首页 > ACM题库 > HDU-杭电 > HDU 4009-Transfer water-图-[解题报告]HOJ
2015
04-14

HDU 4009-Transfer water-图-[解题报告]HOJ

Transfer water

问题描述 :

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.

输入:

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.

输出:

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.

样例输入:

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

样例输出:

30

Hint
In 3�\dimensional space Manhattan distance of point A (x1, y1, z1) and B(x2, y2, z2) is |x2�\x1|+|y2�\y1|+|z2�\z1|.

题意:有一个村庄需要给每户人家提供水,有两个方法:1.自己凿井    2.从有井的人家引过来,不同的方式有不同的花费,问你解决每户人家供水问题的最小花费。

思路:构建一个超级源点,丛源点引一条边到每个点(花费为 home.z * X)),可以引水的两个人家间构建一条边,然后对超级源点跑一次最小树形图即可(引用HH大牛的模板)。

总结:这个题目如果不是看了最小树形图的专题的话,目测我是没有能力做出来的,做出来之后感觉这个题目的有一点对于最小树形图来说是比较敏感的——-解决方案只有一种,也就是说入度为1,注意这个之后或许就能联想到树结构了。继续深化。

/*
每个点找其最小的入边In[v] ? 如果有除跟节点以外的点找不到入边,则无解 : 否则答案累加In[v]
看看有没有环 ? 无环则已经找到解,返回答案 : 将环缩点
重新构图,每条边[u->v]的权值减去In[v],然后重复第一步*/
#include<stdio.h>
#include<string.h>
#include<algorithm>
#define M 1101
#define INF 1000000000
using namespace std;
struct edge
{
    int u , v , w;
}e[M * M];
struct point
{
    int x,y,z;
}home[M];
int pre[M],id[M],vis[M],in[M];
int n,m,X,Y,Z;
int Directed_MST(int root,int NV,int NE)
{
    int ret = 0;
    while(true){
        //步骤1:找到最小边
        for(int i = 0;i < NV;i ++)
            in[i] = INF;
        for(int i = 0;i < NE;i ++){
            int u = e[i].u , v = e[i].v;
            if(e[i].w < in[v] && u != v){
                pre[v] = u;
                in[v] = e[i].w;
            }
        }
        for(int i = 0;i < NV;i ++){
            if(i == root) continue;
            if(in[i] == INF) return -1;//除了根节点以外有点没有入边,则根无法到达他
        }
        int cntnode = 0;
        memset(id,-1,sizeof(id));
        memset(vis,-1,sizeof(vis));
        //找环
        in[root] = 0;
        for(int i = 0;i < NV;i ++){//标记每个环,编号
            ret += in[i];
            int v = i;
            while(vis[v] != i && id[v] == -1 && v != root){
                vis[v] = i;
                v = pre[v];
            }
            if(v != root && id[v] == -1){
                for(int u = pre[v];u != v;u = pre[u]){
                    id[u] = cntnode;
                }
                id[v] = cntnode ++;
            }
        }
        if(cntnode == 0) break;//无环
        for(int i = 0;i < NV;i ++)
            if(id[i] == -1){
                id[i] = cntnode ++;
            }
        //步骤3:缩点,重新标记
        for(int i = 0;i < NE;i ++){
            int v = e[i].v;
            e[i].u = id[e[i].u];
            e[i].v = id[e[i].v];
            if(e[i].u != e[i].v) e[i].w -= in[v];
        }
        NV = cntnode;
        root = id[root];
    }
    return ret;
}
int dis(int a,int b)
{
    return abs(home[a].x - home[b].x) + abs(home[a].y - home[b].y) + abs(home[a].z - home[b].z);
}
void add_edge(int u,int v,int w)
{
    e[m].u = u , e[m].v = v,e[m].w = w;
    m ++;
}
int main()
{
    while(scanf("%d%d%d%d",&n,&X,&Y,&Z) != EOF && n){
        m = 0;
        for(int i = 1;i <= n;i ++){
            scanf("%d%d%d",&home[i].x,&home[i].y,&home[i].z);
            add_edge(0,i,home[i].z * X);
        }
        for(int i = 1;i <= n;i ++){
            int num , u = i , v , w;
            scanf("%d",&num);
            for(int j = 0;j < num;j ++){
                scanf("%d",&v);
                if(v == i) continue;
                w = dis(u,v) * Y;
                if(home[v].z > home[u].z) w += Z;
                add_edge(u,v,w);
            }
        }
        int ans = Directed_MST(0,n + 1,m);
        printf("%d\n",ans);
    }
    return 0;
}

版权声明:本文为博主原创文章,未经博主允许不得转载。

参考:http://blog.csdn.net/yobobobo/article/details/8440712


,
  1. 网站做得很好看,内容也多,全。前段时间在博客园里看到有人说:网页的好坏看字体。觉得微软雅黑的字体很好看,然后现在这个网站也用的这个字体!nice!