hdoj1242(dfs 剪枝解法)

【hdoj1242(dfs 剪枝解法)】的更多相关文章

hdoj1242(dfs 剪枝解法)

题意:拯救行动,天使r有多个朋友a(friends,在这里被坑了几次,没看清题意),天使被关在牢房里,等着朋友来拯救,求拯救天使的最短距离. 解法:如果不剪枝,200*200的地图会超时,可以以天使为起点进行dfs,记录到达map[x][y]的最小值.到达每个a的最小值. #include <iostream> #include <cstring> #include <cstdio> using namespace std; ; char map[M][M]; int…

poj 1564 Sum It Up | zoj 1711 | hdu 1548 (dfs + 剪枝 or 判重)

Sum It Up Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other) Total Submission(s) : 4 Accepted Submission(s) : 1 Font: Times New Roman | Verdana | Georgia Font Size: ← → Problem Description Given a specified total t and…

PKU 2531 Network Saboteur(dfs+剪枝||随机化算法)

题目大意:原题链接给定n个节点,任意两个节点之间有权值,把这n个节点分成A,B两个集合,使得A集合中的每一节点与B集合中的每一节点两两结合(即有|A|*|B|种结合方式)权值之和最大. 标记:A集合:true B集合:false 解法一:dfs+剪枝 #include<iostream> #include<cstring> using namespace std; int n,ans; ]; ][]; void dfs(int i,int cursum) { in[i]=tru…

*HDU1455 DFS剪枝

Sticks Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 9779 Accepted Submission(s): 2907 Problem Description George took sticks of the same length and cut them randomly until all parts became…

POJ 3009 DFS+剪枝

POJ3009 DFS+剪枝原题: Curling 2.0 Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 16280 Accepted: 6725 Description On Planet MM-21, after their Olympic games this year, curling is getting popular. But the rules are somewhat different from our…

poj 1724:ROADS（DFS + 剪枝）

ROADS Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 10777 Accepted: 3961 Description N cities named with numbers 1 ... N are connected with one-way roads. Each road has two parameters associated with it : the road length and the toll…

DFS(剪枝) POJ 1011 Sticks

题目传送门 /* 题意:若干小木棍,是由多条相同长度的长木棍分割而成,问最小的原来长木棍的长度: DFS剪枝:剪枝搜索的好题!TLE好几次,终于剪枝完全! 剪枝主要在4和5:4 相同长度的木棍不再搜索:5 若新的搜索连第一条都没组合出来,直接break: 详细解释:http://blog.csdn.net/lyy289065406/article/details/6647960 http://www.cnblogs.com/devil-91/archive/2012/08/03/2621787.…