Transfer water
Time Limit: 5000/3000 MS (Java/Others) Memory Limit: 65768/65768 K (Java/Others)
Total Submission(s): 3775 Accepted Submission(s): 1356
Problem 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.
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
30HintIn 3‐dimensional space Manhattan distance of point A (x1, y1, z1) and B(x2, y2, z2) is |x2‐x1|+|y2‐y1|+|z2‐z1|.
Source
题解：水源能够看作从虚拟根节点引出来的。这道题必然有解。由于大不了每一个实际点都跟根节点相连嘛，所以ZL_MST函数里的推断非根无入边节点能够忽略掉。
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <limits.h>
#define maxn 1002
#define maxm 1000002
int X, Y, Z;
struct Node{
int x, y, z;
} ver[maxn];
struct Node2{
int u, v, cost;
} E[maxm];
int in[maxn], hash[maxn], vis[maxn], pre[maxn];
int calDist(Node a, Node b){
return abs(a.x - b.x) + abs(a.y - b.y) + abs(a.z - b.z);
}
__int64 ZL_MST(int root, int nv, int ne)
{
__int64 ans = 0;
int u, v, i, cnt;
while(true){
//0.初始化
for(i = 0; i < nv; ++i) in[i] = INT_MAX;
//1.找最小入边集
for(i = 0; i < ne; ++i){
u = E[i].u; v = E[i].v;
if(E[i].cost < in[v] && u != v){
in[v] = E[i].cost; pre[v] = u;
}
}
//2.找非根无入边点(略)。由于必然有解
//3.找环。加权，又一次标号
memset(hash, -1, sizeof(hash));
memset(vis, -1, sizeof(vis));
cnt = in[root] = 0;
for(i = 0; i < nv; ++i){
ans += in[i]; v = i;
while(vis[v] != i && v != root && hash[v] == -1){
vis[v] = i; v = pre[v];
}
if(v != root && hash[v] == -1){
for(u = pre[v]; u != v; u = pre[u])
hash[u] = cnt;
hash[v] = cnt++;
}
}
if(cnt == 0) return ans; //无环，算法完毕
for(i = 0; i < nv; ++i)
if(hash[i] == -1) hash[i] = cnt++;
//4.缩点，遍历每一条边，又一次构图
for(i = 0; i < ne; ++i){
v = E[i].v;
E[i].u = hash[E[i].u];
E[i].v = hash[E[i].v];
if(E[i].u != E[i].v) E[i].cost -= in[v];
}
//顶点数降低
nv = cnt; root = hash[root];
}
return ans;
}
int main()
{
int n, i, a, b, id;
while(scanf("%d%d%d%d", &n, &X, &Y, &Z) != EOF && (n||X||Y||Z)){
for(i = 0; i < n; ++i)
scanf("%d%d%d", &ver[i].x, &ver[i].y, &ver[i].z);
for(i = id = 0; i < n; ++i){
scanf("%d", &a);
while(a--){
scanf("%d", &b);
E[id].cost = calDist(ver[i], ver[--b]) * Y;
if(ver[b].z > ver[i].z) E[id].cost += Z;
E[id].u = i; E[id++].v = b;
}
}
for(i = 0; i < n; ++i){
E[id].u = n; E[id].v = i;
E[id++].cost = ver[i].z * X;
}
printf("%I64d
", ZL_MST(n, n + 1, id));
}
return 0;
}