题目: http://poj.org/problem?id=1129 开始没读懂题,看discuss的做法,都是循环枚举的,很麻烦.然后我就决定dfs,调试了半天终于0ms A了. #include <stdio.h> #include <string.h> ][], vis[][]; int n, ans; void calc() { ; ; i < ; i++) { ; j < n; j++) { if(vis[j][i]) { cnt++; break; } }…

Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 10897   Accepted: 5594 Description When a radio station is broadcasting over a very large area, repeaters are used to retransmit the signal so that every receiver has a strong signal. Howev…

Channel Allocation Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 13357   Accepted: 6836 Description When a radio station is broadcasting over a very large area, repeaters are used to retransmit the signal so that every receiver has a s…

Channel Allocation Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 20000/10000K (Java/Other) Total Submission(s) : 18   Accepted Submission(s) : 7 Problem Description When a radio station is broadcasting over a very large area, repeaters are u…

Channel Allocation Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 13173   Accepted: 6737 Description When a radio station is broadcasting over a very large area, repeaters are used to retransmit the signal so that every receiver has a s…

Description When a radio station is broadcasting over a very large area, repeaters are used to retransmit the signal so that every receiver has a strong signal. However, the channels used by each repeater must be carefully chosen so that nearby repea…

Language: Default Channel Allocation Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 12367   Accepted: 6325 Description When a radio station is broadcasting over a very large area, repeaters are used to retransmit the signal so that ever…

Channel Allocation Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 15546   Accepted: 7871 Description When a radio station is broadcasting over a very large area, repeaters are used to retransmit the signal so that every receiver has a s…

题目:Channel Allocation 题意:要求A:BCD,A与B,C,D都不相同,求不同的值,典型的四色定理: #include <iostream> #include <algorithm> #include <stdlib.h> #include <time.h> #include <cmath> #include <cstdio> #include <string> #include <cstring&…

题目链接:poj1129 Channel Allocation 题意:要求相邻中继器必须使用不同的频道,求需要使用的频道的最少数目. 题解:就是求图的色数,这里采用求图的色数的近似有效算法——顺序着色算法(实质是一种贪心策略:在给任何一个顶点着色时,采用其邻接顶点中没有使用的,编号最小的颜色). 注:中继器网络是一个平面图,即图中不存在相交的边. 看讨论后发现这组数据,AC代码没过orz: 6 A:BEF B:AC C:BD D:CEF E:ADF F:ADE 正确答案应该是3 : A(1)B(…