题目链接 最短路构造题三连:这道题,HDU4903,SRM590 Fox And City. \(Description\) 给定一张\(n\)个点\(m\)条边的有向图,每条边的边权在\([1,n]\)之间.记\(d[i]\)为\(1\)到\(i\)的最短路.你需要对每条边确定一个边权,使得存在一个\(i\in[2,n]\),满足\(d[1]\lt d[2]\lt...d[i]\gt d[i+1]\gt...d[n]\). 输出方案(每条边的边权).输入保证有解. \(n,m\leq10^5\…

http://acm.hdu.edu.cn/showproblem.php?pid=5385 题意: 给定一张n个点m条有向边的图,构造每条边的边权(边权为正整数),令d(x)表示1到x的最短路,使得存在点i(1<=i<=n)满足d(1)<d(2)<…<d(i)>d(i+1)>…>d(n). 从两边向中间构造. 开始L=1,R=n 从L开始bfs,顺次构造L,L+1,L+2…… 构造不动了再从R开始bfs,顺次构造R,R-1,R-2…… 然后在从L开始………

The path Time Limit: 2000ms Memory Limit: 65536KB This problem will be judged on HDU. Original ID: 538564-bit integer IO format: %I64d      Java class name: Main Special Judge   You have a connected directed graph.Let d(x) be the length of the shorte…

题目传送门 /* 题意:给一张图和一些有向边,问如何给边赋值使得d1 < d2 < .. < dx > ,,,< ddn 贪心:左边从2开始,右边从n开始,每次选择未标记且相连的点加入树,加入时间就是d[i],若不在树上的输出n,注意重边和重点 */ /************************************************ * Author :Running_Time * Created Time :2015-8-13 12:09:17 * File…

题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=1973 Prime Path Description The ministers of the cabinet were quite upset by the message from the Chief of Security stating that they would all have to change the four-digit room numbers on their offices…

Binary Tree 题目连接: http://acm.hdu.edu.cn/showproblem.php?pid=5573 Description The Old Frog King lives on the root of an infinite tree. According to the law, each node should connect to exactly two nodes on the next level, forming a full binary tree. S…

题目链接:pid=3631" style="font-size:18px">http://acm.hdu.edu.cn/showproblem.php?pid=3631 Shortest Path Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 3962    Accepted Submission(s): 9…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1973 Prime Path Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Problem Description The ministers of the cabinet were quite upset by the message from the Chief of Secu…

http://acm.hdu.edu.cn/showproblem.php?pid=5015 由于是个二维的递推式,当时没有想到能够这样构造矩阵.从列上看,当前这一列都是由前一列递推得到.依据这一点来构造矩阵.令b[i]代表第i列,是一个(n+2)*1的矩阵,即b[1] = [1,233......],之所以在加了两行,是要从前一个矩阵b[i-1]得到b[i]中的第二个数2333...,再构造一个转换矩阵a,它是一个(n+2)*(n+2)的矩阵,那么a^(m-1) * b就是第m列. /* a矩…

Elegant Construction 题目连接: http://acm.hdu.edu.cn/showproblem.php?pid=5813 Description Being an ACMer requires knowledge in many fields, because problems in this contest may use physics, biology, and even musicology as background. And now in this prob…

B. Fox and Minimal path 题目连接: http://codeforces.com/contest/388/problem/B Description Fox Ciel wants to write a task for a programming contest. The task is: "You are given a simple undirected graph with n vertexes. Each its edge has unit length. You…

Shortest Path 题目连接: http://acm.hdu.edu.cn/showproblem.php?pid=5636 Description There is a path graph G=(V,E) with n vertices. Vertices are numbered from 1 to n and there is an edge with unit length between i and i+1 (1≤i<n). To make the graph more in…

题意: 定义S(N) 为数字N每个位上数字的和.在给两个数a,b,求最小的正整数n,使得 a×S(n)=b×S(2n). 官方题解: 这道题目的结果可能非常大,所以我们直接枚举n是要GG的. 首先可以有这样的基础性结论: 设gcd(a,b)=g, 我们可以先使得b=b/g, a=a/g S(n):S(2n)==b:a,那么我们有S(n):S(2n)=b:a. 然后,一个好的做法是,我们研究本质问题. 我们发现,如果一个digit是0~4,那么*2的效益是完全获得的. 如果一个digit的是5~9…

Triangle Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 0    Accepted Submission(s): 0 Problem Description Mr. Frog has n sticks, whose lengths are 1,2, 3⋯n respectively. Wallice is a bad man,…

floyd算法好像很奇妙的样子.可以做到每次加入一个点再以这个点为中间点去更新最短路,效率是n*n. #include<cstdio> #include<cstring> #include<cmath> #include<vector> #include<algorithm> using namespace std; ; const int INF = 0x7FFFFFFF; int A[maxn][maxn], flag[maxn]; int…

Shortest Path Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others) Total Submission(s): 1146 Accepted Submission(s): 358 Problem Description There is a path graph G=(V,E) with n vertices. Vertices are numbered from 1 to…

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=5636 题解: 1.暴力枚举: #include<cmath> #include<cstdio> #include<iostream> #include<algorithm> using namespace std; typedef long long LL; ; ; int n, m; ], b[]; int main() { int T; scanf(&qu…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5379 题意:一颗n个节点n-1条边的树,现在要给每个节点标号(1~n),要求:(1)每一层的兄弟节点的标号要是连续的(2)每一颗子树的所有节点标号是连续的.问有多少种标号方案. 解法:对于每一层顶多只能存在2个非叶子节点,否则无解:对于每一层有x个叶子节点,y个非叶子节点,那么ans=(ans * x!)%mod,另外如果y!=0,还得ans=2*ans%mod. #include <bits/st…

Dicription You have a connected directed graph.Let $d(x)$ be the length of the shortest path from $1$ to $x$.Specially $d(1)=0$.A graph is good if there exist $x$ satisfy $d(1)<d(2)<....d(x)>d(x+1)>...d(n)$.Now you need to set the length of ev…

Prime Path Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 987    Accepted Submission(s): 635 Problem Description The ministers of the cabinet were quite upset by the message from the Chief of S…

题目链接:HDU - 4823 魔法师百小度也有遇到难题的时候——现在,百小度正在一个古老的石门面前,石门上有一段古老的魔法文字,读懂这种魔法文字需要耗费大量的能量和大量的脑力.过了许久,百小度终于读懂魔法文字的含义:石门里面有一个石盘,魔法师需要通过魔法将这个石盘旋转X度,以使上面的刻纹与天相对应,才能打开石门.但是,旋转石盘需要N点能量值,而为了解读密文,百小度的能量值只剩M点了!破坏石门是不可能的,因为那将需要更多的能量.不过,幸运的是,作为魔法师 的百小度可以耗费V点能量,使得自己的能量…

http://acm.hdu.edu.cn/showproblem.php?pid=1195 这道题虽然只是从四个数到四个数,但是状态很多,开始一直不知道怎么下手,关键就是如何划分这些状态,确保每一个状态都能遍历到. 得到四个数之后,分三种情况处理,每次改变一个数之后都要加入队列,最先输出的就是步数最少. #include <cstdio> #include <cstring> #include <queue> using namespace std; struct p…

Shortest Path Time Limit: 1000MS Memory Limit: 32768KB 64bit IO Format: %I64d & %I64u SubmitStatus Description When YY was a boy and LMY was a girl, they trained for NOI (National Olympiad in Informatics) in GD team. One day, GD team's coach, Prof. G…

其实这个题是抄的题解啦…… 题解给了一个图,按照那个图模拟一遍大概就能理解了. 题意: 有一段程序,给你一个C值(程序中某常量),让你构造一组数据,使程序输出"doge" 那段代码大概是 SPFA的SLF优化.其实题目的意思是让我们构造一组数据,使得总的出队次数大于C.        数据范围 C<=23,333,333.输出的图中最多有100个点,没有重边.自环.负环. 思路: SLF: 设队首元素为 i, 队列中要加入节点 j, 在        时加到队首而不是队尾, 否则…

The Shortest Path in Nya Graph Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Problem Description This is a very easy problem, your task is just calculate el camino mas corto en un grafico, and just solo hay que camb…

Expressions Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 253    Accepted Submission(s): 121 Problem Description Arithmetic expressions are usually written with the operators in between the tw…

题意:bc round 74 分析(官方题解): 你可以选择分类讨论, 但是估计可能会写漏一些地方. 只要抽出新增边的端点作为关键点, 建立一个新图, 然后跑一遍floyd就好了. 复杂度大概O(6^2m) 注:然后我不会这种,这种floyd我觉得复杂度应该是复杂度应该是O(8^3m) 大概在千万级别,其实应该可以过,然后,其实只需要求单元最短路就行,然后是不是可以dij,然后就快一点 反正我也没写 我在比赛的时候写的是分治,考虑走不走新加的边每次走几条,以及走的顺序就好 然后全排列,时间复杂度…

题意:给你2个数n m.从n变成m最少须要改变多少次. 当中: 1.n  m  都是4位数 2.每次仅仅能改变n的一个位数(个位.十位.百位.千位),且每次改变后后的新数为素数 思路:搜索的变形题,这次我们要搜得方向是改变位数中的一位,然后往下搜.直到求出我们须要的那个解 #include<cstdio> #include<cstring> #include<algorithm> #include<queue> #include<cmath> u…

DZY Loves Partition  Accepts: 154  Submissions: 843  Time Limit: 4000/2000 MS (Java/Others)  Memory Limit: 262144/262144 K (Java/Others) 问题描述 DZY喜欢拆分数字.他想知道能否把nn拆成恰好kk个不重复的正整数之和. 思考了一会儿之后他发现这个题太简单,于是他想要最大化这kk个正整数的乘积.你能帮帮他吗? 由于答案可能很大,请模10^9+710​9​​+7输…

f(1) = 1, f(2) = 1, f(n) = (A * f(n - 1) + B * f(n - 2)) mod 7. Sample Input1 1 3 //a b n1 2 100 0 0 Sample Output25 矩阵A  * 矩阵B  = 矩阵C a b        f(n-1)     f(n) 1 0        f(n-2)    f(n-1) # include <iostream> # include <cstdio> # include <…