poj——3177Redundant Paths      洛谷—— P2860 [USACO06JAN]冗余路径Redundant Paths Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 15272   Accepted: 6436 Description In order to get from one of the F (1 <= F <= 5,000) grazing fields (which are nu…

Redundant Paths Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 10798   Accepted: 4626 Description In order to get from one of the F (1 <= F <= 5,000) grazing fields (which are numbered 1..F) to another field, Bessie and the rest of the…

http://poj.org/problem?id=1942 题意 :在一个n*m的矩形上有n*m个网格,从左下角的网格划到右上角的网格,沿着边画,只能向上或向右走,问有多少条不重复的路 . 思路 :这种问题记得高中的时候就做过,学组合数的时候讲的,反正就是向上向右走,加起来要走的路必定为n+m条,选择n条向上,必定剩下的m为向右的,所以这个题就转化求C(n,m+n),或者是C(m,m+n),不过个人建议用m,n中小的那个数去做,因为省时.因为这个数据较大,所以求组合的时候就要注意以防超时,如果…

题目:http://poj.org/problem?id=1942 题意:给定一个矩形网格的长m和高n,其中m和n都是unsigned int32类型,一格代表一个单位,就是一步,求从左下角到右上角有多少种走法,每步只能向上或者向右走 题解:就是 上和 右的排列. 用c(m,n)=n!/m!*(n-m)!; 最重要的是算阶乘. #include<iostream> #include<cstring> #include<cstdio> using namespace st…

Paths on a Grid Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 21297   Accepted: 5212 Description Imagine you are attending your math lesson at school. Once again, you are bored because your teacher tells things that you already mastere…

// n*m 的格子 从左下角走到右上角的种数// 相当于从 n+m 的步数中选 m 步往上走// C(n+m,m) #include <iostream> #include <string> #include<sstream> #include <cmath> #include <map> #include <stdio.h> #include <string.h> #include <algorithm>…

Redundant Paths Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 13717   Accepted: 5824 Description In order to get from one of the F (1 <= F <= 5,000) grazing fields (which are numbered 1..F) to another field, Bessie and the rest of the…

POJ 3177 Redundant Paths Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 12598   Accepted: 5330 Description In order to get from one of the F (1 <= F <= 5,000) grazing fields (which are numbered 1..F) to another field, Bessie and the re…

题目链接 :http://poj.org/problem?id=3177 Description In order to <= F <= ,) grazing fields (which are numbered ..F) to another field, Bessie and the rest of the herd are forced to cross near the Tree of Rotten Apples. The cows are now tired of often bei…

POJ 3177 Redundant Paths POJ 3352 Road Construction 题目链接 题意:两题一样的.一份代码能交.给定一个连通无向图,问加几条边能使得图变成一个双连通图 思路:先求双连通.缩点后.计算入度为1的个数,然后(个数 + 1) / 2 就是答案(这题因为是仅仅有一个连通块所以能够这么搞,假设有多个,就不能这样搞了) 代码: #include <cstdio> #include <cstring> #include <algorithm…