Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 31827 | Accepted: 17443 |
Description
Unfortunately for you, stockbrokers only trust information coming from their “Trusted sources” This means you have to take into account the structure of their contacts when starting a rumour. It takes a certain amount of time for a specific stockbroker to pass the rumour on to each of his colleagues. Your task will be to write a program that tells you which stockbroker to choose as your starting point for the rumour, as well as the time it will take for the rumour to spread throughout the stockbroker community. This duration is measured as the time needed for the last person to receive the information.
Input
Each person is numbered 1 through to the number of stockbrokers. The time taken to pass the message on will be between 1 and 10 minutes (inclusive), and the number of contacts will range between 0 and one less than the number of stockbrokers. The number of stockbrokers will range from 1 to 100. The input is terminated by a set of stockbrokers containing 0 (zero) people.
Output
It is possible that your program will receive a network of connections that excludes some persons, i.e. some people may be unreachable. If your program detects such a broken network, simply output the message “disjoint”. Note that the time taken to pass the message from person A to person B is not necessarily the same as the time taken to pass it from B to A, if such transmission is possible at all.
Sample Input
1 2 3 4 5 6 7 8 9 10 11 |
3 2 2 4 3 5 2 1 2 3 6 2 1 2 2 2 5 3 4 4 2 8 5 3 1 5 8 4 1 6 4 10 2 7 5 2 0 2 2 5 1 5 0 |
Sample Output
1 2 |
3 2 3 10 |
Source
是一道基础图论题, 应该是基本上所有的图论算法都可以用, 因为之前没有用过Floyd-Warshall算法, 所以这次用这道题来练练手.
Floyd-Warshall算法(Floyd-Warshall algorithm)是解决任意两点间的最短路径的一种算法,可以正確處理有向圖或负权的最短路径問題,同时也被用于计算有向图的传递闭包。算法的时间复杂度為O(N^3),空间复杂度为O(N^2)。
这道题的思路就是首先求出任意两点之间的权重, 注意题目中一个人可以同时向多个人传输情报, 比如A到B要10, A到C要6, A到D要2, 那么从A到B到C到D共起来只需要10, 所以在Floyd算法运行完毕过后, 图中初值必须为INF, 然后图的第i行的最大值便是从i点出发遍历全部所需要的时间, 如果最大值是INF, 则说明无法从这i点传递给所有点, 如果所有行的最大值都为INF, 则说明从任意点出发都不能传递给所有点, 应该直接输出 “disjoint”,但是我一开始忘记执行这个情况的判断却AC, 所以这道题数据库略水.
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 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 |
/************************************************************************* > File Name: poj1125.cpp > Author: razrLeLe > Mail: razrlele@outlook.com > Personal homepage: https://yueyu.io > Created Time: Thu 28 Aug 2014 12:08:02 AM CST ************************************************************************/ #include<vector> #include<list> #include<map> #include<set> #include<deque> #include<queue> #include<stack> #include<bitset> #include<algorithm> #include<functional> #include<numeric> #include<utility> #include<sstream> #include<iostream> #include<iomanip> #include<cstdio> #include<cmath> #include<cstdlib> #include<cctype> #include<string> #include<cstring> #include<ctime> #include<string.h> using namespace std; #define INF 0x3f3f3f3f const int MAX = 110; int g[MAX][MAX]; bool vis[MAX]; //int final_time[MAX]; int broker_num; //int countt; int min(int a, int b) { return a<b ?a:b; } /*void output() { for(int i = 1; i <= broker_num; i++) { for(int j = 1; j <= broker_num; j++) { if(g[i][j] > 100) cout < < "INF" << " "; else cout << g[i][j] << " "; if(j != broker_num) cout << "|" << " "; } cout << endl; } return ; }*/ void Floyd() { // countt = 1; int final_time; for(int i = 1; i <= broker_num; i++) { g[i][i] = 0; } for(int t = 1; t <= broker_num; t++) { for(int i = 1; i <= broker_num; i++) { for(int j = 1; j <= broker_num; j++) { if(i == j) continue; g[i][j] = min(g[i][j], g[i][t] + g[t][j]); // cout << countt++ << ":" << endl << endl; // output(); // cout << endl; } } } int ans = INF; for(int i = 1; i <= broker_num; i++) { int temp = 0; for(int j = 1; j <= broker_num; j++) { if(i == j) continue; if(g[i][j] > temp) { temp = g[i][j]; } } if(ans > temp) { ans = temp; // cout < < ans << endl; final_time = i; } } if(ans == INF) { cout << "disjoint" << endl; } else { printf("%d %dn", final_time, ans); } } int main() { int relat; int people, time; while(cin >> broker_num && broker_num) { memset(g, INF, sizeof(g)); memset(vis, false, sizeof(vis)); // memset(final_time, 0x3f, sizeof(final_time)); for(int i = 1; i < = broker_num; i++) { cin >> relat; for(int j = 1; j < = relat; j++) { cin >> people >> time; g[i][people] = time; } } Floyd(); } } |