Channel Allocation Time Limit: 1000 MS Memory Limit: 10000 KB 64-bit integer IO format: %I64d , %I64u Java class name: Main [Submit] [Status] [Discuss] Description When a radio station is broadcasting over a very large area, repeaters are used to ret…

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…

catch that cow POJ 3278 搜索 题意 原题链接 john想要抓到那只牛,John和牛的位置在数轴上表示为n和k,john有三种移动方式:1. 向前移动一个单位,2. 向后移动一个单位,3. 移动到当前位置的二倍处.输出移动的最少次数. 解题思路 使用搜索,准确地说是广搜,要记得到达的位置要进行标记,还有就是减枝. 详情见代码实现. 代码实现 #include<cstdio> #include<cstring> #include<iostream>…

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…

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…

目录 POJ 1426 POJ 1321 POJ 2718 POJ 3414 POJ 1416 POJ 2362 POJ 3126 POJ 3009 个人整了一些搜索的简单题目,大家可以clone来练习,vjudge链接直接公开了.以下部分代码省略了头文件,总之each就是for循环,rush()就是输入T个样例然后while(T--) vjudge练习地址:https://vjudge.net/contest/330414 POJ 1426 二进制的数实在太大了,所以用数组下标直接代表二进制数…

Dungeon Master Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 13923   Accepted: 5424 Description You are trapped in a 3D dungeon and need to find the quickest way out! The dungeon is composed of unit cubes which may or may not be filled…