题目大意建模: 一个有N个节点的无向图,要求对每个节点进行染色,使得相邻两个节点颜色都不同,问最少需要多少种颜色? 那么题目就变成了一个经典的图的染色问题 例如:N=7 A:BCDEFG B:ACDEFG C:ABD D:ABCE E:ABDF F:ABEG G:ABF 画成图就是: 首先考虑四色定理:任何一张地图只用四种颜色就能使具有共同边界的国家着上不同的颜色 judge(int x,int y)枚举判断x的邻接点中是否着色y颜色的 1.正向考虑dfs(int num,int color)从…

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…

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: 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…

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:http://poj.org/problem?id=1129 题意:求最小m,使平面图能染成m色,相邻两块不同色由四色定理可知顶点最多需要4种颜色即可.我们于是从1开始试到3即可. #include <iostream> #include<cstdio> #include<cstring> #include<cstdlib> #include<stack> #include<queue> #include<iomanip&…

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; } }…

http://poj.org/problem?id=1129 import java.util.*; import java.math.*; public class Main { public static boolean flag=false; public static int ans=0; public static void main(String []args) { Scanner cin=new Scanner(System.in); int n; String str; whil…

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…

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(…