http://acm.hdu.edu.cn/showproblem.php?pid=3639 题意: 有一群孩子正在玩老鹰抓小鸡,由于想当老鹰的人不少,孩子们通过投票的方式产生,但是投票有这么一条规则:投票具有传递性,A支持B,B支持C,那么C获得2票(A.B共两票),输出最多能获得的票数是多少张和获得最多票数的人是谁? 思路: 先强连通缩点反向建图,在计算强连通的时候,需要保存每个连通分支的结点个数. 为什么要反向建图呢?因为要寻找票数最多的,那么肯定是入度为0的点,然后dfs计算它的子节点的…

Hawk-and-Chicken Time Limit: 6000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 2409    Accepted Submission(s): 712 Problem Description Kids in kindergarten enjoy playing a game called Hawk-and-Chicken. But t…

Hawk-and-Chicken Time Limit: 6000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3321    Accepted Submission(s): 1041 Problem Description Kids in kindergarten enjoy playing a game called Hawk-and-Chicken. But t…

Problem Description There are N bombs needing exploding. Each bomb has three attributes: exploding radius ri, position (xi,yi) and lighting-cost ci which means you need to pay ci cost making it explode. If a un-lighting bomb is in or on the border th…

http://acm.hdu.edu.cn/showproblem.php?pid=5934 题意:有N个炸弹,每个炸弹有一个坐标,一个爆炸范围和一个爆炸花费,如果一个炸弹的爆炸范围内有另外的炸弹,那么如果该炸弹爆炸,就会引爆所有爆炸范围内的炸弹,求让所有炸弹爆炸的最小花费. 思路:重现的时候来不及做.先n^2的把每个炸弹爆炸范围内的炸弹都连一条有向边,然后再找强连通分量缩点,这样会形成多个DAG,然后对于每个DAG找一个入度为0的点,找这个入度为0的点里面耗费最小的去引爆,就可以了. #inc…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4725 题意:有n个点和n层,m条边,每一层的任意一个点都可以花费固定的值到下一层或者上一层的任意点 然后m条边链接的点可以花费给出的值进行转移,最后问从i点到n点最少要花费多少. 这题点的个数有100000,层数也是100000,不算额外边暴力建边肯定要爆. 所以可以把层数也当成一个点比如说i点在j层于是n+j就与i链接花费0然后i点可以和上下层任意一个点链接 及i与n+j+1链接i与n+j-1链接…

题目 Reward Dandelion's uncle is a boss of a factory. As the spring festival is coming , he wants to distribute rewards to his workers. Now he has a trouble about how to distribute the rewards.  The workers will compare their rewards ,and some one may…

Task Schedule Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 5550    Accepted Submission(s): 1786 Problem Description Our geometry princess XMM has stoped her study in computational geometry t…

逃生 Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 6686 Accepted Submission(s): 1958 Problem Description 糟糕的事情发生啦,现在大家都忙着逃命.但是逃命的通道很窄,大家只能排成一行. 现在有n个人,从1标号到n.同时有一些奇怪的约束条件,每个都形如:a必须在b之前. 同时,社会是不平等的…

主要是建图,建好图之后跑一边dijkstra即可. 一共3N个点,1~N是原图中的点1~N,然后把每层x拆成两个点(N+x)[用于连指向x层的边]和(N+N+x)[用于连从x层指出的边]. 相邻层节点互相可达:AddEdge( N+N+x+1, N+x, C), AddEdge( N+N+x, N+x+1, C); 对于位于x层的节点i,AddEdge(N+x, i, 0), AddEdge(i, N+N+x, 0); #include <cstdio> #include <cstrin…