POJ 3159 Candies、

【POJ 3159 Candies、】的更多相关文章

题意:n个小孩,m个比较(给你两个孩子代号a,b.然后c意味着a比b最多只能少c个糖果),问1和n之间差距最大的糖果数量. 思路:这是一个差分约束思路不懂得:传送门, 转化一下就是一个SPFA求最短路的问题了. 参照了kuangbin大神的模板. #include<cstring> #include<cmath> #include<cstdio> ; const int MAX=0x3f3f3f3f; int head[qq]; //每个结点的头指针. bool vi…

POJ 3159 Candies （图论，差分约束系统，最短路）

POJ 3159 Candies (图论,差分约束系统,最短路) 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…

poj 3159 Candies (dij + heap)

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[…

POJ 3159 Candies（spfa、差分约束）

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…

POJ 3159 Candies（SPFA+栈）差分约束

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>…

POJ 3159 Candies（差分约束，最短路）

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…

POJ 3159 Candies（差分约束）

http://poj.org/problem?id=3159 题意:有向图,第一行n是点数,m是边数,每一行有三个数,前两个是有向边的起点与终点,最后一个是权值,求从1到n的最短路径. 思路:这个题让会神给讲的,用的dijkstra,看的网上很多用SPFA的. 关于SPFA:http://www.cnblogs.com/devtang/archive/2011/08/25/spfa.html http://blog.csdn.net/chenjiang492943457/article/deta…