题意: 给定n个点,m条有向边,邮箱容量. 起点在1,终点在n,開始邮箱满油. 以下m行表示起点终点和这条边的耗油量(就是长度) 再以下给出一个数字m表示有P个加油站,能够免费加满油. 以下一行P个数字表示加油站的点标. 再以下一个整数Q 以下Q行 u v 表示在u点有销售站,能够卖掉邮箱里的随意数量的油,每以单位v元. 问跑到终点能获得最多多少元. 先求个每一个点的最大剩余油量 f[i], 再把边反向,求每一个点距离终点的最短路 dis[i]. 然后枚举一下每一个销售点就可以,( f[i] -…

两次SPFA 第一关找:从1没有出发点到另一个点的多少是留给油箱 把边反过来再找一遍:重每一个点到终点最少须要多少油 Greedy Driver Time Limit: 2 Seconds      Memory Limit: 65536 KB Edward is a truck driver of a big company. His daily work is driving a truck from one city to another. Recently he got a long d…

Burn the Linked Camp Time Limit: 2 Seconds      Memory Limit: 65536 KB It is well known that, in the period of The Three Empires, Liu Bei, the emperor of the Shu Empire, was defeated by Lu Xun, a general of the Wu Empire. The defeat was due to Liu Be…

比赛链接:点击打开链接 上来先搞了f.c,,然后发现状态不正确,一下午都是脑洞大开,, 无脑wa,无脑ce...一样的错犯2次.. 硬着头皮搞了几发,最后20分钟码了一下G,不知道为什么把1直接当成不能加油的站就会wa..太弱.. 唔···太懒第二天才发题解.. B:Gears 并查集 题解:点击打开链接 C:Consecutive Blocks 离散化一下然后模拟 题解:点击打开链接 D:An Easy Game 设dp[i][j]为前i个位置已经匹配了j个位置的方法数. #include <…

A.Another Recurrence Sequence problemId=5287">B.Gears 题目大意:有n个齿轮,一開始各自为一组.之后进行m次操作,包含下面4种类型: 1.合并两组齿轮.合并的两个应该反向旋转 2.把某个齿轮从所在组删除,自为一组.但不影响同组其他齿轮的状态与关系 3.询问两个齿轮是同向.反向或无关系(即不在同一组) 4.询问某个齿轮所在组的齿轮总数 分析:典型的并查集操作,可是注意两点: 1.因为操作3要询问两个齿轮的相对状态.因此对并查集中每一个元素应…

============================================================================== 深入linux kernel内核配置选项 ============================================================================== 如果自己不亲自实践配置的话,你永远也体会不到内核的真实所在. 使用过linux的发行版之一gentoo的话,你应该知道这是一个彻头彻尾的自定义…

ZOJ Problem Set - 3080 ChiBi Time Limit: 5 Seconds      Memory Limit: 32768 KB watashi's mm is so pretty as well as smart. Recently, she has watched the movie Chibi. So she knows more about the War of ChiBi. In the war, Cao Cao had 800,000 soldiers,…

POJ 1860 Currency Exchange / ZOJ 1544 Currency Exchange (最短路径相关,spfa求环) Description Several currency exchange points are working in our city. Let us suppose that each point specializes in two particular currencies and performs exchange operations onl…

ZOJ Problem Set - 3946 Highway Project Time Limit: 2 Seconds      Memory Limit: 65536 KB Edward, the emperor of the Marjar Empire, wants to build some bidirectional highways so that he can reach other cities from the capital as fast as possible. Thus…

Beer Problem Time Limit: 2 Seconds      Memory Limit: 32768 KB Everyone knows that World Finals of ACM ICPC 2004 were held in Prague. Besides its greatest architecture and culture, Prague is world famous for its beer. Though drinking too much is prob…

题目链接: POJ:http://poj.org/problem?id=1201 HDU:http://acm.hdu.edu.cn/showproblem.php? pid=1384 ZOJ:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=508 Description You are given n closed, integer intervals [ai, bi] and n integers c1, ..., cn.…

//差分约束 >=求最长路径 <=求最短路径 结果都一样//spfa#include<stdio.h> #include<string.h> #include<limits.h> #include<queue> using namespace std; #define N 1010 #define M 1010*1010//注意边和点集的数组大小 struct edge { int to,value,next; }; struct edge ed…

从点(n,1)到点(1,m)的最短路径,可以转换地图成从(1,1)到(n,m)的最短路,因为有负权回路,所以要用spfa来判负环, 注意一下如果负环把终点包围在内的话, 如果用负环的话会输出无穷,但是有答案的,所以从终点出去的点不要,还有就是负环习惯用-1判断,但是可能会有负值,所以要用-inf #include<map> #include<set> #include<ctime> #include<cmath> #include<stack>…

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=5718   Highway Project Time Limit: 2 Seconds      Memory Limit: 65536 KB Edward, the emperor of the Marjar Empire, wants to build some bidirectional highways so that he can reach other cities…

人老了就比较懒,故意挑了到看起来很和蔼的题目做,然后套个spfa和dinic的模板WA了5发,人老了,可能不适合这种刺激的竞技运动了…… 题目链接:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=2760 Description Given a weighted directed graph, we define the shortest path as the path who has the smallest leng…

Idiomatic Phrases Game Tom is playing a game called Idiomatic Phrases Game. An idiom consists of several Chinese characters and has a certain meaning. This game will give Tom two idioms. He should build a list of idioms and the list starts and ends w…

http://poj.org/problem?id=1932 http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=1935 题目大意: 看到XYZZY可不要以为是在玩扫雷哦. 给你一张图,初始你在房间1,初始生命值为100,进入每个房间会加上那个房间的生命(可能为负),要你进入房间n,问是否可能.(要求进入每个房间后生命值都大于0) 思路: 1.SPFA求最长路径,如果路径存在(即无环),那么肯定可以. 2.存在负环,不管她,…

Tram Time Limit: 1000MS Memory Limit: 30000K Total Submissions: 14630 Accepted: 5397 Description Tram network in Zagreb consists of a number of intersections and rails connecting some of them. In every intersection there is a switch pointing to the o…

ZOJ 2734 Exchange Cards 题目大意: 给定一个值N,以及一堆卡片,每种卡片有一个值value和数量number.求使用任意张卡片组成N的方式. 例如N = 10 ,cards(10,2)(7,2)(5,3)(2,2)(1,5),则10 = 10,10 =7+3+1,10=5+5… 思路分析: 由于之前做过1204,知道这题就是赤裸裸的搜索,直接用dfs暴力即可求得. 可做的优化处理就是——这个过程是贪心Greedy的,先从大到小这样取过来,所以,可以做一步降序排列的预处理.…

转自——http://blog.csdn.net/qwe20060514/article/details/8112550 =============================以下是最小生成树+并查集======================================[HDU]1213   How Many Tables   基础并查集★1272   小希的迷宫   基础并查集★1325&&poj1308  Is It A Tree?   基础并查集★1856   More i…

Tram Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 12005   Accepted: 4365 Description Tram network in Zagreb consists of a number of intersections and rails connecting some of them. In every intersection there is a switch pointing to t…

[热烈庆祝ZOJ回归] P1002:简单的DFS #include <cstdio> #include <cstring> #include <algorithm> ][]; int N; int input() { scanf("%d",&N); ;i<N;i++) scanf("%s",map[i]); return N; } ][]; ,sizeof(disable)); } int dfs(int cur)…

我觉得他整理的有一些乱,我都改成插入代码了,看的顺眼一些 转载自http://blog.csdn.net/juststeps/article/details/8772755 下面的都是原文: 最短路径 之 SPFA算法 http://hi.baidu.com/southhill/item/ab26a342590a5aae60d7b967 求最短路径的算法有许多种,除了排序外,恐怕是OI界中解决同一类问题算法最多的了.最熟悉的无疑是Dijkstra,接着是Bellman-Ford,它们都可以求出由…

HDU 1069 Monkey and Banana / ZOJ 1093 Monkey and Banana (最长路径) Description A group of researchers are designing an experiment to test the IQ of a monkey. They will hang a banana at the roof of a building, and at the mean time, provide the monkey with…

POJ 1511 Invitation Cards / UVA 721 Invitation Cards / SPOJ Invitation / UVAlive Invitation Cards / SCU 1132 Invitation Cards / ZOJ 2008 Invitation Cards / HDU 1535 (图论,最短路径) Description In the age of television, not many people attend theater perfor…

POJ 2240 Arbitrage / ZOJ 1092 Arbitrage / HDU 1217 Arbitrage / SPOJ Arbitrage(图论,环) Description Arbitrage is the use of discrepancies in currency exchange rates to transform one unit of a currency into more than one unit of the same currency. For exa…

POJ 2235 Frogger / UVA 534 Frogger /ZOJ 1942 Frogger(图论,最短路径) Description Freddy Frog is sitting on a stone in the middle of a lake. Suddenly he notices Fiona Frog who is sitting on another stone. He plans to visit her, but since the water is dirty a…

B - Benny's Compiler Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%lld & %llu Submit Status Practice ZOJ 2475 Description These days Benny has designed a new compiler for C programming language. His compilation system provides a com…

Treasure Hunt Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 20000/10000K (Java/Other) Total Submission(s) : 3   Accepted Submission(s) : 2 Problem Description Archeologists from the Antiquities and Curios Museum (ACM) have flown to Egypt to…

A.ZOJ 3666 Alice and Bob 组合博弈,SG函数应用 #include<vector> #include<cstdio> #include<cstring> #include<algorithm> using namespace std; const int maxn = 10000 + 100; int SG[maxn]; vector<int> g[maxn]; int mex(int u) { //minimal exc…