FZU 2271 X（Floyd）

最新推荐文章于 2019-05-16 20:56:34 发布

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本文介绍了一个算法问题，即在一个完全连通的网络中确定最多能删除多少条边，而不会改变任意两城市间的最短路径长度。通过使用Floyd算法并结合标记已访问边的方法，有效地解决了这一问题。

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Problem 2271 X

Accept: 166 Submit: 584
Time Limit: 1500 mSec Memory Limit : 32768 KB

Problem Description

X is a fully prosperous country, especially known for its complicated transportation networks. But recently, for the sake of better controlling by the government, the president Fat Brother thinks it’s time to close some roads in order to make the transportation system more effective.

Country X has N cities, the cities are connected by some undirected roads and it’s possible to travel from one city to any other city by these roads. Now the president Fat Brother wants to know that how many roads can be closed at most such that the distance between any two cities in country X does not change. Note that the distance between city A and city B is the minimum total length of the roads you need to travel from A to B.

Input

The first line of the date is an integer T (1 <= T <= 50), which is the number of the text cases.

Then T cases follow, each case starts with two numbers N, M (1 <= N <= 100, 1 <= M <= 40000) which describe the number of the cities and the number of the roads in country X. Each case goes with M lines, each line consists of three integers x, y, s (1 <= x, y <= N, 1 <= s <= 10, x is not equal to y), which means that there is a road between city x and city y and the length of it is s. Note that there may be more than one roads between two cities.

Output

For each case, output the case number first, then output the number of the roads that could be closed. This number should be as large as possible.

See the sample input and output for more details.

Sample Input

22 31 2 11 2 11 2 23 31 2 12 3 11 3 1

Sample Output

Case 1: 2Case 2: 0

题意：求可以删除的边数

思路：利用floyd来判断该边是否可以被替换，用visit数组来记录是否被访问

上代码

#include<iostream>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<stack>
#include<queue>
#include<vector>
#include<cstdio>
#include<string>
#include<map>
using namespace std;
#define ll long long
#define inf 0x3f3f3f3f
#define Max 105
int M[Max][Max],MM[Max][Max];
bool visit[Max][Max];
int n,sum;
void floyd()
{
	for (int k = 1; k <= n; k++)
		for (int i = 1; i <= n; i++)
			for (int j = 1; j <= n; j++)
				if (M[i][k] + M[k][j] <= M[i][j])
					M[i][j] = M[i][k] + M[k][j], visit[i][j] = true;
}
int main()
{
	//freopen("Text.txt","r",stdin);
	int t,m,u,v,c,k=1;
	cin >> t;
	while (t--)
	{
		sum = 0;
		scanf("%d%d", &n, &m);
		memset(M, inf, sizeof(M));
		memset(visit, false, sizeof(visit));
		while (m--)
		{
			scanf("%d%d%d", &u, &v, &c);
			if (M[u][v] != inf)//重边删除
				sum++;
			if (M[u][v] > c)
			{
				M[u][v] = c;
				M[v][u] = c;
			}
		}
		memcpy(MM, M, sizeof(MM));
		floyd();
		for(int i=1;i<=n;i++)
			for (int j = 1; j < i; j++)//矩阵的一半就可以，不然除二
			{
				if (MM[i][j] == inf)
					continue;
				if (M[i][j] <= MM[i][j] && visit[i][j])//这条边可以被替换，替换无非就是某两条组成的距离大于等于原来的距离
					sum++;
			}
		printf("Case %d: %d\n", k++, sum);
	}
	return 0;
}