洛谷P1550 USACO08OCT]打井Watering Hole
題目背景
John的農場缺水了!!!
題目描述
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
農民John 決定將水引入到他的n(1<=n<=300)個牧場。他準備通過挖若
乾井,並在各塊田中修築水道來連通各塊田地以供水。在第i 號田中挖一口井需要花費W_i(1<=W_i<=100,000)元。連線i 號田與j 號田需要P_ij (1 <= P_ij <= 100,000 , P_ji=P_ij)元。
請求出農民John 需要為使所有農場都與有水的農場相連或擁有水井所需要的錢數。
輸入輸出格式
輸入格式:
第1 行為一個整數n。
第2 到n+1 行每行一個整數,從上到下分別為W_1 到W_n。
第n+2 到2n+1 行為一個矩陣,表示需要的經費(P_ij)。
輸出格式:
只有一行,為一個整數,表示所需要的錢數。
輸入輸出樣例
輸入樣例#1:
4 5 4 4 3 0 2 2 2 2 0 3 3 2 3 0 4 2 3 4 0
輸出樣例#1:
9
#include <iostream> #include <cstdio> #include <algorithm> using namespace std; int n,m; long long ans; struct node{ int x; int y; int z; }a[100005]; int s[100055]; int fa[100005]; int cnt=0; bool cmp(const node &x,const node &y) { return x.z<y.z; } int find(int x) { if(fa[x]==x) { return x; } return fa[x]=find(fa[x]); } void kkk() { int f1; int f2; int k=0; for(int i=1;i<=n;i++) fa[i]=i; for(int i=1;i<=cnt;i++) { f1=find(a[i].x); f2=find(a[i].y); if(f1!=f2) { ans=ans+a[i].z; fa[f1]=f2; k++; if(k==n-1) break; } } } int main(){ cin>>n; for(int i=1;i<=n;i++) { scanf("%d",&s[i]); } for(int i=1;i<=n;i++) { for(int j=1;j<=n;j++) { scanf("%d",&a[++cnt].z); a[cnt].x=i; a[cnt].y=j; } a[++cnt].x=i; a[cnt].y=n+1; a[cnt].z=s[i]; } n++; sort(a+1,a+1+cnt,cmp); kkk(); cout<<ans<<endl; }