FZU 2107 Hua Rong Dao Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u  Practice  Description Cao Cao was hunted down by thousands of enemy soldiers when he escaped from Hua Rong Dao. Assuming Hua Rong Dao is a narrow aisl…

Problem 2107 Hua Rong Dao Accept: 318 Submit: 703 Time Limit: 1000 mSec Memory Limit : 32768 KB Problem Description Cao Cao was hunted down by thousands of enemy soldiers when he escaped from Hua Rong Dao. Assuming Hua Rong Dao is a narrow aisle (one…

Problem 2107 Hua Rong Dao Accept: 106    Submit: 197 Time Limit: 1000 mSec    Memory Limit : 32768 KB  Problem Description Cao Cao was hunted down by thousands of enemy soldiers when he escaped from Hua Rong Dao. Assuming Hua Rong Dao is a narrow ais…

dfs暴力回溯,这个代码是我修改以后的,里面的go相当简洁,以前的暴力手打太麻烦,我也来点技术含量.. #include<iostream> #include<cstring> #include<cstdio> using namespace std; #define m 4 ][] = {,,,,,,,}; ][][] = {{{,},{,}},{{,},{,}},{{,}}}; ] = {,,}; ][]; int n,tmp; void work_caocao(i…

Problem 2107 Hua Rong Dao Accept: 503    Submit: 1054Time Limit: 1000 mSec    Memory Limit : 32768 KB Problem Description Cao Cao was hunted down by thousands of enemy soldiers when he escaped from Hua Rong Dao. Assuming Hua Rong Dao is a narrow aisl…

                                                                                                                                 Problem 2107 Hua Rong Dao Accept: 401    Submit: 853Time Limit: 1000 mSec    Memory Limit : 32768 KB Problem Description…

题目:跟N皇后问题一样,不考虑对角冲突,但考虑墙的存在,只要中间有墙就不会冲突. N皇后一行只能放一个,而这题不行,所以用全图暴力放棋,回溯dfs即可,题目最多就到4*4,范围很小. 刚开始考虑放一个棋子后就把其他不能放的地方标记下,然后再暴力,后来发现如果一个点重复标记在去标记时就会把点标成合法的,于是改用放棋子是进行检查,由于数据量小,也不会占用多少时间. 之后才想到,在标记时可以用累加的,去标记时再一个一个减下来即可... 代码: #include <cstdio> const int…

思路:得到输入得到mj[]的各个牌的数量,还差最后一张牌.直接暴力枚举34张牌就可以了. 当假设得到最后一张牌,则得到了的牌看看是不是可以胡,如果可以胡的话,就假设正确.否者假设下一张牌. 关键还是如何判断这组牌是不是可以胡.怎么判断呢?因为胡牌的条件是:nABC+mBBB+1DD, 而且n+m=4,其中n,m可以为0. 下面是一张:递归DD的牌的递归情况: 然后:依次是找顺子和刻子也是相同的算法 #include<iostream> #include<cstring> using…

Kim is a magician, he can use n kinds of magic, number from 1 to n. We use string Si to describe magic i. Magic Si will make Wi points of damage. Note that Wi may change over time. Kim obey the following rules to use magic: Each turn, he picks out on…

题意:给定一个 10*10的矩阵,每一个W可以跳过一个B向对角走到#并把B吃掉,并且可以一直跳直到不能动为止,现在是W走的时候,问你最多吃几个B. 析:直接暴力+回溯,深搜就好. 代码如下: #pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <cstdlib> #include <cmath&g…