fzu2271 X 图论

xiaoxiao2021-02-28  9

 Problem 2271 X

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

import java.io.BufferedReader; import java.io.InputStream; import java.io.InputStreamReader; import java.io.PrintWriter; import java.math.BigInteger; import java.util.Arrays; import java.util.StringTokenizer; public class Main { public static void main(String[] args) { new Task().solve(); } } class Task { InputReader in = new InputReader(System.in) ; PrintWriter out = new PrintWriter(System.out); final int inf = 10000000 ; void solve(){ int t = in.nextInt() ; for(int cas = 1 ; cas <= t ; cas++){ int n = in.nextInt() ; int[][] minLen = new int[n][n] ; int[][] dp = new int[n][n] ; boolean[][] slack = new boolean[n][n] ; for(int i = 0 ; i < n ; i++){ Arrays.fill(minLen[i], inf) ; Arrays.fill(dp[i] , inf) ; Arrays.fill(slack[i], false); } int sum = 0 ; int m = in.nextInt() ; for(int i = 0 ; i < m ; i++){ int u = in.nextInt() - 1 ; int v = in.nextInt() - 1 ; int s = in.nextInt() ; if(minLen[u][v] != inf){ sum++ ; } minLen[u][v] = minLen[v][u] = dp[u][v] = dp[v][u] = Math.min(s , minLen[u][v]) ; } for(int k = 0 ; k < n ; k++){ for(int i = 0 ; i < n ; i++){ for(int j = 0 ; j < n ; j++){ if(i == j || dp[i][k] == inf || dp[k][j] == inf){ continue ; } if(dp[i][j] >= dp[i][k] + dp[k][j]){ dp[i][j] = dp[i][k] + dp[k][j] ; slack[i][j] = true ; } } } } for(int i = 0 ; i < n ; i++){ for(int j = i+1 ; j < n ; j++){ if(minLen[i][j] == inf){ continue ; } if(dp[i][j] <= minLen[i][j] && slack[i][j]){ sum++ ; } } } out.println("Case " + cas + ": " + sum) ; } out.flush(); } } class InputReader { public BufferedReader reader; public StringTokenizer tokenizer; public InputReader(InputStream stream) { reader = new BufferedReader(new InputStreamReader(stream), 32768); tokenizer = new StringTokenizer(""); } private void eat(String s) { tokenizer = new StringTokenizer(s); } public String nextLine() { try { return reader.readLine(); } catch (Exception e) { return null; } } public boolean hasNext() { while (!tokenizer.hasMoreTokens()) { String s = nextLine(); if (s == null) return false; eat(s); } return true; } public String next() { hasNext(); return tokenizer.nextToken(); } public int nextInt() { return Integer.parseInt(next()); } public int[] nextInts(int n) { int[] nums = new int[n]; for (int i = 0; i < n; i++) { nums[i] = nextInt(); } return nums; } public long nextLong() { return Long.parseLong(next()); } public double nextDouble() { return Double.parseDouble(next()); } public BigInteger nextBigInteger() { return new BigInteger(next()); } }

转载请注明原文地址: https://www.6miu.com/read-450204.html

最新回复(0)