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An abandoned country has n(n≤100000)n(n≤100000) villages which are numbered from 1 to nn. Since abandoned for a long time, the roads need to be re-built. There are m(m≤1000000)m(m≤1000000) roads to be

一个国家有N个公民，标记为0,1,2，…，N-1，每个公民有一个存款额。已知每个公民有一些朋友，同时国家有一条规定朋友间的存款额之差不能大于d。也就是说，a和b是朋友的话，a有x元的存款，b有y元，那么|x-y|<=d。给定d值与N个人的朋友关系，求这个国家最富有的人和最贫穷的人的存款相差最大的可能值是多少？即求贫富差距的最大值的下界。若这个值为无穷大，输出-1. Input 多组测试数据，第

Anya loves to fold and stick. Today she decided to do just that.Anya has n cubes lying in a line and numbered from 1 to n from left to right, with natural numbers written on them. She also has k sticke

Today on a lecture about strings Gerald learned a new definition of string equivalency. Two strings a and b of equal length are called equivalent in one of the two cases:They are equal. If we split st

再用邻接矩阵存图直播吃屎的#include<iostream>#include<cmath>#include<string>#include<queue>#include<stack>#include<cstring>#include<vector>using namespace std;int gaga=0x3f3f3f3f;int n,m,zouguo[11];vector<

The Vampires and Lykans are fighting each other to death. The war has become so fierce that, none knows who will win. The humans want to know who will survive finally. But humans are afraid of going to

#include<iostream>#include<string>#include<memory.h>#include<algorithm>using namespace std;int n;string tut[5];int biaoji[5][5];int inf=0x3f3f3f3f;int dfs(int x,int y){ for(int s=x+1;s<n;

#include<iostream>#include<cstdio>#include<string>#include<algorithm>#include<queue>#include<string>using namespace std;string w[30];int xingzou[4][2]={{0,1},{1,0},{-1,0},{0,-1}};struct p{

Description 《Journey to the West》(also 《Monkey》) is one of the Four Great Classical Novels of Chinese literature. It was written by Wu Cheng’en during the Ming Dynasty. In this novel, Monkey King Sun

There are 2 special dices on the table. On each face of the dice, a distinct number was written. Consider a1.a2,a3,a4,a5,a6 to be numbers written on top face, bottom face, left face, right face, front

在m<0的时候就不要了… 多复制粘贴几次 总有一波好使的#include<iostream>#include<algorithm>#include<cstdio>#include<memory.h>using namespace std;int n, m;int biaoji[270];int jilu[300];int qw[27];void dfs(int zuobiao,

别人都用的dfs，并查集 我用的bfs也过了#include<iostream>#include<algorithm>#include<queue>#include<functional>#include<vector>#include<string>#include<cstdlib>#include<stack>using namespace std;int zujian[11

广搜

其实这里边就一个奇偶剪枝值得看 再就是注意全局变量的多组数据的初始化 没啥好说的了… 不过需要注意的是这个题不能用bfs bfs只能求出最短的来 求出最短的就停了，问题是这个东西最短的不一定就是答案 所以用dfs bfs我跪了无数次 Problem DescriptionThe doggie found a bone in an ancient maze, which fascina

都很水…#include<iostream>#include<string>#include<cmath>#include<algorithm>#include<queue>#include<stack>#include<cstring>using namespace std;int prime[38] = { 0,0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,

一个有N个点M条边的图，球其中由S个点构成的团的个数。一个团是一个完全子图。 Input 第一行是测试点数。对于每个测试点，第一行包含3个整数N,M和S (N ≤ 100,M ≤ 1000,2 ≤ S ≤ 10),接下来的M行的每一行包含2整数u和v(1 ≤ u < v ≤ N)，那就是说那里有一条边在u和v之间。保证顶点最大的度数不超过20。 Output 对于每个测试点，输出一个数表示