NYOJ 1249 物资调度(DFS+剪枝)】的更多相关文章

题目链接: http://acm.nyist.net/JudgeOnline/problem.php?pid=1249 描述 某地区发生了地震,灾区已经非常困难,灾民急需一些帐篷.衣物.食品和血浆等物资.可通往灾区的道路到处都是塌方,70%以上的路面损坏,桥梁全部被毁.国家立即启动应急预案,展开史上最大强度非作战空运行动,准备向灾区空投急需物资. 一方有难,八方支援.现在已知有N个地方分别有A1,A2,-.,An个物资可供调配.目前灾区需要物资数量为M. 现在,请你帮忙算一算,总共有多少种物质调…
10401: A.物资调度 Time Limit: 2 Sec  Memory Limit: 128 MB Submit: 95  Solved: 54 [Submit][Status][Web Board] Description 某地区发生了地震,灾区已经非常困难,灾民急需一些帐篷.衣物.食品和血浆等物资.可通往灾区的道路到处都是塌方,70%以上的路面损坏,桥梁全部被毁.国家立即启动应急预案,展开史上最大强度非作战空运行动,准备向灾区空投急需物资. 一方有难,八方支援.现在已知有N个地方分别…
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…
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…
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…
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.…
题目传送门 /* 题意:告诉一个区间[L,R],问根节点的n是多少 DFS+剪枝:父亲节点有四种情况:[l, r + len],[l, r + len - 1],[l - len, r],[l - len -1,r]; */ #include <cstdio> #include <algorithm> #include <cstring> #include <cmath> #include <queue> using namespace std;…
Counting Cliques Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 539    Accepted Submission(s): 204 Problem Description A clique is a complete graph, in which there is an edge between every pair…
Equation Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 92    Accepted Submission(s): 24 Problem Description Little Ruins is a studious boy, recently he learned addition operation! He was rewa…
题目链接 Solution DFS+剪枝 对于一个走过点k,如果有必要再走一次,那么一定是走过k后在k点的最大弹药数增加了.否则一定没有必要再走. 记录经过每个点的最大弹药数,对dfs进行剪枝. #include <iostream> #include <cstring> #include <algorithm> #include <cstdio> #include <map> using namespace std; map<string…