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…

FZU 2150 Fire Game Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u   Description Fat brother and Maze are playing a kind of special (hentai) game on an N*M board (N rows, M columns). At the beginning, each grid of this bo…

对于一颗树,dfs遍历为每个节点标号,在进入一个树是标号和在遍历完这个树的子树后标号,那么子树所有的标号都在这两个数之间,是一个连续的区间.(好神奇~~~) 这样每次操作一个结点的子树时,在每个点的开始结束两个点标记一下就可以,用树状数组求前缀和就可知道每个点的值. 这道题虽然很麻烦(dep[y]-dep[x])%k .但是注意到K很小(1<=k<=5),可以维护k个树状数组. 提交时编译器选GUN C++迷之RE...换Visual C++ #include <cstdio> #…

Problem Description Fat brother and Maze are playing a kind of special (hentai) game on an N*M board (N rows, M columns). At the beginning, each grid of this board is consisting of grass or just empty and then they start to fire all the grass. Firstl…

题解:图的深度遍历就是顺着一个最初的结点开始,把与它相邻的结点都找到,也就是一直往下搜索直到尽头,然后在顺次找其他的结点. #include <bits/stdc++.h> using namespace std; int gra[200][200]; //储存图的大小 int vis[200]; // 标记数组 void dfs(int x,int k) // 有k个点,x代表目前搜到了那个点. { int i; for(i = 0; i < k; i ++) { if(!vis[i]…