A group of commandos were assigned a critical task. They are to destroy an enemy head quarter. The enemy head quarter consists of several buildings and the buildings are connected by roads. The commandos must visit each building and place a bomb at the base of each building. They start their mission at the base of a particular building and from there they disseminate to reach each building. The commandos must use the available roads to travel between buildings. Any of them can visit one building after another, but they must all gather at a common place when their task in done. In this PRoblem, you will be given the description of different enemy headquarters. Your job is to determine the minimum time needed to complete the mission. Each commando takes exactly one unit of time to move between buildings. You may assume that the time required to place a bomb is negligible. Each commando can carry unlimited number of bombs and there is an unlimited supply of commando troops for the mission.
Input Input starts with an integer T (≤50), denoting the number of test cases.
The first line of each case starts with a positive integer N (1 ≤ N ≤ 100), where N denotes the number of buildings in the head quarter. The next line contains a positive integer R, where R is the number of roads connecting two buildings. Each of the next R lines contain two distinct numbers u v (0 ≤ u, v < N), this means there is a road connecting building u to building v. The buildings are numbered from 0 to N-1. The last line of each case contains two integers s d (0 ≤ s, d < N). Where s denotes the building from where the mission starts and d denotes the building where they must meet. You may assume that two buildings will be directly connected by at most one road. The input will be given such that, it will be possible to go from any building to another by using one or more roads.
Output For each case, print the case number and the minimum time required to complete the mission.
Sample Input 2 4 3 0 1 2 1 1 3 0 3 2 1 0 1 1 0 Sample Output Case 1: 4 Case 2: 1
題目：有一個(gè)敢死隊(duì)，要摧毀一群建筑，他們從一個(gè)特定的建筑出發(fā)，最后到一個(gè)特定的建筑集合；
現(xiàn)在給你各個(gè)建筑之間的連接路線，在建筑中穿梭需要1個(gè)單位時(shí)間，問集合的最早時(shí)間。分析：圖論，最短路徑。直接計(jì)算起點(diǎn)s和終點(diǎn)e到那個(gè)其他所有點(diǎn)的最短路徑； floyd比較簡單 也可以用兩個(gè)dijk 因?yàn)閿?shù)據(jù)小。
#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>using namespace std;const int maxn = 101000;#define inf 0x3f3f3f3fint d[200];int s[200];int e[200][200];int main(){ int t; cin>>t; int cc=1; while(t--) { int n,m; cin>>n>>m; memset(e,0x3f,sizeof(e)); for(int i=0;i<n;i++) { for(int j=0;j<n;j++) e[i][j]=inf; e[i][i]=0; } for(int i=0;i<m;i++) { int a,b; cin>>a>>b; e[a][b]=1; e[b][a]=1; } int si,di; cin>>si>>di; for(int i=0;i<n;i++) for(int j=0;j<n;j++) for(int k=0;k<n;k++) { if(e[j][k]>e[j][i]+e[i][k]) e[j][k]=e[j][i]+e[i][k]; } int maxx=0; for(int i=0;i<n;i++) maxx=max(e[si][i]+e[i][di],maxx); printf("Case %d: %d/n",cc++,maxx); }}新聞熱點(diǎn)
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