Terrorist’s destroy Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)Total Submission(s): 24    Accepted Submission(s): 6 Problem Description There is a city which is built like a tree.A terrorist wants to destroy th…

Integer Partition Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 22    Accepted Submission(s): 15 Problem Description Given n, k, calculate the number of different (unordered) partitions of n s…

Editor Time Limit: 3000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 118    Accepted Submission(s): 38 Problem Description   Sample Input 8 I 2 I -1 I 1 Q 3 L D R Q 2   Sample Output 2 3 Hint The following…

Mutiples on a circle Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)Total Submission(s): 171    Accepted Submission(s): 28 Problem Description Tom has a necklace with n jewels. There is a number on each jewel. Now…

Front compression Time Limit: 5000/5000 MS (Java/Others)    Memory Limit: 102400/102400 K (Java/Others)Total Submission(s): 158    Accepted Submission(s): 63 Problem Description Front compression is a type of delta encoding compression algorithm wher…

Backup Plan Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)Total Submission(s): 93    Accepted Submission(s): 36Special Judge Problem Description Makomuno has N servers and M databases. All databases are synchroniz…

Building Fence Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)Total Submission(s): 171    Accepted Submission(s): 25Special Judge Problem Description Long long ago, there is a famous farmer named John. He owns a bi…

Divide Groups Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 423    Accepted Submission(s): 161 Problem Description   This year is the 60th anniversary of NJUST, and to make the celebration mor…

链接:http://acm.hdu.edu.cn/showproblem.php?pid=4679 题意:给定一颗树,每条边有一个权值w,问切掉哪条边之后,分成的两颗树的较大的直径*切掉边的权值最小?如果存在多条边使得结果相同,输出边id最小的 思路: dept一次找出最深的节点,之后以最深的节点出发(rt1)dept找到树的直径(即找到rt2):将路径保存在f[]中: 之后分别从rt1/rt2进行深搜,找到以一个节点为根的树的直径,这样在每次dfs之后,可以求出每条边的一边的最值,这样两次取m…

如果不在最长路的边,那么肯定是w*最长路. 如果在最长路,那么把最长路分成两段,左边树的最长路就是左段+左边点的次短路(不包含最长路上的点的最长路) ,右边同理. 还有就是更新,经过左端点的最长路,不一定是整颗左边树的最长路,乱搞一下就可以了.我是搞成一条链,写的很麻烦..从一点搞到了快四点.. #include <iostream> #include <cstdio> #include <cstring> #include <cmath> #include…