UVa - 11283 - PLAYING BOGGLE
先上题目
Problem F
PLAYING BOGGLE
Boggle® is a classic word game played on a 4 by 4 grid of letters. The letter grid is randomly generated by shaking 16 cubes labeled with a distribution of letters similar to that found in English words. Players try to find words hidden within the grid.
Words are formed from letters that adjoin horizontally, vertically, or diagonally. However, no letter may be used more than once within a single word.

An example Boggle® letter grid, showing the formation of the words "taxes" and "rise".
The score awarded for a word depends on its length, with longer words being worth more points. Exact point values are shown in the table below. A word is only ever scored once, even if it appears multiple times in the grid.
| No. of letters: | 3 | 4 | 5 | 6 | 7 | 8 or more |
| Points: | 1 | 1 | 2 | 3 | 5 | 11 |
In this problem, your task is to write a program that plays Boggle®.
Given a letter grid and a dictionary of words, you are to calculate the
total score of all the words in the dictionary that can be found in the
grid.
Input
The first line of the input file contains a number N, the number of Boggle® games that follow.
Each Boggle® game begins with 16 capital letters arranged in a 4 by 4 grid,
representing the board configuration for that game.
A blank line always precedes the letter grid.
Following the letter grid is a single number M (1 ≤ M ≤ 100), the number of words in your dictionary
for that game. The next M lines contain the dictionary words, one per line, in no particular order.
Each word consists of between 3 and 16 capital letters.
No single word will appear in the dictionary more than once for a given Boggle® game.
Output
For each Boggle® game in the input, your program should output the total score for that game.
Follow the format given in the sample output.
Sample Input
2 TNXO
AAEI
IOSR
BFRH
8
TAXES
RISE
ANNEX
BOAT
OATS
FROSH
HAT
TRASH FNEI
OBCN
EERI
VSIR
1
BEER
Output for the Sample Input
Score for Boggle game #1: 6
Score for Boggle game #2: 1 排位赛时这一题没有做出来,因为题意理解错了= =,以为是找到一个单词以后,这个单词用过的格子全部都不可以再用了,但其实不是这样。这一题的题意是不同长度的单词有不同的得分,给出字符矩阵让你在其中找一系列单词,找到一个单词就得到特定的分数,同一个单词得分只计算
一次,当然,也有可能里面找不到给出的单词,如果是这样就加0分,最后问你可以得到多少得分。由于这一题给的矩阵是4*4,完全就是一个裸的dfs,只需跑一边dfs就出结果。 上代码:
#include <stdio.h>
#include <string.h>
#include <map>
#include <iostream>
#include <algorithm>
using namespace std; map<string,int> M;
char s[][],c[];
bool mark[][]; bool dfs(int i,int j,int n)
{
if(i< || j<) return ;
if(mark[i][j]) return ;
if(s[i][j]==c[n])
{
if(c[n+]=='\0') return ;
mark[i][j]=;
if(dfs(i-,j-,n+)) return ; if(dfs(i-,j,n+)) return ; if(dfs(i-,j+,n+)) return ;
if(dfs(i,j-,n+)) return ; if(dfs(i,j+,n+)) return ;
if(dfs(i+,j-,n+)) return ; if(dfs(i+,j,n+)) return ; if(dfs(i+,j+,n+)) return ;
mark[i][j]=;
}
return ; } bool check()
{
int i,j;
for(i=;i<;i++)
{
for(j=;j<;j++)
{
if(s[i][j]==c[])
{
memset(mark,,sizeof(mark));
if(dfs(i,j,)) return ;
}
}
}
return ;
} int main()
{
int m,t,i,j,maxn,da,k;
//freopen("data.txt","r",stdin);
scanf("%d",&t);
for(k=;k<=t;k++)
{
M.clear();
memset(s,,sizeof(s));
for(i=;i<;i++)
{
scanf("%s",s[i]);
}
scanf("%d",&m);
maxn=;
for(j=;j<m;j++)
{
scanf("%s",c);
if(M.count(c)>) continue;
M[c]=;
if(!check()) continue;
da=strlen(c);
if(da<) continue;
switch(da)
{
case :
case : da=;break;
case : da=;break;
case : da=;break;
case : da=;break;
default :da=;
}
maxn=da+maxn;
}
printf("Score for Boggle game #%d: %d\n",k,maxn);
}
return ;
}
11283
UVa - 11283 - PLAYING BOGGLE的更多相关文章
- UVA 1482 - Playing With Stones(SG打表规律)
UVA 1482 - Playing With Stones 题目链接 题意:给定n堆石头,每次选一堆取至少一个.不超过一半的石子,最后不能取的输,问是否先手必胜 思路:数值非常大.无法直接递推sg函 ...
- [uva] 10067 - Playing with Wheels
10067 - Playing with Wheels 题目页:https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Ite ...
- uva 1482 - Playing With Stones
对于组合游戏的题: 首先把问题建模成NIM等经典的组合游戏模型: 然后打表找出,或者推出SG函数值: 最后再利用SG定理判断是否必胜必败状态: #include<cstdio> #defi ...
- uva 6757 Cup of Cowards(中途相遇法,貌似)
uva 6757 Cup of CowardsCup of Cowards (CoC) is a role playing game that has 5 different characters (M ...
- 09_Sum游戏(UVa 10891 Game of Sum)
问题来源:刘汝佳<算法竞赛入门经典--训练指南> P67 例题28: 问题描述:有一个长度为n的整数序列,两个游戏者A和B轮流取数,A先取,每次可以从左端或者右端取一个或多个数,但不能两端 ...
- UVA 1292 十二 Strategic game
Strategic game Time Limit:3000MS Memory Limit:0KB 64bit IO Format:%lld & %llu Submit Sta ...
- 【UVa 10881】Piotr's Ants
Piotr's Ants Porsition:Uva 10881 白书P9 中文改编题:[T^T][FJUT]第二届新生赛真S题地震了 "One thing is for certain: ...
- 容斥原理--计算错排的方案数 UVA 10497
错排问题是一种特殊的排列问题. 模型:把n个元素依次标上1,2,3.......n,求每一个元素都不在自己位置的排列数. 运用容斥原理,我们有两种解决方法: 1. 总的排列方法有A(n,n),即n!, ...
- uva 1354 Mobile Computing ——yhx
aaarticlea/png;base64,iVBORw0KGgoAAAANSUhEUgAABGcAAANuCAYAAAC7f2QuAAAgAElEQVR4nOy9XUhjWbo3vu72RRgkF5
随机推荐
- oc46--nonatomic, retain
// // Person.h #import <Foundation/Foundation.h> #import "Room.h" #import "Car. ...
- Linux Framebuffer 驱动框架之一概念介绍及LCD硬件原理【转】
本文转载自:http://blog.csdn.net/liuxd3000/article/details/17464779 一.基本概念 帧缓冲(Framebuffer)是Linux系统为显示设备提供 ...
- matlab中s函数编写心得-转自水木
S函数是system Function的简称,用它来写自己的simulink模块.(够简单吧,^_^, 详细的概念介绍大伙看帮助吧)可以用matlab.C.C++.Fortran.Ada等语言来写, ...
- [python 基础]python装饰器(一)添加functools获取原函数信息以及functools.partial分析
python装饰器学习的时候有两点需要注意一下 1,被装饰器装饰的函数取其func.__name__和func.func_doc的时候得到的不是被修饰函数的相关信息而是装饰器wrapper函数的doc ...
- sublime -text 删除已安装插件
按ctr+shift +p然后输入remove 回车,再输入要删除的插件名
- 认识JS的基础对象,定义对象的方法
JS的基础对象: 1.window //窗口对象 2.document //文档对象 3.document.documentElement //html对象 4.docume ...
- A - HQ9+
Problem description HQ9+ is a joke programming language which has only four one-character instructio ...
- iframe弹出窗体丢失焦点的问题
好像在不同的浏览器都有这个现象,用javascript弹出一个iframe的窗口,第一次input的焦点是正常的, 然后弹出第二次的时候,选择,按钮都可以获取到,但是input无法获得焦点,而且页面不 ...
- buf.readInt16LE函数详解
offset {Number} 0 noAssert {Boolean} 默认:false 返回:{Number} 从该 Buffer 指定的带有特定尾数格式(readInt16BE() 返回一个较大 ...
- Unity引擎GUI之Image
UGUI的Image等价于NGUI的Sprite组件,用于显示图片. 一.Image组件: Source Image(图像源):纹理格式为Sprite(2D and UI)的图片资源(导入图片后选择T ...