John的农场缺水了！！！
Farmer John has decided to bring water to his N (1 <= N <= 300) pastures which are conveniently numbered 1..N. He may bring water to a pasture either by building a well in that pasture or connecting the pasture via a pipe to another pasture which already has water.
Digging a well in pasture i costs W_i (1 <= W_i <= 100,000).
Connecting pastures i and j with a pipe costs P_ij (1 <= P_ij <= 100,000; P_ij = P_ji; P_ii=0).
Determine the minimum amount Farmer John will have to pay to water all of his pastures.
POINTS: 400
农民John 决定将水引入到他的n(1<=n<=300)个牧场。他准备通过挖若
干井，并在各块田中修筑水道来连通各块田地以供水。在第i 号田中挖一口井需要花费W_i(1<=W_i<=100,000)元。连接i 号田与j 号田需要P_ij (1 <= P_ij <= 100,000 , P_ji=P_ij)元。
请求出农民John 需要为连通整个牧场的每一块田地所需要的钱数。
输入格式：
第1 行为一个整数n。
第2 到n+1 行每行一个整数，从上到下分别为W_1 到W_n。
第n+2 到2n+1 行为一个矩阵，表示需要的经费（P_ij）。
输出格式：
只有一行，为一个整数，表示所需要的钱数。
输出样例#1： 9<br>
John等着用水，你只有1s时间！！！
好吧，原谅我太弱了，我也是看的题解，直接被泥萌的脑洞给吓着了。
$$%%%$$
居然是把打井的点看做是与农夫山泉农夫三拳连起来，就是和0号节点连边，然后直接就是最小生成树的模板题了。
脑洞脑洞
上代码
#include <algorithm> #include <iostream> #include <cstring> #include <cstdio> using namespace std; int n, xxx, cnt, ans, tot, f[307]; bool vis[307][307]; struct edge{int u, v, w;}ed[301*301]; int find(int x) { if(f[x] == x) return x; else return f[x] = find(f[x]); } bool cmp(edge a, edge b) { return a.w < b.w; } int main() { scanf("%d", &n); for(int i=1; i<=n; i++) {scanf("%d", &ed[++cnt].w); ed[cnt].u = 0, ed[cnt].v = i;} for(int i=1; i<=n; i++) f[i] = i; for(int i=1; i<=n; i++) { for(int j=1; j<=n; j++) { scanf("%d", &xxx); if(i != j&&!vis[i][j]&&!vis[j][i]) { ed[++cnt].u = i, ed[cnt].v = j, ed[cnt].w = xxx; vis[i][j] = 1, vis[j][i] = 1; } } } sort(ed+1, ed+1+cnt, cmp); for(int i=1; i<=cnt; i++) { int x = find(ed[i].u), y = find(ed[i].v); if(x != y) { f[x] = find(y); ans += ed[i].w; tot++; } if(tot == n) break; } printf("%d", ans); return 0; }
转载于:https://www.cnblogs.com/bljfy/p/8946750.html
相关资源：数据结构—成绩单生成器