题目链接:https://icpcarchive.ecs.baylor.edu/external/66/6665.pdf 题目大意: 有一个3 * 3 的格子: 每个格子上面的数字能够朝上下左右四个方向移动.假设移出范围.则到与其边界上字母相应的还有一边. 例如以下图所看到的: 空白部分分别向上下左右移动之后的情况. 如今.给你左右移动的费用ch,上下移动cv.给你一个初始状态(9个数字,当中0代表该空格为空),一个结束状态,问从初始状态移动到结束状态的最小花费. 解题思路: 事实上这道题想法非…

题意大概就是八数码问题,只不过把空格的移动方式改变了:空格能够向前或向后移动一格或三格(循环的). 分析:其实跟八数码问题差不多,用康托展开记录状态,bfs即可. 代码: #include <iostream> #include <cstdio> #include <cstring> #include <cmath> #include <algorithm> #include <vector> #include <queue&g…

UVALive - 4108 SKYLINE Time Limit: 3000MS     64bit IO Format: %lld & %llu Submit Status uDebug Description   The skyline of Singapore as viewed from the Marina Promenade (shown on the left) is one of the iconic scenes of Singapore. Country X would a…

UVALive - 3942 Remember the Word A potentiometer, or potmeter for short, is an electronic device with a variable electric resistance. It has two terminals and some kind of control mechanism (often a dial, a wheel or a slide) with which the resistance…

UVALive - 3942 Remember the Word Neal is very curious about combinatorial problems, and now here comes a problem about words. Know- ing that Ray has a photographic memory and this may not trouble him, Neal gives it to Jiejie. Since Jiejie can’t remem…

题目传送门 /* 题意:本来有n个雕塑,等间距的分布在圆周上,现在多了m个雕塑,问一共要移动多少距离: 思维题:认为一个雕塑不动,视为坐标0,其他点向最近的点移动,四舍五入判断,比例最后乘会10000即为距离: 详细解释:http://www.cnblogs.com/zywscq/p/4268556.html */ #include <cstdio> #include <iostream> #include <algorithm> #include <cmath&…

题目链接:https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=4156 题目拷贝难度大我就不复制了. 题目大意:维护一个字符串,要求支持插入.删除操作,还有输出第 i 次操作后的某个子串.强制在线. 思路1:使用可持久化treap可破,详细可见CLJ的<可持久化数据结构的研究>. 思路2:rope大法好,详见:http…

Permutation Graphs Time Limit:3000MS     Memory Limit:0KB     64bit IO Format:%lld & %llu Submit Status Practice UVALive 6508 #include<stdio.h> #include<string.h> ],a[],b[],c[],b1[]; long long num; void merg_sort(int a[],int l,int r) { int…

Boxes Time Limit:3000MS     Memory Limit:0KB     64bit IO Format:%lld & %llu Submit Status Practice UVALive 6500 #include<stdio.h> #include<string.h> int main() { int T,m,n; int i,j,s; ][]; scanf("%d",&T); while(T--) { s=; sc…

题目链接:UVALive 6948  Jokewithpermutation 题意:给一串数字序列,没有空格,拆成从1到N的连续数列. dfs. 可以计算出N的值,也可以直接检验当前数组是否合法. #include <stdio.h> #include <iostream> #include <string.h> #define maxn 100 using namespace std; char str[maxn]; int num[maxn]; bool vis[m…