poj3159 Candies 这题实质为裸的差分约束. 先看最短路模型:若d[v] >= d[u] + w, 则连边u->v,之后就变成了d[v] <= d[u] + w , 即d[v] – d[u] <= w. 再看题目给出的关系:b比a多的糖果数目不超过c个,即d[b] – d[a] <= c ,正好与上面模型一样, 所以连边a->b,最后用dij+heap求最短路就行啦. ps:我用vector一直TLE,后来改用链式前向星才过了orz... #include&…

分析:设每个人的糖果数量是a[i] 最终就是求a[n]-a[1]的最大值 然后给出m个关系 u,v,c 表示a[u]+c>=a[v] 就是a[v]-a[u]<=c 所以对于这种情况,按照u,v,c建单向边,一条从1到n的路径就是一个关于1和n的推广不等式a[n]-a[1]<=k(k为这条路的权) 所以找到所有不等式中最小k,就是求1到n的最短路,这就是差分约束 然后上代码: #include<cstdio> #include<cstring> #include&l…

Candies Time Limit: 1500MS   Memory Limit: 131072K Total Submissions: 40407   Accepted: 11367 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…

题意:n个人,m个信息,每行的信息是3个数字,A,B,C,表示B比A多出来的糖果不超过C个,问你,n号人最多比1号人多几个糖果 解题关键:差分约束系统转化为最短路,B-A>=C,建有向边即可,与dijkstra中的d[v]>=d[u]+C相同,即可求解. 最小值用最长路,最大值用最短路. 技巧:有一个超级源点向所有点连边,来固定每个点的约束. #include<cstdio> #include<cstring> #include<algorithm> #in…

转载请注明出处: http://www.cnblogs.com/fraud/          ——by fraud Candies 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’…

题目链接:http://poj.org/problem?id=3159 Candies Time Limit: 1500MS   Memory Limit: 131072K Total Submissions: 33576   Accepted: 9422 Description During the kindergarten days, flymouse was the monitor of his class. Occasionally the head-teacher brought th…

Candies POJ-3159 这里是图论的一个应用,也就是差分约束.通过差分约束变换出一个图,再使用Dijikstra算法的链表优化形式而不是vector形式(否则超时). #include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<string> #include<vector> #include<queue>…

题意 编号为 1..N 的人, 每人有一个数; 需要满足 dj - di <= c 求1号的数与N号的数的最大差值.(略坑: 1 一定要比 N 大的...difference...不是"差别", 而是"做差"....) 思路 差分约束 差分约束顾名思义就是以"差值"作为约束条件的规划问题. 这个"差值"的特点使得这个问题可以转化为最短路问题(或最长路?) 由于SFPA(或Dijkstra)中的松弛操作: d[v] <…

Candies Time Limit: 1500MS   Memory Limit: 131072K Total Submissions: 22177   Accepted: 5936 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 b…

http://poj.org/problem?id=3159 题意: flymouse是幼稚园班上的班长,一天老师给小朋友们买了一堆的糖果,由flymouse来分发,在班上,flymouse和snoopy是死对头,两人势如水火,不能相容,因此fly希望自己分得的糖果数尽量多于snoopy,而对于其他小朋友而言,则只希望自己得到的糖果不少于班上某某其他人就行了. 比如A小朋友强烈希望自己的糖果数不能少于B小朋友m个,即B- A<=m,A,B分别为A.B小朋友的分得的糖果数.这样给出若干组这样的条件…

Candies Time Limit: 1500MS   Memory Limit: 131072K Total Submissions: 33283   Accepted: 9334 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 b…

Candies Time Limit: 1500MS   Memory Limit: 131072K Total Submissions: 39666   Accepted: 11168 题目链接:http://poj.org/problem?id=3159 Description: During the kindergarten days, flymouse was the monitor of his class. Occasionally the head-teacher brought…

3159 -- Candies 明明找的是差分约束,然后就找到这题不知道为什么是求1~n的最短路的题了.然后自己无聊写了一个heap,518ms通过. 代码如下: #include <cstdio> #include <cstring> #include <iostream> #include <algorithm> using namespace std; ; ; struct Edge { int t, nx, c; } edge[M]; int eh[…

题意:现在需要分糖果,有n个人,现在有些人觉得某个人的糖果数不能比自己多多少个,然后问n最多能在让所有人都满意的情况下比1多多少个. 这道题其实就是差分约束题目,根据题中给出的 a 认为 b 不能比 a 多 c 个,也就是 d[b] - d[a] ≤ c,就可以建立 value 值为 c 的单向边 e(a,b) ,然后先定d[1] = 0 ,用最短路跑完得到的 d[n] 就是所求答案. #include<stdio.h> #include<string.h> #include<…

Candies Time Limit: 1500MS   Memory Limit: 131072K Total Submissions: 20067   Accepted: 5293 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 b…

题目大意:给N个小屁孩分糖果,每个小屁孩都有一个期望,比如A最多比B多C个,再多了就不行了,会打架的,求N最多比1多几块糖 分析:就是求一个极小极大值...试试看 这里需要用到一个查分约束的东西 下面是查分约束详解: 一直不知道差分约束是什么类型题目,最近在写最短路问题就顺带看了下,原来就是给出一些形如x-y<=b不等式的约束,问你是否满足有解的问题 好神奇的是这类问题竟然可以转换成图论里的最短路径问题,下面开始详细介绍下 比如给出三个不等式,b-a<=k1,c-b<=k2,c-a<…

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 the kids loved candies very much and ofte…

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 the kids loved candies very much and often compared t…

<题目链接> 题目大意: 给n个人派糖果,给出m组数据,每组数据包含A,B,c 三个数,意思是A的糖果数比B少的个数不多于c,即B的糖果数 - A的糖果数<= c .最后求n 比 1 最多多多少糖果. 解题分析: 这是一题典型的差分约束题.不妨将糖果数当作距离,把相差的最大糖果数看成有向边AB的权值,我们得到 dis[B]-dis[A]<=w(A,B).看到这里,我们联想到求最短路时的松弛技术,即if(dis[B]>dis[A]+w(A,B), dis[B]=dis[A]+w…

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 the kids loved candies very much and often compared t…

Language:Default Candies Time Limit: 1500MS   Memory Limit: 131072K Total Submissions: 43021   Accepted: 12075 Description During the kindergarten days, flymouse was the monitor of his class. Occasionally the head-teacher brought the kids of flymouse…

//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…

题目链接:http://poj.org/problem?id=3159 题意:给出m给 x 与y的关系.当中y的糖数不能比x的多c个.即y-x <= c  最后求fly[n]最多能比so[1] 多多少糖? 差分约束问题, 就是求1-n的最短路,  队列实现spfa 会超时了,改为栈实现,就可以 有负环时,用栈比队列快 数组开小了,不报RE,报超时 ,我晕 #include <iostream> #include <cstdlib> #include <cstdio>…

原题地址:http://poj.org/problem?id=3159 题意大概是班长发糖果,班里面有不良风气,A希望B的糖果不比自己多C个.班长要满足小朋友的需求,而且要让自己的糖果比snoopy的尽量多. 比如现在ABCD四个小朋友,B的糖果不能超过A的5个,如果A的史努比,D是班长,那么班长最多比史努比多7个糖果,而不是5+4+1=9个. 因为如果是9个,就不满足D-A<=(D-C)+(C-A)<=7的条件. 不懂的可以翻一下算法导论,上面有差分约束的定义和证明,总之这是一个求最短路的问…

题意: 就是分糖果 然后A觉得B比他优秀  所以分的糖果可以比他多 但最多不能超过c1个, B又觉得A比他优秀.... 符合差分约束的条件 设A分了x个  B分了y个  则x-y <= c1 , 根据其它的关系可以找出c2 c3 ···· 如果不懂差分约束的请  点击 所以构成不等式组:x-y <= c1   x-y <= c2    x-y <=c3 因为这些条件要都符合 所以取最小的c  即最短路 #include <iostream> #include <c…

题意:给a b c要求,b拿的比a拿的多但是不超过c,问你所有人最多差多少 思路:在最短路专题应该能看出来是差分约束,条件是b - a <= c,也就是满足b <= a + c,和spfa的松弛条件相对应,所以我们建一条a到b的边,权值c,然后跑最短路,求出所有差值最大的那个即为答案.应该算是基础的线性差分约束题. ps:队列超时,这里用栈. 关于差分约束可以点这里 #include<cstdio> #include<set> #include<map> #…

题目链接:http://poj.org/problem?id=3159 题目大意:给n个人派糖果,给出m组数据,每组数据包含A,B,C三个数,意思是A的糖果数比B少的个数不多于C,即B的糖果数 - A的糖果数<=C . 最后求n 比 1 最多多多少颗糖果. 解题思路:经典差分约束的题目,具体证明看这里<数与图的完美结合——浅析差分约束系统>. 不妨将糖果数当作距离,把相差的最大糖果数看成有向边AB的权值,我们得到 dis[B]-dis[A]<=w(A,B).看到这里,我们可以联想到…

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…

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…

http://poj.org/problem?id=1201 题意:给出N个整数区间[ai,bi],并且给出一个约束ci,( 1<= ci <= bi-ai+1),使得数组Z在区间[ai,bj]的个数>= ci个,求出数组Z的最小长度. 思路:建立差分约束系统.因为这里要求数组长度的最小值,要变为 x-y>=k的标准形式. 设数组 s[j] 表示数组 Z 区间[0,j]里包含的元素个数.所以 s[bi+1] - s[ai] >= ci,注意是 j+1, 隐含条件   0 &l…