AcWing 858. Prim演算法求最小生成樹
阿新 • • 發佈:2020-08-03
AcWing 858. Prim演算法求最小生成樹
//y總做法 #include <bits/stdc++.h> using namespace std; const int N = 510,INF=0x3f3f3f3f; int n, m; int g[N][N]; bool st[N]; int dist[N]; int prim() { memset(dist, 0x3f, sizeof dist); int res = 0; dist[n]=0; for (int i = 0; i < n; i++) { int t = -1; for (int j = 1; j <= n; j++) if (!st[j] && (t == -1 || dist[t] > dist[j])) t = j; if (i && dist[t] == INF) return INF; if (i) res += dist[t]; for (int j = 1; j <= n; j++) dist[j] = min(dist[j], g[t][j]); st[t] = true; } return res; } int main() { scanf("%d%d", &n, &m); memset(g, 0x3f, sizeof g); while (m--) { int a, b, c; scanf("%d%d%d", &a, &b, &c); g[a][b] = g[b][a] = min(g[a][b], c); } int t = prim(); if (t == INF) printf("impossible"); else printf("%d", t); return 0; } /* //演算法競賽進階指南 #include <bits/stdc++.h> using namespace std; const int N=510,INF=0x3f3f3f3f; int n,m; int g[N][N],ans; //dist表示點到集合的距離 int dist[N]; bool st[N]; void prim(){ memset(dist,0x3f,sizeof dist); dist[1]=0; for(int i=1;i<n;i++){ int t=-1; for(int j=1;j<=n;j++) if(!st[j]&&(t==-1||dist[t]>dist[j])) t=j; st[t]=true; for(int j=1;j<=n;j++) if(!st[j]) dist[j]=min(dist[j],g[t][j]); } } int main(){ scanf("%d%d",&n,&m); memset(g,0x3f,sizeof g); while(m--){ int a,b,c; scanf("%d%d%d",&a,&b,&c); g[a][b]=g[b][a]=min(g[a][b],c); } prim(); for(int i=2;i<=n;i++) ans+=dist[i]; if(ans>INF/2) printf("impossible"); else printf("%d",ans); return 0; } */