Partial Tree Problem Description In mathematics, and more specifically in graph theory, a tree is an undirected graph in which any two nodes are connected by exactly one path. In other words, any connected graph without simple cycles is a tree. You f…

题意:求给出图的表面积,不包括底面 #include<bits/stdc++.h> using namespace std ; typedef long long ll; #define mem(a) memset(a,0,sizeof(a)) #define meminf(a) memset(a,127,sizeof(a)); #define TS printf("111111\n"); #define FOR(i,a,b) for( int i=a;i<=b;i+…

Dancing Stars on Me Problem Description The sky was brushed clean by the wind and the stars were cold in a black sky. What a wonderful night. You observed that, sometimes the stars can form a regular polygon in the sky if we connect them properly. Yo…

Travel Time Limit: 1500/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 2404    Accepted Submission(s): 842 Problem Description Jack likes to travel around the world, but he doesn’t like to wait. Now, he is tr…

Chip Factory Problem Description John is a manager of a CPU chip factory, the factory produces lots of chips everyday. To manage large amounts of products, every processor has a serial number. More specifically, the factory produces n chips today, th…

题目链接: H - Partial Tree  HDU - 5534 题目大意:首先是T组测试样例,然后n个点,然后给你度数分别为(1~n-1)对应的不同的权值,然后问你在这些点形成树的前提下的所能形成的最大权值. 具体思路: 这个题是学长做的,我记录一下思路. 有点背包的感觉,但是和之前我做过的有点不同,原来的背包是互相不会影响的.但是对于这个题,我们需要的前提是形成一棵树,所以为了解决这个问题,我们先给每个点分配一个度,然后再去把剩余的n-2的度再分配下去就可以了. AC代码: #inclu…

Partial Tree http://acm.hdu.edu.cn/showproblem.php?pid=5534 Time Limit: / MS (Java/Others) Memory Limit: / K (Java/Others) Total Submission(s): Accepted Submission(s): Problem Description In mathematics, and more specifically in graph theory, a tree…

求区间最值,数据范围也很小,因为只会线段树,所以套了线段树模板=.= Sample Input3110011 151 2 3 4 551 21 32 43 43 531 999999 141 11 22 33 3 Sample Output1002344519999999999991 # include <iostream> # include <cstdio> # include <cstring> # include <algorithm> # incl…

题意:M=p1*p2*...pk:求C(n,m)%M,pi小于10^5,n,m,M都是小于10^18. pi为质数 M不一定是质数 所以只能用Lucas定理求k次 C(n,m)%Pi最后会得到一个同余方程组x≡B[0](mod p[0])x≡B[1](mod p[1])x≡B[2](mod p[2])......解这个同余方程组 用中国剩余定理 Sample Input19 5 23 5 Sample Output6 # include <iostream> # include <cst…

设定每个节点的上限和下限,之后向上更新,判断是否出现矛盾 #include<cstdio> #include<iostream> #include<algorithm> #include<cstring> #include<cmath> #include<queue> #include<map> using namespace std; #define MOD 1000000007 ; ; typedef long lon…