概况
今天下午组队赛，队伍一共A了5道题，我A了两道题，一道是水题，还有一道dfs的题目。
题目
L - Huatuo’s Medicine
Huatuo was a famous doctor. He use identical bottles to carry the medicine. There are different types of medicine. Huatuo put medicines into the bottles and chain these bottles together.
However, there was a critical problem. When Huatuo arrived the patient’s home, he took the chain out of his bag, and he could not recognize which bottle contains which type of medicine, but he remembers the order of the bottles on the chain.
Huatuo has his own solution to resolve this problem. When he need to bring 2 types of medicines, E.g. A and B, he will put A into one bottle and put B into two bottles. Then he will chain the bottles in the order of ′−B−A−B−′. In this way, when he arrived the patient’s home, he knew that the bottle in the middle is medicine A and the bottle on two sides are medicine B.
Now you need to help Huatuo to work out what’s the minimal number of bottles needed if he want to bring N
types of medicine.
Input
The first line of the input gives the number of test cases, T(1≤T≤100). T lines follow. Each line consist of one integer N(1≤N≤100)
, the number of types of the medicine.
Output
For each test case, output one line containing Case #x: y, where x is the test case number (starting from 1) and y
is the minimal number of bottles Huatuo needed.
Sample Input
1
2
Sample Output
Case #1: 3
水题一枚，意思是让你帮助华佗分配药物让他可以能快速找到药物。
题目已经给出思路-A-B-A-,接下来就是-A-B-C-B-A-故可知需要的瓶子数为药物种类数的二倍减一，属于一个简单的递推。
AC代码
#include <stdio.h>
#include<stdlib.h>
#include <string.h>
#include <math.h>
int main()
{
int t,n,i=0;
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
printf("Case #%d: %d\n",++i,2*n-1);
}
return 0;
}
H - Sudoku
Yi Sima was one of the best counselors of Cao Cao. He likes to play a funny game himself. It looks like the modern Sudoku, but smaller.
Actually, Yi Sima was playing it different. First of all, he tried to generate a 4×4 board with every row contains 1 to 4, every column contains 1 to 4. Also he made sure that if we cut the board into four 2×2
pieces, every piece contains 1 to 4.
Then, he removed several numbers from the board and gave it to another guy to recover it. As other counselors are not as smart as Yi Sima, Yi Sima always made sure that the board only has one way to recover.
Actually, you are seeing this because you’ve passed through to the Three-Kingdom Age. You can recover the board to make Yi Sima happy and be promoted. Go and do it!!!
Input
The first line of the input gives the number of test cases, T(1≤T≤100).
T test cases follow. Each test case starts with an empty line followed by 4 lines.
Each line consist of 4 characters. Each character represents the number in the corresponding cell (one of ‘1’, ‘2’, ‘3’, ‘4’). ‘*’ represents that number was removed by Yi Sima.
It’s guaranteed that there will be exactly one way to recover the board.
Output
For each test case, output one line containing Case #x:, where x
is the test case number (starting from 1). Then output 4 lines with 4 characters each. indicate the recovered board.
Sample Input
3
****
2341
4123
3214
*243
*312
*421
*134
*41*
**3*
2*41
4*2*
Sample Output
Case #1:
1432
2341
4123
3214
Case #2:
1243
4312
3421
2134
Case #3:
3412
1234
2341
4123
司马懿喜欢玩数独，最后还统一了天下，所以喜欢玩数独==能统一天下。（开玩笑【笑哭】）
这是一道dfs的题，就像经典问题马踏棋盘一样，需要把符合条件的一个一个的试，可以的话就继续，不行就回溯。
输出的条件为你从（1,1）遍历到了（4,4）。
AC代码
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
char s[10];
int f;
int hang[10][10], lie[10][10], m[10][10];
int g(int x, int y, int a)//g函数为判断一个点是否符合要求的函数
{
for (int i = 1; i <= 4; i++)//行不重
{
if (m[x][i] == a)
return 0;
}
for (int i = 1; i <= 4; i++)//列不重
{
if (m[i][y] == a)
return 0;
}
if(x>0&&x<3)//左上、左下、右上、右下四个小正方形不重
{
if(y>0&&y<3)
{
if(m[1][1]==a||m[1][2]==a||m[2][1]==a||m[2][2]==a)
return 0;
}
else if(m[1][3]==a||m[1][4]==a||m[2][3]==a||m[2][4]==a)
return 0;
}
else
{
if(y>0&&y<3)
{
if(m[3][1]==a||m[3][2]==a||m[4][1]==a||m[4][2]==a)
return 0;
}
else if(m[3][3]==a||m[3][4]==a||m[4][3]==a||m[4][4]==a)
return 0;
}
return 1;
}
void dfs(int x, int y)//深度优先搜索
{
if (f)
return;//f==1,逐层退出递归
else if (y > 4)
{
for (int i = 1; i <= 4; i++)
{
for (int j = 1; j <= 4; j++)
{
printf("%d", m[i][j]);
}
printf("\n");
}
f = 1;
return;
}
else if (x > 4)//x第一次大于4，即第一列遍历完成
dfs(1, y + 1);
else if (!m[x][y])
{
for (int i = 1; i <= 4; i++)
{
if (!hang[x][i] && !lie[y][i] && g(x, y, i))
{
hang[x][i] = lie[y][i] = 1;
m[x][y] = i;
dfs(x + 1, y);
m[x][y] = 0;
hang[x][i] = lie[y][i] = 0;
}
}
}
else dfs(x + 1, y);
}
int main()
{
int t, i, j, k = 1;
scanf("%d", &t);
while (t--)
{
memset(hang, 0, sizeof(hang));
memset(lie, 0, sizeof(lie));
for (i = 1; i <= 4; i++)
{
scanf("%s", s);
for (j = 1; j <= 4; j++)
{
if (s[j-1] == '*')
{
m[i][j] = 0;//标记该位置有没有数字
}
else
{
m[i][j] = s[j-1] - '0';
hang[i][m[i][j]] = 1;
lie[j][m[i][j]] = 1;
}
}
}
f = 0;
printf("Case #%d:\n", k++);
dfs(1, 1);
}
return 0;
}
总结
对于dfs的题目还是不够熟悉，导致该题目浪费了较长的时间，dfs和bfs还需要加强练习。
加油！！！！