题意:输入x,y,t.以及一个x行y列的地图,起点‘S’终点‘D’地板‘.’墙壁‘X’:判断能否从S正好走t步到D. 题解:dfs,奇偶性减枝,剩余步数剪枝. ps:帮室友Debug的题:打错了两个字母.题目看错.没有初始化map.orz ...不过我dfs也不熟,更别说剪枝了. //#include <bits/stdc++.h> #define _CRT_SECURE_NO_WARNINGS #include <iostream> #include <cstdio>…

Tempter of the Bone Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 88774    Accepted Submission(s): 24159 Problem Description The doggie found a bone in an ancient maze, which fascinated him a…

Tempter of the Bone Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other) Total Submission(s) : 15   Accepted Submission(s) : 9 Font: Times New Roman | Verdana | Georgia Font Size: ← → Problem Description The doggie found a…

Tempter of the Bone Time Limit: / MS (Java/Others) Memory Limit: / K (Java/Others) Total Submission(s): Accepted Submission(s): Problem Description The doggie found a bone in an ancient maze, which fascinated him a lot. However, when he picked it up,…

Sea and Sky are the most favorite things of iSea, even when he was a small child.  Suzi once wrote: white dew fly over the river, water and light draw near to the sky. What a wonderful scene it would be, connecting the two charming scenery. But iSea…

题目链接:poj1190 生日蛋糕 解题思路: 深搜,枚举:每一层可能的高度和半径 确定搜索范围:底层蛋糕的最大可能半径和最大可能高度 搜索顺序:从底层往上搭蛋糕,在同一层尝试时,半径和高度都是从大到小试 剪枝: ①已建好的面积已经超过目前求得的最优表面积,或者预见到搭完后面积一定会超过目前最优表面积,则停止搭建(最优性剪枝) ②预见到再往上搭,高度已经无法安排,或者半径无法安排,则停止搭建(可行性剪枝) ③还没搭的那些层的体积,一定会超过还缺的体积,则停止搭建(可行性剪枝) ④还没搭的那些层的…

Problem D: Servicing stations A company offers personal computers for sale in N towns (3 <= N <= 35). The towns are denoted by 1, 2, ..., N. There are direct routes connecting M pairs from among these towns. The company decides to build servicing st…

"我是要成为海贼王的男人!" 路飞他们伟大航路行程的起点是罗格镇,终点是拉夫德鲁(那里藏匿着"唯一的大秘宝"--ONE PIECE).而航程中间,则是各式各样的岛屿. 因为伟大航路上的气候十分异常,所以来往任意两个岛屿之间的时间差别很大,从A岛到B岛可能需要1天,而从B岛到A岛则可能需要1年.当然,任意两个岛之间的航行时间虽然差别很大,但都是已知的. 现在假设路飞一行从罗格镇(起点)出发,遍历伟大航路中间所有的岛屿(但是已经经过的岛屿不能再次经过),最后到达拉夫德鲁…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1518 题目大意:根据题目所给的几条边,来判断是否能构成正方形,一个很好的深搜应用,注意剪枝,以防超时! #include <iostream> #include <cstdio> #include<algorithm> #include <cstring> using namespace std; ],visit[]; int l,n; int dfs(int…

这道题目如果数据很小的话.我们通过这个dfs就可以完成深搜: void dfs(int s) { if (s==N) { minLen=min(minLen,totalLen); return ; } for (int i=0;i<G[s].size();i++) { Road r=G[s][i]; if (r.t+totalCost>K) continue; if (!visited[r.d]) { visited[r.d]=1; totalLen+=r.L; totalCost+=r.t;…