题意:一棵有n个结点的树,要取其中的一个结点,使得该结点到其他所有结点的距离和dis最小,即损耗I * I * R * dis最小,输出最小损耗和该结点(有多个的话按结点编号从小到大输出)(3 <= n <= 50000, 1 <= I <= 10, 1 <= R <= 50). 题目链接:http://poj.org/problem?id=4045 ——>>怒刷树状dp... 设cnt[i]为以i为根的子树的结点数,d[i]为以i为根的子树中所有结点到i的…

题目连接 http://poj.org/problem?id=1459 Power Network Description A power network consists of nodes (power stations, consumers and dispatchers) connected by power transport lines. A node u may be supplied with an amount s(u) >= 0 of power, may produce an…

POJ 1459 Power Network / HIT 1228 Power Network / UVAlive 2760 Power Network / ZOJ 1734 Power Network / FZU 1161 (网络流,最大流) Description A power network consists of nodes (power stations, consumers and dispatchers) connected by power transport lines. A…

题目地址:http://poj.org/problem?id=2109 /* 题意:k ^ n = p,求k 1. double + pow:因为double装得下p,k = pow (p, 1 / n); 基础知识: 类型 长度 (bit) 有效数字 绝对值范围 float 32 6~7 10^(-37) ~ 10^38 double 64 15~16 10^(-307) ~ 10^308 long double 128 18~19 10^(-4931) ~ 10 ^ 4932 2. 二分查找…

点击打开链接 Power Network Time Limit: 2000MS   Memory Limit: 32768K Total Submissions: 20903   Accepted: 10960 Description A power network consists of nodes (power stations, consumers and dispatchers) connected by power transport lines. A node u may be su…

Power Network Time Limit: 2000MS   Memory Limit: 32768K Total Submissions: 25514   Accepted: 13287 Description A power network consists of nodes (power stations, consumers and dispatchers) connected by power transport lines. A node u may be supplied…

Power Network Time Limit: 2000MS   Memory Limit: 32768K Total Submissions: 22987   Accepted: 12039 Description A power network consists of nodes (power stations, consumers and dispatchers) connected by power transport lines. A node u may be supplied…

Power Network Time Limit: 2000MS Memory Limit: 32768K Description A power network consists of nodes (power stations, consumers and dispatchers) connected by power transport lines. A node u may be supplied with an amount s(u) >= 0 of power, may produc…

连续重复子串问题 poj 2406 Power Strings http://poj.org/problem?id=2406 问一个串能否写成a^n次方这种形式. 虽然这题用kmp做比较合适,但是我们还是用后缀数组做一做,巩固后缀数组的能力. 对于一个串,如果能写出a^n这种形式,我们可以暴力枚举循环节长度L,那么后缀suffix(1)和suffix(1 + L)的LCP应该就是 lenstr - L.如果能满足,那就是,不能,就不是. 这题的话da算法还是超时,等我学了DC3再写上来. 其实这…

题目传送门 /* 题意:一个串有字串重复n次产生,求最大的n KMP:nex[]的性质应用,感觉对nex加深了理解 */ /************************************************ * Author :Running_Time * Created Time :2015-8-10 10:51:54 * File Name :POJ_2406.cpp ************************************************/ #incl…

#include<cstdio> #include<cstring> #include<algorithm> #include<queue> #include<vector> #define INF 1e9 using namespace std; const int maxn=100+5; struct Edge { int from,to,cap,flow; Edge(){} Edge(int f,int t,int c,int fl):fr…

题意:给你一棵树,让你求一点,使该点到其余各点的距离之和最小.如果这样的点有多个,则按升序依次输出. 树型dp #include <cstdio> #include <cstring> #include <vector> #include <set> using namespace std; const int maxn=50010; typedef __int64 LL; vector<int>tree[maxn];// to save the…

妈妈呀....这简直是目前死得最惨的一次. 贴题目: http://poj.org/problem?id=3233 Matrix Power Series Time Limit: 3000MS Memory Limit: 131072K Total Submissions: 19128 Accepted: 8068 Description Given a n × n matrix A and a positive integer k, find the sum S = A + A2 + A3 +…

F - Power Strings Time Limit:3000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u Submit Status Practice POJ 2406 Description Given two strings a and b we define a*b to be their concatenation. For example, if a = "abc" and b = &…

http://poj.org/problem?id=2406 Power Strings Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 27003   Accepted: 11311 Description Given two strings a and b we define a*b to be their concatenation. For example, if a = "abc" and b = &q…

题目链接: http://poj.org/problem?id=2607 Description A city is served by a number of fire stations. Some residents have complained that the distance from their houses to the nearest station is too far, so a new station is to be built. You are to choose t…

Power Strings Problem's Link: http://poj.org/problem?id=2406 Mean: 给你一个字符串,让你求这个字符串最多能够被表示成最小循环节重复多少次得到. analyse: KMP之next数组的运用.裸的求最小循环节. Time complexity: O(N) Source code:  ;;      ;);      ) ;}/* */…

Matrix Power Series Time Limit: 3000MS   Memory Limit: 131072K Total Submissions: 20309   Accepted: 8524 Description Given a n × n matrix A and a positive integer k, find the sum S = A + A2 + A3 + … + Ak. Input The input contains exactly one test cas…

Description Roger Wilco is in charge of the design of a low orbiting space station for the planet Mars. To simplify construction, the station is made up of a series of Airtight Cubical Modules (ACM's), which are connected together once in space. One…

Power Strings Time Limit: 3000MSMemory Limit: 65536K Total Submissions: 29663Accepted: 12387 Description Given two strings a and b we define a*b to be their concatenation. For example, if a = "abc" and b = "def" then a*b = "abcdef…

Power Strings Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 30069   Accepted: 12553 Description Given two strings a and b we define a*b to be their concatenation. For example, if a = "abc" and b = "def" then a*b = "…

点击打开链接 Power of Cryptography Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 16388   Accepted: 8285 Description Current work in cryptography involves (among other things) large prime numbers and computing powers of numbers among these pr…

点击打开链接 Power Strings Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 27368   Accepted: 11454 Description Given two strings a and b we define a*b to be their concatenation. For example, if a = "abc" and b = "def" then a*b =…

题目:http://poj.org/problem?id=1459 题意:有一些发电站,消耗用户和中间线路,求最大流.. 加一个源点,再加一个汇点.. 其实,过程还是不大理解.. #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <queue> using namespace std; <<; ][],flow[…

题目:http://poj.org/problem?id=2109 题意:求一个整数k,使得k满足kn=p. 思路:exp()用来计算以e为底的x次方值,即ex值,然后将结果返回.log是自然对数,就是e为底计算的.换底公式 log<a>(b) = log<c>(b) / log<c>(a). float 的范围为-2^128 ~ +2^127,也即-3.40E+38 ~ +3.40E+38: double 的范围为-2^1024 ~ +2^1023,也即-1.79E+…

题目: http://poj.org/problem?id=2406 跟1961差不多,题解就不写了,一开始理解错题了,导致WA一次. #include <stdio.h> #include <string.h> #include <algorithm> ]; ]; void kmp_init() { ; next[] = -; ; i < n; i++) { && s[j+] != s[i]) j = next[j]; ] == s[i]) j+…

题目链接: http://poj.org/problem?id=1459 因为发电站有多个,所以需要一个超级源点,消费者有多个,需要一个超级汇点,这样超级源点到发电站的权值就是发电站的容量,也就是题目中的pmax,消费者到超级汇点的权值就是消费者的容量,也就是题目中的cmax.初学网络流,第一眼看到这个题还以为应该先做一遍EK算法,然后减去max(p-pmax, c-cmax)呢..没想到这个题的难点就是建图而已.. #include <stdio.h> #include <string…

Power Strings Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 36926   Accepted: 15254 Description Given two strings a and b we define a*b to be their concatenation. For example, if a = "abc" and b = "def" then a*b = "…

Description Starting with x and repeatedly multiplying by x, we can compute x31 with thirty multiplications: x2 = x × x, x3 = x2 × x, x4 = x3 × x, …, x31 = x30 × x. The operation of squaring can be appreciably shorten the sequence of multiplications.…

Power Strings Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 28102   Accepted: 11755 Description Given two strings a and b we define a*b to be their concatenation. For example, if a = "abc" and b = "def" then a*b = "…