差分约数: 求满足不等式条件的尽量小的值---->求最长路---->a-b>=c----> b->a (c) Schedule Problem Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 1503    Accepted Submission(s): 647 Special Judge Problem Descr…

结题报告合集请戳:http://972169909-qq-com.iteye.com/blog/1185527 /*题意:求符合题意的最小集合的元素个数 题目要求的是求的最短路, 则对于 不等式 f(b)-f(a)>=c,建立 一条 b 到 a 的边 权值为 c,则求的最长路 即为 最小值(集合) 并且有隐含条件:0<=f(a)-f(a-1)<=1 则有边权关系(a,a-1,0)以及(a-1,a,-1); 将源点到各点的距离初始化为INF(无穷大),其中之1为0,最终求出的最短路满足 它…

Problem Description Ali has taken the Computer Organization and Architecture course this term. He learned that there may be dependence between instructions, like WAR (write after read), WAW, RAW. If the distance between two instructions is less than…

转自:http://blog.csdn.net/shahdza/article/details/7779273 最短路 [HDU] 1548 A strange lift基础最短路(或bfs)★2544 最短路 基础最短路★3790 最短路径问题基础最短路★2066 一个人的旅行基础最短路(多源多汇,可以建立超级源点和终点)★2112 HDU Today基础最短路★1874 畅通工程续基础最短路★1217 Arbitrage 货币交换 Floyd (或者 Bellman-Ford 判环)★124…

出处:http://blog.csdn.net/shahdza/article/details/7779273 最短路 [HDU] 1548    A strange lift基础最短路(或bfs)★ 2544    最短路  基础最短路★ 3790    最短路径问题基础最短路★ 2066    一个人的旅行基础最短路(多源多汇,可以建立超级源点和终点)★ 2112    HDU Today基础最短路★ 1874    畅通工程续基础最短路★ 1217    Arbitrage   货币交换…

Time Limit: 1500MS Memory Limit: 131072K Description During the kindergarten days, flymouse was the monitor of his class. Occasionally the head-teacher brought the kids of flymouse's class a large bag of candies and had flymouse distribute them. All…

//Accepted 2692 KB 1282 ms //差分约束 -->最短路 //TLE到死,加了输入挂,手写queue #include <cstdio> #include <cstring> #include <iostream> #include <queue> #include <cmath> #include <algorithm> using namespace std; /** * This is a docu…

偶尔做了一下差分约束. 题目大意:给出n个军营,每个军营最多有ci个士兵,且[ai,bi]之间至少有ki个士兵,问最少有多少士兵. --------------------------------------------------- 差分约束:就是利用多个不等式来推导另一个不等式. 由于不等式a-b<=c和求最短路径时的三角形不等式相同,就变成了求最短路. 所有不等式化为a-b<=c的形式,则建造b到a的边,权为c. 求a到b的最短距离,则转化为b-a<=c,距离的值为c. 该题中:…

裸差分约束. //#pragma comment(linker, "/STACK:1024000000,1024000000") #include<cstdio> #include<cstring> #include<cstdlib> #include<algorithm> #include<iostream> #include<sstream> #include<cmath> #include<…

原题:ZOJ 3668 http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3668 典型差分约束题. 将sum[0] ~ sum[n] 作为节点,A<=sum[R]-sum[L-1]<=B可以分别建边: x(L-1)-->xR  w = B xR --> x(L-1)  w = -A 注意还有 -10000<=sum[i]-sum[i-1]<=10000 (for i in (2,n)),所以相邻点…