UVALive 6257 Chemist's vows --一道题的三种解法(模拟,DFS,DP)
题意:给一个元素周期表的元素符号(114种),再给一个串,问这个串能否有这些元素符号组成(全为小写)。
解法1:动态规划
定义:dp[i]表示到 i 这个字符为止,能否有元素周期表里的符号构成。
则有转移方程:dp[i] = (dp[i-1]&&f(i-1,1)) || (dp[i-2]&&f(i-2,2)) f(i,k):表示从i开始填入k个字符,这k个字符在不在元素周期表中。 dp[0] = 1
代码:
//109ms 0KB #include <iostream> #include <cstdio> #include <cstring> #include <cmath> #include <algorithm> #include <string> using namespace std; #define N 50007string single[25] = {"h","b","c","n","o","f","k","p","s","y","i","w","u","v"}; string ss[130] = {"he","li","be","ne","na","mg", "al","si","cl","ar","ca","sc","ti","cr","mn", "fe","co","ni","cu","zn","ga","ge","as","se", "br","kr","rb","sr","zr","nb","mo","tc","ru", "rh","pd","ag","cd","in","sn","sb","te","xe", "cs","ba","hf","ta","re","os","ir","pt","au", "hg","tl","pb","bi","po","at","rn","fr","ra", "rf","db","sg","bh","hs","mt","ds","rg","cn", "fl","lv","la","ce","pr","nd","pm","sm","eu", "gd","tb","dy","ho","er","tm","yb","lu","ac", "th","pa","np","pu","am","cm","bk","cf","es", "fm","md","no","lr"};int vis[30][30],tag[30]; int dp[N]; char st[N];void init() {memset(vis,0,sizeof(vis));memset(tag,0,sizeof(tag));for(int i=0;i<14;i++)tag[single[i][0]-'a'] = 1;for(int i=0;i<100;i++)vis[ss[i][0]-'a'][ss[i][1]-'a'] = 1; }int main() {int t,len,i;init();scanf("%d",&t);while(t--){scanf("%s",st+1);len = strlen(st+1);memset(dp,0,sizeof(dp));dp[0] = 1;for(i=0;i<len;i++){if(dp[i]){if(tag[st[i+1]-'a'])dp[i+1] = 1;dp[i+2] |= vis[st[i+1]-'a'][st[i+2]-'a'];}}if(dp[len])puts("YES");elseputs("NO");}return 0; }
View Code
解法2:DFS
搜索时循环的是元素周期表的符号个数。详见代码
代码: (306ms)
#include <iostream> #include <cstdio> #include <cstring> #include <cmath> #include <algorithm> #include <string> using namespace std; #define N 50007string ss[130] = {"h","b","c","n","o","f","k","p","s","y","i","w","u","v","he","li","be","ne","na","mg", "al","si","cl","ar","ca","sc","ti","cr","mn", "fe","co","ni","cu","zn","ga","ge","as","se", "br","kr","rb","sr","zr","nb","mo","tc","ru", "rh","pd","ag","cd","in","sn","sb","te","xe", "cs","ba","hf","ta","re","os","ir","pt","au", "hg","tl","pb","bi","po","at","rn","fr","ra", "rf","db","sg","bh","hs","mt","ds","rg","cn", "fl","lv","la","ce","pr","nd","pm","sm","eu", "gd","tb","dy","ho","er","tm","yb","lu","ac", "th","pa","np","pu","am","cm","bk","cf","es", "fm","md","no","lr"};int vis[N]; int len[140]; char st[N]; int Length; bool Tag;void init() {int i;for(i=0;i<14;i++)len[i] = 1;for(i=14;i<114;i++)len[i] = 2; }void dfs(int u) {if(u == Length)Tag = 1;if(Tag)return;for(int i=0;i<114;i++){int flag = 1;if(u+len[i] <= Length && !vis[u+len[i]]){for(int j=0;j<len[i];j++){if(ss[i][j] != st[u+j]){flag = 0;break;}}if(flag){vis[u+len[i]] = 1;dfs(u+len[i]);}}} }int main() {init();int t,i;scanf("%d",&t);while(t--){scanf("%s",st);Length = strlen(st);memset(vis,0,sizeof(vis));Tag = 0;dfs(0);if(Tag)puts("YES");elseputs("NO");}return 0; }
View Code
解法3:乱搞,模拟。
分成: 单个元素存在与否,与前面匹不匹配,与后面匹不匹配,总共2^3 = 8种情况,然后O(n)扫过去,代码很长。。。
代码:(586ms)
#include <iostream> #include <cstdio> #include <cstring> #include <cmath> #include <algorithm> #include <string> using namespace std; #define N 50007string single[25] = {"h","b","c","n","o","f","k","p","s","y","i","w","u","v"}; string ss[130] = {"he","li","be","ne","na","mg", "al","si","cl","ar","ca","sc","ti","cr","mn", "fe","co","ni","cu","zn","ga","ge","as","se", "br","kr","rb","sr","zr","nb","mo","tc","ru", "rh","pd","ag","cd","in","sn","sb","te","xe", "cs","ba","hf","ta","re","os","ir","pt","au", "hg","tl","pb","bi","po","at","rn","fr","ra", "rf","db","sg","bh","hs","mt","ds","rg","cn", "fl","lv","la","ce","pr","nd","pm","sm","eu", "gd","tb","dy","ho","er","tm","yb","lu","ac", "th","pa","np","pu","am","cm","bk","cf","es", "fm","md","no","lr"};char st[N]; int vis[N];int main() {int t,len,i,j,k;scanf("%d",&t);while(t--){scanf("%s",st);len = strlen(st);int flag = 1;memset(vis,0,sizeof(vis));for(i=0;i<len;i++){if(vis[i])continue;string S = "";S += st[i];for(j=0;j<14;j++){if(single[j] == S)break;}if(j == 14) //not single {if(i > 0 && !vis[i-1]){S = st[i-1]+S;for(j=0;j<100;j++){if(ss[j] == S)break;}if(j != 100) //pre match {if(i < len-1){string ks = "";ks += st[i];ks += st[i+1];for(k=0;k<100;k++){if(ss[k] == ks)break;}if(k != 100) //back matchvis[i] = 0;else //back not matchvis[i] = 1;}}else //pre not match {if(i < len-1){string ks = "";ks += st[i];ks += st[i+1];for(k=0;k<100;k++){if(ss[k] == ks)break;}if(k != 100) //back matchvis[i+1] = 1;else //back not match {flag = 0;break;}}else{flag = 0;break;}}}else{if(i < len-1){string ks = "";ks += st[i];ks += st[i+1];for(k=0;k<100;k++){if(ss[k] == ks)break;}if(k != 100) //back matchvis[i+1] = 1;else //back not match {flag = 0;break;}}else{flag = 0;break;}}}else //single {if(i > 0 && !vis[i-1]){S = st[i-1]+S;for(j=0;j<100;j++){if(ss[j] == S)break;}if(j != 100) //pre match {if(i < len-1){string ks = "";ks += st[i];ks += st[i+1];for(k=0;k<100;k++){if(ss[k] == ks)break;}if(k != 100) //back matchvis[i] = 0;else //back not matchvis[i] = 1;}}else //pre not match {if(i < len-1){string ks = "";ks += st[i];ks += st[i+1];for(k=0;k<100;k++){if(ss[k] == ks)break;}if(k != 100) //back matchvis[i] = 0;else //back not matchvis[i] = 1;}elsevis[i] = 1;}}else{if(i < len-1){string ks = "";ks += st[i];ks += st[i+1];for(k=0;k<100;k++){if(ss[k] == ks)break;}if(k != 100) //back matchvis[i] = 0;else //back not matchvis[i] = 1;}elsevis[i] = 1;}}}if(flag)puts("YES");elseputs("NO");}return 0; }
View Code
转载于:https://www.cnblogs.com/whatbeg/p/3876636.html
UVALive 6257 Chemist's vows --一道题的三种解法(模拟,DFS,DP)相关推荐
- Central Europe Regional Contest 2012 Problem c: Chemist’s vows
字符串处理的题目: 学习了一下string类的一些用法: 这个代码花的时间很长,其实可以更加优化: 代码: 1 #include<iostream> 2 #include<strin ...
- 测试抑郁症的20道题(三)
问题一:什么是"软件测试"? 1.出自(IEEE, 1986; IEEE, 1990). Software testing is the process of analyzing ...
- 【蓝桥杯】搜索专题总结——真题讲解
写在前面 小伙伴们我们又见面啦~这篇文章拖了又拖,终于写完啦.这篇讲了几道蓝桥杯中考察DFS和BFS的真题,大家可以去看看前面两篇文章,对搜索讲的很详细.[一万字]蓝桥杯算法竞赛备考(一)--搜索专题 ...
- 青蛙跳台阶c语言递归函数,青蛙跳台阶问题的四种解法
http://raychase.iteye.com/blog/1337359 题目:一只青蛙一次可以跳1级台阶,也可以跳2级.求该青蛙跳上一个n级的台阶总共有多少种跳法. 这道题还被ITEye放在了博 ...
- cvsdfgdfdf
数组-136. 只出现一次的数字 题目描述 题目样例 Java方法:位运算 算法思路 代码 复杂度 题目描述 给定一个非空整数数组,除了某个元素只出现一次以外,其余每个元素均出现两次.找出那个只出现了 ...
- B.Sport Mafia
题意是有两种操作,一种是放蛋糕,一种是吃蛋糕,放蛋糕放的数量是递增的,而吃蛋糕则是固定的每次吃一个 然后题目输入操作步数和最后剩余蛋糕的数量,要求输出吃了多少蛋糕 这道题有两种解法,首先第一种是最容易 ...
- 删除链表的倒数第n个节点 python_19.leetcode题目讲解(Python):删除链表的倒数第N个节点...
题目如下: 题目 这道题说明了给定的删除位置肯定是可行的,所以就不用对n进行判断.这道题提供两种解法,第一种,利用python的 list 来把删除节点,参考答案如下: class Solution: ...
- js 数字递增递减_数字推理满分技巧 !不是干货,请批我!
公众号:兼得公考 数字推理不是广东的特色考点,但广东的数字推理是比较有"特色"的,他的特色就是简单,特别是2019年的数字推理.但我们不能因为某一两年考得简单而简单地学,为了让大家 ...
- LeetCode算法题-Minimum Depth of Binary Tree(Java实现)
这是悦乐书的第168次更新,第170篇原创 01 看题和准备 今天介绍的是LeetCode算法题中Easy级别的第27题(顺位题号是111).给定二叉树,找到它的最小深度.最小深度是沿从根节点到最近的 ...
最新文章
- NLP顶级专家Dan Roth :自然语言处理领域近期的任务和主要应用
- 如何防范SpringBoot 相关漏洞?
- Linux上操作Docker守护态运行
- 1. Linux内核的配置与裁减:
- php实现二叉搜索树,二叉搜索树有几种实现方式
- php注释验证,注解验证 · ThinkPHP6.0完全开发手册 · 看云
- 一文讲透 Serverless Kubernetes 容器服务
- 用户奖励体系有哪些反作弊的机制?
- Windows Server2012 64位安装OpenSSH服务
- 【优化模型】逐步回归算法
- 嵩天老师python123测验_嵩天老师python123测验4: 程序的控制结构 (第4周)
- Python模块查询
- 《你心柔软,却有力量》-林清玄--读书笔记
- 学堂云 减脂与运动塑形
- wps和office有什么区别?
- markdown插入本地图片小技巧
- 什么是高防服务器,高防服务器的原理
- 黎曼积分并非战无不胜
- CSS 导入方式 选择器
- 中机云告诉你,云计算有这10大好处|中机智库