这题给的一个教训:Codeforces没有超时这个概念.本来以为1000*(1000+1)/2*10*10要超时的.结果我想多了. 这题由于k层都可能有关系,所以建一个图,每两个点之间连边,边权为n*m和他们之间的差值*w的最小值,然后求一个最小生成树就可以得出结果.且可以证明不会存在环.由于边比较稠密,用Prim算法求最小生成树. 代码: #include <iostream> #include <cstdio> #include <cstring> #include…

Zepto Code Rush 2014:http://codeforces.com/problemset/problem/436/C 题意:k个点,每个点都是一个n * m的char型矩阵.对与每个点,权值为n * m或者找到一个之前的点,取两个矩阵对应位置不同的字符个数乘以w.找到一个序列,使得所有点的权值和最小. 题解:很明显的最小生成树.但是要加入一个0点,边权为n*m,其余k个点两两建立一条边,边权是diff[i][j]*w,最后这一题,我要死掉的地方就是输出,不仅要输出费用,还要输出…

题目链接 题意: k个点,每一个点都是一个n * m的char型矩阵.对与每一个点,权值为n * m或者找到一个之前的点,取两个矩阵相应位置不同的字符个数乘以w.找到一个序列,使得全部点的权值和最小 分析: 首先,这个图是一个无向图.求权值和最小,每一个权值相应的是一条边,且每一个点仅仅能有一个权值即一条边,一个k个边,和生成树非常像,可是须要证明不能有环形.最好还是如果如今有三个点,每一个点的最小边成环,这时候是不能找到一个序列使得每一个点都取到它的最小边值的,所以,k个点k个边不能有环且边值…

题目链接 C. Dungeons and Candies time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output During the loading of the game "Dungeons and Candies" you are required to get descriptions of k levels fro…

Time Limit: 5000MS Memory Limit: 131072K Case Time Limit: 2000MS Description N children are sitting in a circle to play a game. The children are numbered from 1 to N in clockwise order. Each of them has a card with a non-zero integer on it in his/her…

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…

Who Gets the Most Candies? Time Limit:5000MS     Memory Limit:131072KB     64bit IO Format:%I64d & %I64u Submit Status Description N children are sitting in a circle to play a game. The children are numbered from 1 to N in clockwise order. Each of th…

Who Gets the Most Candies? Time Limit: 5000MS   Memory Limit: 131072K Total Submissions: 10373   Accepted: 3224 Case Time Limit: 2000MS Description N children are sitting in a circle to play a game. The children are numbered from 1 to N in clockwise…

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&…

Dogs' Candies Time Limit: 30000/30000 MS (Java/Others) Memory Limit: 512000/512000 K (Java/Others) Total Submission(s): 1701 Accepted Submission(s): 404 Problem Description Far far away, there live a lot of dogs in the forest. Unlike other dogs, thos…