HDU4009：Transfer water（有向图的最小生成树）

Transfer water

Time Limit: 5000/3000 MS (Java/Others)    Memory Limit: 65768/65768 K (Java/Others)
Total Submission(s): 6126    Accepted Submission(s): 2181

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4009

Description:

XiaoA lives in a village. Last year flood rained the village. So they decide to move the whole village to the mountain nearby this year. There is no spring in the mountain, so each household could only dig a well or build a water line from other household. If the household decide to dig a well, the money for the well is the height of their house multiplies X dollar per meter. If the household decide to build a water line from other household, and if the height of which supply water is not lower than the one which get water, the money of one water line is the Manhattan distance of the two households multiplies Y dollar per meter. Or if the height of which supply water is lower than the one which get water, a water pump is needed except the water line. Z dollar should be paid for one water pump. In addition,therelation of the households must be considered. Some households may do not allow some other households build a water line from there house. Now given the 3‐dimensional position (a, b, c) of every household the c of which means height, can you calculate the minimal money the whole village need so that every household has water, or tell the leader if it can’t be done.

Input:

Multiple cases. 
First line of each case contains 4 integers n (1<=n<=1000), the number of the households, X (1<=X<=1000), Y (1<=Y<=1000), Z (1<=Z<=1000). 
Each of the next n lines contains 3 integers a, b, c means the position of the i‐th households, none of them will exceeded 1000. 
Then next n lines describe the relation between the households. The n+i+1‐th line describes the relation of the i‐th household. The line will begin with an integer k, and the next k integers are the household numbers that can build a water line from the i‐th household. 
If n=X=Y=Z=0, the input ends, and no output for that. 

Output:

One integer in one line for each case, the minimal money the whole village need so that every household has water. If the plan does not exist, print “poor XiaoA” in one line. 

Sample Input:

2 10 20 30
1 3 2
2 4 1
1 2
2 1 2
0 0 0 0

Sample Output:

30

题意：

现在给出n户人家，每户人家都有对应的海拔高度，现在每户人家需要水，获得水有两个来源：自己挖井，从其它人家修建水渠。

假设从u到v修建水渠，如果u的海拔较高，那么只需要支付水渠的费用；否则还要加上水泵的费用；如果自己挖井费用只和海拔有关。

问当所有人家都有水时，最小花费为多少。

题解：

这个题可以看成是有向图的最小生成树模型，毕竟是要用有向边把图连通嘛，这个题不存在不成功的情况（天灾人祸除外 = =）。

还是建立一个虚点，然后直接向每户人家连边，边权为打井的费用；之和再根据题目描述构造其它边。

最后从虚点出发跑朱刘算法就行了。

代码如下：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <queue>
#include <cmath>
#define INF 0x3f3f3f3f
using namespace std;
typedef long long ll;
const int N = 1005;
int n,x,y,z,tot;
struct Point{
    int x,y,z;
}p[N];
int dis(int a,int b){
    return abs(p[a].x-p[b].x)+abs(p[a].y-p[b].y)+abs(p[a].z-p[b].z);
}
struct Edge{
    int u,v,w;
}e[N*N];
int pre[N]; //记录前驱.
int id[N],vis[N],in[N];
int dirMst(int root){
    int ans=0;
    while(1){
        memset(in,INF,sizeof(in));
        memset(id,-1,sizeof(id));
        memset(vis,-1,sizeof(vis));
        for(int i=1;i<=tot;i++){
            int u=e[i].u,v=e[i].v,w=e[i].w;
            if(w<in[v] && v!=u){
                pre[v]=u;
                in[v]=w;
            }
        }           //求最小入边集
        in[root]=0;
        pre[root]=root;
        for(int i=0;i<n;i++){
            if(in[i]==INF) return -1;
            ans+=in[i];
        }
        int idx = 0; //新标号
        for(int i=0;i<n;i++){
            if(vis[i] == -1 ){
                int u = i;
                while(vis[u] == -1){
                    vis[u] = i;
                    u = pre[u];
                }
                if(vis[u]!=i || u==root) continue;     //判断是否形成环
                for(int v=pre[u];v!=u;v=pre[v] )
                    id[v]=idx;
                id[u] = idx++;
            }
        }
        if(idx==0) break;
        for(int i=0;i<n;i++){
            if(id[i]==-1) id[i]=idx++;
        }
        for(int i=1;i<=tot;i++){
            e[i].w-=in[e[i].v];
            e[i].u=id[e[i].u];
            e[i].v=id[e[i].v];
        }
        n = idx;
        root = id[root];//给根新的标号
    }
    return ans;
}
int main(){
    while(scanf("%d%d%d%d",&n,&x,&y,&z)!=EOF){
        if(n+x+y+z<=0) break ;
        tot=0;
        for(int i=1;i<=n;i++) scanf("%d%d%d",&p[i].x,&p[i].y,&p[i].z);
        for(int i=1;i<=n;i++){
            int k;
            scanf("%d",&k);
            for(int j=1;j<=k;j++){
                int id;
                scanf("%d",&id);
                e[++tot].u=i;e[tot].v=id;
                if(p[i].z>=p[id].z) e[tot].w=dis(i,id)*y;
                else e[tot].w=dis(i,id)*y+z;
            }
        }
        for(int i=1;i<=n;i++){
            e[++tot].u=0;e[tot].v=i;e[tot].w=p[i].z*x;
        }
        n++;
        int ans = dirMst(0);
        cout<<ans<<endl;
    }
    return 0;
}