题目链接:http://poj.org/problem?id=3159 题目很容易理解 就是简单的SPFA算法应用 刚开始用STL里的队列超时了,自己写了个栈,果断过,看来有时候栈还是快啊.... 代码: #include<iostream> #include<cstdlib> #include<cstdio> #include<cstring> #define INF 0x3f3f3f3f #define maxn 30010 #define max_e…

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&…

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…

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:https://cn.vjudge.net/problem/POJ-3159 题意 给出一组不等式 求第一个变量和最后一个变量可能的最大差值 数据保证有解 思路 一个不等式a-b<=c,通过移项,实际上就是满足了a<=b+c 发现在整个约束系统中,a在下满足不等式的情况下求最大值,就是在求最短路 然而如果直接用BellmanFord(spfa)的话,还是会超时 这时得对Bellman做第二次优化,用stack代替queue 但是对于更多的图中,Dijsktra依然更优,所以没有必要太…

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…

#include <iostream> #include <queue> #include <cstring> #define maxn 30005 #define scanf(x) scanf("%d", &x) using namespace std; struct CNode { int next; // index of the next vector from the same vertex int k; // the end ve…

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…

差分约束系统 2008-11-28 20:53:25|  分类: 算法与acm|举报|字号 订阅     出处:http://duanple.blog.163.com/blog/static/709717672008102885325526/ poj 1275 3159 1364 1716 1201 3169这类问题,就像网络流,图论,dp,关键在列出满足的表达式,建立好数学模型,剩下的过程就很简单了.所以主要难点在于构图,从实际的描述抽象成模型.准确找到约束条件. 关于基础知识可以查看clrs…