来源:HDU1233
Prim裸题
代码:
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
int n,m;
const int MAXN = 105;
const int INF = 0x3f3f3f3f;
int edge[MAXN][MAXN];
long long ans;
int d[MAXN];
int nxt[MAXN];
void prim(){
ans=0;
for(int i=1;i<=n;i++){
nxt[i]=1;
d[i]=edge[1][i];
}
nxt[1]=-1;
int dv=-1;
for(int i=0;i<n;i++){
int dm = INF;
dv=-1;
for(int j=1;j<=n;j++){
if(nxt[j]!=-1&&dm>d[j]){
dm=d[j];
dv=j;
}
}
if(dm==INF) return ;
ans+=(long long)dm;
nxt[dv]=-1;
for(int j=1;j<=n;j++){
if(nxt[j]!=-1&&d[j]>edge[dv][j])
d[j]=edge[dv][j];
}
}
}
int main(){
int x,y,z;
while(scanf("%d",&n)!=EOF&&n) {
m = n*(n-1)/2;
for(int i=0;i<m;i++){
scanf("%d%d%d",&x,&y,&z);
edge[x][y]=edge[y][x]=z;
}
prim();
printf("%lld\n",ans);
}
return 0;
}