标签：scanf for can include iostream prime 城市 code 生成
最小生成树。
题意就是有N座城市，每个城市有一定的幸福值a[i]。对于任意两个城市i和j，如果a[i]，a[j]，a[i]+a[j]中任意一者的值为素数，那么他们的边权就是min(min(a[i],a[j]),abs(a[i]-a[j]))。问题就是这一幅图的最小生成树。
显然，边一旦建出来了，这就是一道裸题。
#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <queue> using namespace std; typedef pair<int, int> pii; const int N = 600 + 10, maxn = 2000000 + 10, inf = 0x3f3f3f3f; int graph[N][N]; int mincost[N]; int vis[N]; int prime[maxn]; int prim(int n) { memset(mincost, 0x3f, sizeof mincost); memset(vis, 0, sizeof vis); mincost[1] = 0; int ans = 0, num = 0; while(true) { int v = -1; for(int i = 1; i <= n; i++) if(! vis[i] && (v == -1 || mincost[i] < mincost[v])) v = i; if(v == -1 || mincost[v] == inf) break; vis[v] = true; ans += mincost[v]; num++; for(int i = 1; i <= n; i++) mincost[i] = min(mincost[i], graph[v][i]); } return num == n ? ans : -1; } void filter() { prime[0] = prime[1] = 1; for(int i = 2; i * i < maxn; i++) if(! prime[i]) for(int j = i + i; j < maxn; j += i) prime[j] = 1; } int main() { filter(); int t, n, a[N]; scanf("%d", &t); while(t--) { memset(graph, 0x3f3f3f, sizeof(graph)); scanf("%d", &n); for(int i = 1; i <= n; i++) scanf("%d", &a[i]); for(int i = 1; i <= n; i++) for(int j = i + 1; j <= n; j++) { if(! prime[a[i]] || ! prime[a[j]] || ! prime[a[i]+a[j]]) { int cost = min(min(a[i], a[j]), abs(a[i] - a[j])); graph[i][j] = graph[j][i] = cost; } } printf("%d\n", prim(n)); } return 0; }
标签：scanf for can include iostream prime 城市 code 生成
原文地址：https://www.cnblogs.com/Vikyanite/p/11385567.html
