Transfer waterTime Limit: 5000/3000 MS (Java/Others) Memory Limit: 65768/65768 K (Java/Others)Total Submission(s): 5902 Accepted Submission(s): 2098 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. 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 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|.Source The 36th ACM/ICPC Asia Regional Dalian Site —— Online Contest Recommend lcy |
山上有一些村民, 每家可以自己挖井取水, 有一个花费(和高度有关), 也可以选择从比他们高的村民家中引水, 建造管道也有一个花费, 那么问最少的花费是多少.
这道题的话，其实是一个最小树形图的裸题，不过建图的时候需要思考一下，把水源建为0，分别与各个家建立边（就是打井这步操作）就好。然后以0为root跑一边最小树形图，不过要注意的地方是即使两个家的高度一样，在建水管的时候也是需要z这个额外费用的~
#include<cstdio>
#include<iostream>
#include<cmath>
using namespace std;
#define inf 0x3f3f3f3f
struct fuck
{
int from, to, len;
}ed[1500000];
struct fucker
{
int x, y, z;
}spot[1005];
int pre[1005];
int in[1005];
int vis[1005];
int id[1005];
int zhuliu(int root, int n, int m)
{
int res = 0;
while (1){
for (int s = 0; s < n; s++)
in[s] = inf;
for (int s = 0; s < m; s++)
if (ed[s].from != ed[s].to&&in[ed[s].to] > ed[s].len){
in[ed[s].to] = ed[s].len;
pre[ed[s].to] = ed[s].from;
}
for (int s = 0; s < n; s++)
if (in[s] == inf&&s != root)
return -1;
int sum_huan = 0;
in[root] = 0;
memset(id, -1, sizeof(id));
memset(vis, -1, sizeof(vis));
for (int s = 0; s < n; s++) {
res += in[s];
int v = s;
while(vis[v]!=s&&id[v]==-1&&v!=root){
vis[v] = s;
v = pre[v];
}
if (v != root&&id[v] == -1){
for (int w = pre[v]; w != v; w = pre[w])
id[w] = sum_huan;
id[v] = sum_huan++;
}
}
if (sum_huan == 0) {
break;
}
for (int s = 0; s < n; s++) {
if (id[s] == -1) {
id[s] = sum_huan++;
}
}
for (int s = 0; s < m; s++) {
int v = ed[s].to;
ed[s].from = id[ed[s].from];
ed[s].to = id[ed[s].to];
if (ed[s].from != ed[s].to)
ed[s].len -= in[v];
}
n = sum_huan;
root = id[root];
}
return res;
}
int main()
{
int n, x, y, z;
while (~scanf("%d%d%d%d", &n, &x, &y, &z) && n && x && y && z)
{
int cnt = 0, sum, t;
for (int s = 1; s <= n; s++)
{
scanf("%d%d%d", &spot[s].x, &spot[s].y, &spot[s].z);
}
for (int s = 1; s <= n; s++)
{
scanf("%d", &sum);
while (sum--)
{
scanf("%d", &t);
ed[cnt].from = s;
ed[cnt].to = t;
int len = abs(spot[s].x - spot[t].x) + abs(spot[s].y - spot[t].y) + abs(spot[s].z - spot[t].z);
ed[cnt++].len = spot[s].z >= spot[t].z ? (len)*y : (len)*y + z;
}
}
for (int s = 1; s <= n; s++)
{
ed[cnt].from = 0;
ed[cnt].to = s;
ed[cnt++].len = x*spot[s].z;
}
int ans = zhuliu(0, n + 1, cnt);
cout << ans << endl;
}
}