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.
输入
- Line 1: A single integer: N
- Lines 2..N + 1: Line i+1 contains a single integer: W_i
- Lines N+2..2N+1: Line N+1+i contains N space-separated integers; the j-th integer is P_ij
输出
- Line 1: A single line with a single integer that is the minimum cost of providing all the pastures with water.
样例输入
4
5
4
4
3
0 2 2 2
2 0 3 3
2 3 0 4
2 3 4 0
样例输出
9
提示
INPUT DETAILS:
There are four pastures. It costs 5 to build a well in pasture 1,4 in pastures 2 and 3, 3 in pasture 4. Pipes cost 2, 3, and 4depending on which pastures they connect.
OUTPUT DETAILS:
Farmer John may build a well in the fourth pasture and connect each pasture to the first, which costs 3 + 2 + 2 + 2 = 9.
题意
农夫要打若干井，和挖连通两个田的水路，求所有田都有水的最小花费
题解
由于有一个挖井系统且必须要挖1个井，所以可以把井看成1个图上的点，然后跑一遍最小生成树即可
代码
1 #include<stdio.h> 2 #include<string.h> 3 #include<algorithm> 4 using namespace std; 5 6 #define MAXN 300+5 7 #define MAXM 90000 8 int n,m,F[MAXN],Vis[MAXN]; 9 struct edge 10 { 11 int u,v,w; 12 }edges[MAXM]; 13 int Find(int x) 14 { 15 return F[x]==-1?x:F[x]=Find(F[x]); 16 } 17 bool cmp(edge a,edge b) 18 { 19 return a.w<b.w; 20 } 21 int Kruskal() 22 { 23 memset(F,-1,sizeof(F)); 24 sort(edges,edges+m,cmp); 25 int ans=0,cnt=0,u,v,w,fx,fy; 26 for(int i=0;i<m;i++) 27 { 28 u=edges[i].u; 29 v=edges[i].v; 30 w=edges[i].w; 31 fx=Find(u); 32 fy=Find(v); 33 if(fx!=fy) 34 { 35 ans+=w; 36 cnt++; 37 F[fx]=fy; 38 } 39 if(cnt==n)break;//连通田n-1条边，加井1条边 40 } 41 return ans; 42 } 43 void addedges(int u,int v,int w) 44 { 45 edges[m].u=u; 46 edges[m].v=v; 47 edges[m++].w=w; 48 } 49 int main() 50 { 51 int w; 52 scanf("%d",&n); 53 for(int i=1;i<=n;i++) 54 { 55 scanf("%d",&w); 56 addedges(0,i,w); 57 } 58 for(int i=1;i<=n;i++) 59 { 60 for(int j=1;j<=n;j++) 61 { 62 scanf("%d",&w); 63 if(i>=j)continue; 64 addedges(i,j,w); 65 } 66 } 67 int ans=Kruskal(); 68 printf("%d ",ans); 69 return 0; 70 }