首页 > ACM题库 > HDU-杭电 > HDU 1853 Cyclic Tour-二分图-[解题报告] C++
2013
12-23

HDU 1853 Cyclic Tour-二分图-[解题报告] C++

Cyclic Tour

问题描述 :

There are N cities in our country, and M one-way roads connecting them. Now Little Tom wants to make several cyclic tours, which satisfy that, each cycle contain at least two cities, and each city belongs to one cycle exactly. Tom wants the total length of all the tours minimum, but he is too lazy to calculate. Can you help him?

输入:

There are several test cases in the input. You should process to the end of file (EOF).
The first line of each test case contains two integers N (N ≤ 100) and M, indicating the number of cities and the number of roads. The M lines followed, each of them contains three numbers A, B, and C, indicating that there is a road from city A to city B, whose length is C. (1 ≤ A,B ≤ N, A ≠ B, 1 ≤ C ≤ 1000).

输出:

Output one number for each test case, indicating the minimum length of all the tours. If there are no such tours, output -1.

样例输入:

6 9
1 2 5
2 3 5
3 1 10
3 4 12
4 1 8
4 6 11
5 4 7
5 6 9
6 5 4
6 5
1 2 1
2 3 1
3 4 1
4 5 1
5 6 1

样例输出:

42
-1

Hint
In the first sample, there are two cycles, (1->2->3->1) and (6->5->4->6) whose length is 20 + 22 = 42.

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1853

题目大意:

给你n个城市和m条路,现在汤姆想要旅游所有的城市,而且每个城市只能经过一次,当然,旅游路上有一点的花费,现在问汤姆怎么走才能使总花费最小。

解题思路:

二分图最优匹配的最小权问题。KM取负数实现。。经典思路。

代码如下:

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<climits>
using namespace std;
#define N 505
#define MAXN 1<<28
int map[N][N];
int lx[N], ly[N];
int slack[N];
int match[N];
bool visitx[N], visity[N];
int n;

bool Hungary(int u)
{
	visitx[u] = true;
	for(int i = 1; i <= n; ++i)
	{
		if(visity[i])
			continue;
		else
		{
			if(lx[u] + ly[i] == map[u][i])
			{
				visity[i] = true;
				if(match[i] == -1 || Hungary(match[i]))
				{
					match[i] = u;
					return true;
				}
			}
			else
				slack[i] = min(slack[i], lx[u] + ly[i] - map[u][i]);
		}
	}
	return false;
}

void KM_perfect_match()
{
	int temp;
	for(int i = 1; i <= n; ++i)
		lx[i] = -MAXN;
	memset(ly, 0, sizeof(ly));
	for(int i = 1; i <= n; ++i)
		for(int j = 1; j <= n; ++j)
			lx[i] = max(lx[i], map[i][j]);
	for(int i = 1; i <= n; ++i)
	{
		for(int j = 1; j <= n; ++j)
			slack[j] = MAXN;
		while(1)
		{
			memset(visitx, false, sizeof(visitx));
			memset(visity, false, sizeof(visity));
			if(Hungary(i))
				break;
			else
			{
				temp = MAXN;
				for(int j = 1; j <= n; ++j)
					if(!visity[j])
						temp = min(temp, slack[j]);
				for(int j = 1; j <= n; ++j)
				{
					if(visitx[j])
						lx[j] -= temp;
					if(visity[j])
						ly[j] += temp;
					else
						slack[j] -= temp;
				}
			}
		}
	}
}

int main()
{
	int m;
	int a, b, cost;
	int ans;
	bool flag;
	while(scanf("%d%d", &n, &m) != EOF)
	{
		ans = 0;
		flag = true;
		memset(match, -1, sizeof(match));
		for(int i = 1; i <= n; ++i)
			for(int j = 1; j <= n; ++j)
				map[i][j] = -MAXN;
		for(int i = 1; i <= m; ++i)
		{
			scanf("%d%d%d", &a, &b, &cost); //防止有重边。取较小值(负数实现)
			if(-cost > map[a][b])
				map[a][b] = -cost;
		}
		KM_perfect_match();
		for(int i = 1; i <= n; ++i) //是否有完美匹配
		{
			if(match[i] == -1 || map[ match[i] ][i] == -MAXN)
			{
				flag = false;
				break;
			}
			ans += map[match[i]][i];
		}
		if(flag)
			printf("%d\n", -ans);
		else
			printf("-1\n");
	}
	return 0;
}

解题报告转自:http://blog.csdn.net/niushuai666/article/details/7172960


  1. 第二个方法挺不错。NewHead代表新的头节点,通过递归找到最后一个节点之后,就把这个节点赋给NewHead,然后一直返回返回,中途这个值是没有变化的,一边返回一边把相应的指针方向颠倒,最后结束时返回新的头节点到主函数。

  2. 约瑟夫也用说这么长……很成熟的一个问题了,分治的方法解起来o(n)就可以了,有兴趣可以看看具体数学的第一章,关于约瑟夫问题推导出了一系列的结论,很漂亮