Crawling in process... Crawling failed Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u
Description
We know that there are already some roads between some villages and your job is the build some roads such that all the villages are connect and the length of all the roads built is minimum.
Input
Then there is an integer Q (0 <= Q <= N * (N + 1) / 2). Then come Q lines, each line contains two integers a and b (1 <= a < b <= N), which means the road between village a and village b has been built.
Output
Sample Input
Sample Output
#include"cstdio"
#include"cstring"
#include"iostream"
using namespace std;
#define INF 0x7f7f7f7f
int n;
int maze[105][105];
int dis[105];
bool vis[105];
void Prim()
{
int ans = 0;
dis[1] = 0;
vis[1] = true;
for(int i = 1;i <= n;i++)
{
int minn = INF;
int mark = INF;
for(int j = 1;j <= n;j++)
{
if(!vis[j] && dis[j] < minn)
{
minn = dis[j];
mark = j;
}
}
if(minn == INF)
{
break;
}
ans += dis[mark];
vis[mark] = true;
for(int j = 1;j <= n;j++)
{
if(!vis[j] && maze[mark][j] < dis[j])
{
dis[j] = maze[mark][j];
}
}
}
printf("%d\n",ans);
}
int main()
{
while(~scanf("%d",&n))
{
for(int i = 1;i <= n;i++)
{
for(int j = 1;j <= n;j++)
{
scanf("%d",&maze[i][j]);
}
}
int q;
scanf("%d",&q);
while(q--)
{
int a,b;
scanf("%d%d",&a,&b); //已經修建的道路
maze[a][b] = 0;
maze[b][a] = 0;
}
for(int i = 1;i <= n;i++)
{
dis[i] = maze[1][i];
}
memset(vis,false,sizeof(vis));
Prim();
}
return 0;
}