Black Spot Time limit: 1.0 secondMemory limit: 64 MB Bootstrap: Jones's terrible leviathan will find you and drag the Pearl back to the depths and you along with it. Jack: Any idea when Jones might release said terrible beastie? Bootstrap: I already…

右侧是1.维护的同时保持最短路p值至少,我有直接存款(1-p).该概率不满足,为了使这个值极大. #include <iostream> #include <cstdlib> #include <cstring> #include <string> #include <cstdio> #include <cmath> #include <algorithm> #include <vector> #include…

D - Black Spot Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u Submit Status Practice URAL 1934 Description Bootstrap: Jones's terrible leviathan will find you and drag the Pearl back to the depths and you along with it.…

DES:给出起点和终点.给出所有小岛的编号.所有路径的起始点.和遇到怪物的概率.要求在最短路的条件下维护遇见怪物的概率最小的路径.就是用 SPFA算法.每条路的权值设为1.最短路即为途径的岛数最少.同时要用pre数组维护每个点的前驱.最后递归输出所走路径.把p变为不遇见怪物的概率, 即为维护p最大.就是把原来的SPFA里的判断多加一条,如果权值相等判断概率,选择概率大的一条. 注意.这是无向图.本来觉得没有影响.很幸福的WA了.确实是.给你a->b的,就代表可以从b->a.而不是简单的可以往返…

Meeting Time limit: 2.0 secondMemory limit: 64 MB K friends has decided to meet in order to celebrate their victory at the programming contest. Unfortunately, because of the tickets rise in price there is a problem: all of them live in different part…

1355. Bald Spot Revisited Time limit: 1.0 secondMemory limit: 64 MB A student dreamt that he walked along the town where there were lots of pubs. He drank a mug of ale in each pub. All the pubs were numbered with positive integers and one could pass…

题意:给定 n + m 个街道,问你从左上角走到右下角的所有路的权值最小的中的最大的. 析:我们只要考虑几种情况就好了,先走行再走列和先走列再走行差不多.要么是先横着,再竖着,要么是先横再竖再横,要么是先横再竖再横再竖,全考虑一下就好了. 代码如下: #pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <cs…

1934 水题 RE了N久 后来发现是无向图 #include <iostream> #include<cstring> #include<algorithm> #include<stdlib.h> #include<vector> #include<cstdio> #include<queue> using namespace std; #define N 2000010 #define INF 0xfffffff st…

题目链接 题意 : 一个学生梦到自己在一条有很多酒吧的街上散步.他可以在每个酒吧喝一杯酒.所有的酒吧有一个正整数编号,这个人可以从n号酒吧走到编号能整除n的酒吧.现在他要从a号酒吧走到b号,请问最多能喝到多少酒. 思路 :因为b肯定要是a的倍数,是a从头开始乘下去的,实际上就是找构成b/a的素数划分,有多少个素数划分就可以喝到多少的酒. #include <stdio.h> #include <string.h> #include <iostream> using na…

1741. Communication Fiend Time limit: 1.0 second Memory limit: 64 MB Kolya has returned from a summer camp and now he's a real communication fiend. He spends all his free time on the Web chatting with his friends via ICQ. However, lately the protocol…