Time Limit:2000MS Memory Limit:65536KB 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
3
0 990 692
990 0 179
692 179 0
1
1 2
Sample Output
179 刚开始用kruskal，结果tle 了，可能是我的代码写挫了吧。改为prim，wa了，改了cost[u][v]=cost[v][u]
#include<cstdio>
#include<cstring>
const int INF=0x3f3f3f;
const int MAXN=110;//看清上限
bool vis[MAXN];
int lowc[MAXN];
int cost[MAXN][MAXN];
int prim(int n)
{
memset(vis,false,sizeof(vis));
int ans=0;
vis[1]=true;
for(int i=2;i<=n;i++)
lowc[i]=cost[1][i];
for(int i=2;i<=n;i++)//我开始这里写成了i=1；当然一直是return -1；
{
int minc=INF;
int p=-1;
for(int j=1;j<=n;j++)
if(!vis[j]&&lowc[j]<minc)
{
minc=lowc[j];
p=j;
}
if(minc==INF)
return -1;
ans+=minc;
vis[p]=true;
for(int j=1;j<=n;j++)
if(!vis[j]&&cost[p][j]<lowc[j])
lowc[j]=cost[p][j];
}
return ans;
}
int main()
{
int n;
while(scanf("%d",&n)!=EOF)
{
int w;
for(int i=1;i<=n;i++)
for(int j=1;j<=n;j++)
cost[i][j]=INF;
for(int i=1;i<=n;i++)
{
lowc[i]=INF;
for(int j=1;j<=n;j++)
{
scanf("%d",&w);
if(w<cost[i][j])//只保存最小权值
cost[i][j]=w;
}
}
int m;
scanf("%d",&m);
int u,v;
for(int i=1;i<=m;i++)
{
scanf("%d%d",&u,&v);
cost[u][v]=cost[v][u]=0;//这里不能只写cost[u][v]=0;
}
int ans=prim(n);
printf("%d\n",ans);//我忘了要\n
}
return 0;
}