06月
19

HDU_4081 Qin Shi Huangs National Road System

发表于 • 本文总阅读数 次
字数统计: 1.2k

题目

  http://acm.hdu.edu.cn/showproblem.php?pid=4081

题目描述

  During the Warring States Period of ancient China(476 BC to 221 BC), there were seven kingdoms in China —- they were Qi, Chu, Yan, Han, Zhao, Wei and Qin. Ying Zheng was the king of the kingdom Qin. Through 9 years of wars, he finally conquered all six other kingdoms and became the first emperor of a unified China in 221 BC. That was Qin dynasty —- the first imperial dynasty of China(not to be confused with the Qing Dynasty, the last dynasty of China). So Ying Zheng named himself “Qin Shi Huang” because “Shi Huang” means “the first emperor” in Chinese.
  Qin Shi Huang undertook gigantic projects, including the first version of the Great Wall of China, the now famous city-sized mausoleum guarded by a life-sized Terracotta Army, and a massive national road system. There is a story about the road system:
  There were n cities in China and Qin Shi Huang wanted them all be connected by n-1 roads, in order that he could go to every city from the capital city Xianyang.
  Although Qin Shi Huang was a tyrant, he wanted the total length of all roads to be minimum,so that the road system may not cost too many people’s life. A daoshi (some kind of monk) named Xu Fu told Qin Shi Huang that he could build a road by magic and that magic road would cost no money and no labor. But Xu Fu could only build ONE magic road for Qin Shi Huang. So Qin Shi Huang had to decide where to build the magic road. Qin Shi Huang wanted the total length of all none magic roads to be as small as possible, but Xu Fu wanted the magic road to benefit as many people as possible —- So Qin Shi Huang decided that the value of A/B (the ratio of A to B) must be the maximum, which A is the total population of the two cites connected by the magic road, and B is the total length of none magic roads.
  Would you help Qin Shi Huang?
  A city can be considered as a point, and a road can be considered as a line segment connecting two points.

输入

  The first line contains an integer t meaning that there are t test cases(t <= 10).
  For each test case:
  The first line is an integer n meaning that there are n cities(2 < n <= 1000).
  Then n lines follow. Each line contains three integers X, Y and P ( 0 <= X, Y <= 1000, 0 < P < 100000). (X, Y) is the coordinate of a city and P is the population of that city.
  It is guaranteed that each city has a distinct location.

输出

  For each test case, print a line indicating the above mentioned maximum ratio A/B. The result should be rounded to 2 digits after decimal point.

样例输入

2
4
1 1 20
1 2 30
200 2 80
200 1 100
3
1 1 20
1 2 30
2 2 40

样例输出

65.00
70.00

题意

  有n个城市，给出每个城市的坐标和人口数，秦始皇想修一些路，使得这n个城市之间可以互相连通，同时希望花费最少的劳动力（修的路的长度总和越大，花费的劳动力越多）。有一个叫徐福的道士可以用魔法帮秦始皇修一条路，并且不需要额外的劳动力，不过徐福希望他用魔法修的路能帮到更多的人。所以他们最终决定采集A/B最小的方案，A是通过魔道连接的两个城市的总人口，而B是不用魔法修的路的总长度。求A/B的最小值。

题目解析

  次小生成树的变形。先求出最小生成树的总长度mst_l和maxd数组，然后枚举图上的每一条边，作为用魔法修的路，那么不用魔法修的路的总长度为mst_l-maxd[i][j]，再根据两个城市的总人口，就可以算出A/B的最小值。

代码

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
/* http://acm.hdu.edu.cn/showproblem.php?pid=4081 */
/* AC 109MS 16968K */
#include <stdio.h>
#include <string.h>
#include <math.h>

const int MAX_N = 1000 + 10;

typedef struct node
{
	int x, y, p;
} node;

int n;
node nd[MAX_N];
int mst[MAX_N];
double mp[MAX_N][MAX_N];
double maxd[MAX_N][MAX_N];

double prim()
{
	int i, j, k;
	double min, mst_l;
	double dist[MAX_N];

	for(i = 0; i < n; i++)
	{
		dist[i] = mp[0][i];
		mst[i] = 0;
	}
	dist[0] = -1;
	memset(maxd, 0, sizeof(maxd));

	mst_l = 0;
	for(i = 1; i < n; i++)
	{
		min = 1e10;
		for(j = 1; j < n; j++)
		{
			if(dist[j] != -1 && min > dist[j])
			{
				min = dist[j];
				k = j;
			}
		}
		if(min == 1e10)
		{
			return -1;
		}
		mst_l += min;
		dist[k] = -1;
		for(j = 0; j < n; j++)
		{
			if(dist[j] == -1 && j != k)
			{
				int pre = mst[k];
				maxd[j][k] = (maxd[j][pre] > mp[pre][k]) ? maxd[j][pre] : mp[pre][k];
				maxd[k][j] = maxd[j][k];
			}
			if(dist[j] != -1 && dist[j] > mp[k][j])
			{
				dist[j] = mp[k][j];
				mst[j] = k;
			}
		}
	}
	return mst_l;
}

double solve()
{
	int i, j;
	double mst_l, mst_sl, temp, ans;

	mst_l = prim();
	ans = 0;
	for(i = 0; i < n; i++)
	{
		for(j = 0; j < n; j++)
		{
			if(i == j)
			{
				continue;
			}
			temp = mst_l - maxd[i][j];
			if(ans < (nd[i].p + nd[j].p) / temp)
			{
				ans = (nd[i].p + nd[j].p) / temp;
			}
		}
	}
	return ans;
}

int main()
{
	int t, i, j;
	double ans;

	scanf("%d", &t);
	while(t--)
	{
		scanf("%d", &n);
		for(i = 0; i < n; i++)
		{
			scanf("%d %d %d", &nd[i].x, &nd[i].y, &nd[i].p);
		}
		for(i = 0; i < n; i++)
		{
			for(j = 0; j < n; j++)
			{
				mp[i][j] = pow(nd[i].x - nd[j].x, 2) + pow(nd[i].y - nd[j].y, 2);
				mp[i][j] = sqrt(mp[i][j]);
				mp[j][i] = mp[i][j];
			}
		}
		ans = solve();
		printf("%.2f\n", ans);
	}
	return 0;
}
分享到：
0%