Transfer water
Time Limit: 5000/3000 MS (Java/Others) Memory Limit: 65768/65768 K (Java/Others)
Total Submission(s): 1770 Accepted Submission(s): 650
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.
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
Recommend
lcy
今年大連賽區(qū)網(wǎng)絡(luò)賽一道題.....我怎么一點(diǎn)印象也沒有呢.....
一道赤裸的最小樹形圖，除了數(shù)據(jù)有點(diǎn)大而已.....
思路：因?yàn)闆]有根，所以虛擬一個(gè)根，所有點(diǎn)和這個(gè)根連線，權(quán)值是該點(diǎn)造井的價(jià)格，這樣以這個(gè)根出發(fā)，構(gòu)造出來的最小樹形圖就是最小的費(fèi)用了。
代碼：
#include <cstdio>
#include <cstring>
#include <cmath>
#include <iostream>
using namespace std;
const int maxn = 1100;
const int maxm = 1100000;
const int maxint = 0x3fffffff;
struct edge
{
int u, v, w;
edge(){}
edge(int u1, int v1, int w1) : u(u1), v(v1), w(w1){}
} e[maxm];
int root, n, edgeNum, vis[maxn], pre[maxn], belong[maxn], in[maxn];
int a[maxn], b[maxn], c[maxn];
int Abs(int a)
{
return a > 0 ? a : -a;
}
int Dis(int i, int j)
{
return Abs(a[i] - a[j]) + Abs(b[i] - b[j]) + Abs(c[i] - c[j]);
}
int solve()
{
int i, j, k, num, sum = 0;
n++;
while (1)
{
for (i = 0; i < n; ++i)
in[i] = maxint;
for (i = 0; i < edgeNum; ++i)
{
if (in[e[i].v] > e[i].w && e[i].u != e[i].v)
{
pre[e[i].v] = e[i].u;
in[e[i].v] = e[i].w;
}
}
memset(vis, -1, sizeof(vis));
memset(belong, -1, sizeof(belong));
in[root] = 0;
for (num = 0, i = 0; i < n; ++i)
{
sum += in[i];
j = i;
while (vis[j] != i && belong[j] == -1 && j != root)
{
vis[j] = i;
j = pre[j];
}
if (vis[j] == i)
{
for (k = pre[j]; k != j; k = pre[k])
belong[k] = num;
belong[j] = num++;
}
}
if (!num) return sum;
for (i = 0; i < n; ++i)
if (belong[i] == -1)
belong[i] = num++;
for (i = 0; i < edgeNum; ++i)
{
int j = e[i].v;
e[i].u = belong[e[i].u];
e[i].v = belong[e[i].v];
e[i].w -= in[j];
}
n = num;
root = belong[root];
}
return sum;
}
int main()
{
int x, y, z, i, j, k, ii, d;
while (scanf("%d%d%d%d", &n, &x, &y, &z) != EOF)
{
if (!n && !x && !y && !z) break;
for (edgeNum = 0, i = 1; i <= n; ++i)
{
scanf("%d%d%d", &a[i], &b[i], &c[i]);
e[edgeNum++] = edge(0, i, c[i] * x);
}
for (i = 1; i <= n; ++i)
{
scanf("%d", &k);
while (k--)
{
scanf("%d", &ii);
if (i == ii) continue;
d = Dis(i, ii);
if (c[i] >= c[ii])
e[edgeNum++] = edge(i, ii, d * y);
else
e[edgeNum++] = edge(i, ii, d * y + z);
}
}
root = 0;
printf("%d\n", solve());
}
return 0;
}
posted on 2011-10-15 22:20
LLawliet 閱讀(238)
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