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这题我用prim算法和kruskal算法都提交了一下
从代码量来看,prim算法更简洁一些,从时间复杂度来看,prim:O(n^2),kruskal:O(mlogm),kruskal更优
kruskal:
#include <iostream> #include <algorithm> #include <cstring> using namespace std; const int Max = 2000000; struct ed { int u; int v; int w; }; ed e[Max]; int f[Max]; int x[Max]; bool cmp(ed a,ed b) { if(a.w<b.w) return true; else return false; } int getf(int v) { if(f[v] == v) return v; else { f[v] = getf(f[v]); return f[v]; } } int merge(int u,int v) { int t1,t2; t1 = getf(f[u]); t2 = getf(f[v]); if(t1 != t2) { f[t2] = t1; return 1; } return 0; } int main() { int n; cin >> n; int vn,en; while(n--) { int sum = 0; int count = 0; int minele; cin >> vn >> en; for(int i = 0; i < en; i++) cin >> e[i].u >> e[i].v >> e[i].w; for(int i = 0; i < vn; i++) cin >> x[i]; minele = *min_element(x,x+vn); sort(e,e+en,cmp); //按权值排序 for(int i = 1; i <= vn; i++) f[i] = i; for(int i = 0; i < en; i++) { if(merge(e[i].u,e[i].v)) { //若u和v不连通 count ++; sum = sum + e[i].w; } if(count == vn-1) { break; } } cout << sum + minele << endl; } return 0; }
prim:
#include <iostream> #include <algorithm> #include <cstring> using namespace std; const int Max = 550; const int inf = 999999999; int dis[Max] ; int e[Max][Max]; int book[Max]; int x[Max]; int main() { int n; int vn,en; cin >> n; while(n--) { cin >> vn >> en; for(int i =1; i <= vn ;i++) for(int j = 1; j <= vn ;j++) e[i][j] = inf; memset(book,0,sizeof(book)); int a,b,c; int sum = 0; int count = 1; for(int i = 0;i < en;i++) { cin >> a >> b >> c; e[a][b] = c; e[b][a] = c; } for(int i = 0; i < vn; i++) cin >> x[i]; int minele = *min_element(x,x+vn); for(int i =1 ; i <= vn;i++) dis[i] = e[1][i]; book[1] = 1; while(count < vn) { int mini = inf; int j; for(int i = 1; i <= vn;i ++) if(book[i]==0&&dis[i] < mini) { mini = dis[i]; j = i; } sum = sum + dis[j]; book[j] = 1; count++; for(int i = 1; i <= vn; i++) if(dis[i] > e[j][i]) //更新顶点i到树的最短距离 dis[i] = e[j][i]; } cout << sum + minele << endl; } return 0; }
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原文地址:http://www.cnblogs.com/ekinzhang/p/4396010.html