Constructing Roads
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 12591 Accepted Submission(s): 4792
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.
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.
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<string>
#include<stack>
#include<queue>
#include<vector>
#include<algorithm>
#include<iostream>
using namespace std;
#ifdef __int64
typedef __int64 LL;
#else
typedef long long LL;
#endif
#define maxn 120
#define inf 0x3fffffff
int n,q;
int _map[maxn][maxn];
int lowc[maxn];
int main()
{
while(~scanf("%d",&n))
{
for(int i=0;i<n;i++)
{
lowc[i]=inf;
for(int j=0;j<n;j++)
{
scanf("%d",&_map[i][j]);
_map[j][i]=_map[i][j];
}
}
scanf("%d",&q);
while(q--)
{
int a,b;
scanf("%d%d",&a,&b);
_map[a-1][b-1]=_map[b-1][a-1]=0;
}
int next;
int start=0;
int ans=0;
while(1)//核心代碼
{
lowc[start]=-1;
next=-1;
int dis=inf;
for(int i=0;i<n;i++)
{
lowc[i]=min(lowc[i],_map[i][start]);
if(lowc[i]!=-1&&lowc[i]<dis)
{
dis=lowc[i];
next=i;
}
}
if(next==-1)
break;
start=next;
ans+=dis;
}
printf("%d\n",ans);
}
return 0;
}