poj 1129(dfs+图的四色定理)

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poj 1129(dfs+图的四色定理)

题目链接:http://poj.org/problem?id=1129 思路:根据图的四色定理,最多四种颜色就能满足题意,使得相邻的两部分颜色不同.而最多又只有26个点,因此直接dfs即可. #include<iostream> #include<cstdio> #include<cstring> #include<algorithm> using namespace std; ][]; ]; ]; int n,ans; bool Judge(int x,i…

Channel Allocation (poj 1129 dfs)

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…

POJ 1129：Channel Allocation 四色定理+暴力搜索

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…

迭代加深搜索 POJ 1129 Channel Allocation

POJ 1129 Channel Allocation Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 14191 Accepted: 7229 Description When a radio station is broadcasting over a very large area, repeaters are used to retransmit the signal so that every receive…

POJ 1129 Channel Allocation 四色定理dfs

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; } }…

四色定理+dfs（poj 1129）

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&…

poj 1129 Channel Allocation（图着色，DFS）

题意: N个中继站,相邻的中继站频道不得相同,问最少需要几个频道. 输入输出: Sample Input 2 A: B: 4 A:BC B:ACD C:ABD D:BC 4 A:BCD B:ACD C:ABD D:ABC 0 Sample Output 1 channel needed. 3 channels needed. 4 channels needed. 题意抽象+思路: 一张有N个点的无向图,对每个点进行染色,相邻的点颜色不得一致,最少需多少种颜色.DFS即可. 代码: <span s…