:2000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u
Description
Input
接下來的M行裏,每行包括3個整數a,b,c.代表a和b之間有一條通路,並且需要花費c元(c <= 100)。
Output
Sample Input
Sample Output
對於i到j的最短路徑有兩種形式,①經過k,最短路徑就是i到k加上k到j的距離;②不過k,最短路徑就是i 到j的距離。
求無向圖最小環,枚舉最大環的連接點,更新環,用dist[i][j]表示i到j的最短路徑,當前最小環的權用minn表示,則對於每一個接點k (1到k - 1中間結點的最短路徑都已經確定), 有minn = ( dist[i][j] + maze[j][k] + maze[k][i] , minn ),即環的權 = i到j的最短路徑(1 < i,j <= k - 1) + jk邊 + ki邊。
代碼:
#include"cstdio"
#include"cstring"
#include"iostream"
#include"algorithm"
using namespace std;
#define INF 100000000
int n,m;
int maze[105][105];
int dist[105][105];
void initial()
{
for(int i = 0;i < 105;i++)
{
for(int j = 0;j < 105;j++)
{
maze[i][j] = INF;
dist[i][j] = INF;
}
}
}
void Floyd()
{
int minn = INF;
for(int k = 1;k <= n;k++) //對於每一個k值,1到k - 1中間結點的最短路徑都已經確定
{
for(int i = 1;i < k;i++)
{
for(int j = i + 1;j < k;j++)
{
minn = min(dist[i][j] + maze[j][k] + maze[k][i],minn); //更新環的權值
}
}
for(int i = 1;i <= n;i++)
{
for(int j = 1;j <= n;j++)
{
dist[i][j] = min(dist[i][j],dist[i][k] + dist[k][j]); //保存i到j之間最短路徑
}
}
}
if(minn == INF)
{
printf("It's impossible.\n");
}
else
{
printf("%d\n",minn);
}
}
int main()
{
while(~scanf("%d%d",&n,&m))
{
initial();
int a,b,c;
while(m--)
{
scanf("%d%d%d",&a,&b,&c);
if(a != b && maze[a][b] > c)
{
maze[a][b] = c;
maze[b][a] = c;
dist[a][b] = c;
dist[b][a] = c;
}
}
Floyd();
}
return 0;
}