2013
12-21

# The Sultan’s Successors

The Sultan of Nubia has no children, so she has decided that the country will be split into up to k separate parts on her death and each part will be inherited by whoever performs best at some test. It is possible for any individual to inherit more than one or indeed all of the portions. To ensure that only highly intelligent people eventually become her successors, the Sultan has devised an ingenious test. In a large hall filled with the splash of fountains and the delicate scent of incense have been placed k chessboards. Each chessboard has numbers in the range 1 to 99 written on each square and is supplied with 8 jewelled chess queens. The task facing each potential successor is to place the 8 queens on the chess board in such a way that no queen threatens another one, and so that the numbers on the squares thus selected sum to a number at least as high as one already chosen by the Sultan. (For those unfamiliar with the rules of chess, this implies that each row and column of the board contains exactly one queen, and each diagonal contains no more than one.)

Write a program that will read in the number and details of the chessboards and determine the highest scores possible for each board under these conditions. (You know that the Sultan is both a good chess player and a good mathematician and you suspect that her score is the best attainable.)

Input will consist of k (the number of boards), on a line by itself, followed by k sets of 64 numbers, each set consisting of eight lines of eight numbers. Each number will be a positive integer less than 100.

Output will consist of k numbers consisting of your k scores, each score on a line by itself and right justified in a field 5 characters wide.

1
1  2  3  4  5  6  7  8
9 10 11 12 13 14 15 16
17 18 19 20 21 22 23 24
25 26 27 28 29 30 31 32
33 34 35 36 37 38 39 40
41 42 43 44 45 46 47 48
48 50 51 52 53 54 55 56
57 58 59 60 61 62 63 64

  260

# The Sultan’s Successors

The Sultan of Nubia has no children, so she has decided that the country will be split into up to k separate parts on her death and each part will be inherited by whoever performs best at some test. It is possible for any individual to inherit more
than one or indeed all of the portions. To ensure that only highly intelligent people eventually become her successors, the Sultan has devised an ingenious test. In a large hall filled with the splash of fountains and the delicate scent of incense have been
placed k chessboards. Each chessboard has numbers in the range 1 to 99 written on each square and is supplied with 8 jewelled chess queens. The task facing each potential successor is to place the 8 queens on the chess board in such a way that no
queen threatens another one, and so that the numbers on the squares thus selected sum to a number at least as high as one already chosen by the Sultan. (For those unfamiliar with the rules of chess, this implies that each row and column of the board contains
exactly one queen, and each diagonal contains no more than one.)

Write a program that will read in the number and details of the chessboards and determine the highest scores possible for each board under these conditions. (You know that the Sultan is both a good chess player and a good mathematician and you suspect that
her score is the best attainable.)

## Input

Input will consist of k (the number of boards), on a line by itself, followed by k sets of 64 numbers, each set consisting of eight lines of eight numbers. Each number will be a positive integer less than 100. There will never be more than
20 boards.

## Output

Output will consist of k numbers consisting of your k scores, each score on a line by itself and right justified in a field 5 characters wide.

## Sample input

1
1  2  3  4  5  6  7  8
9 10 11 12 13 14 15 16
17 18 19 20 21 22 23 24
25 26 27 28 29 30 31 32
33 34 35 36 37 38 39 40
41 42 43 44 45 46 47 48
48 50 51 52 53 54 55 56
57 58 59 60 61 62 63 64

## Sample output

  260

vis[0]是列。vis[1]是副斜线。vis[2]是主斜线。

#include <stdio.h>
#include <string.h>
#define N 8
int t;
int map[10][10];
int vis[3][20];
int max;

void dfs(int cur, int sum)
{
if (cur == N + 1)
{
if (max < sum)
max = sum;
return;
}
for (int i = 1; i <= N; i ++)
{
if (!vis[0][i] && !vis[1][cur + i] && !vis[2][cur - i + N])
{
vis[0][i] = vis[1][cur + i] = vis[2][cur - i + N] = 1;
dfs(cur + 1, sum + map[cur][i]);
vis[0][i] = vis[1][cur + i] = vis[2][cur - i + N] = 0;
}
}
}
int main()
{
int t;
scanf("%d", &t);
while (t --)
{
max = 0;
memset(vis, 0, sizeof(vis));
memset(map, -1, sizeof(map));
for (int i = 1; i <= N; i ++)
for (int j = 1; j <= N; j ++)
{
scanf("%d", &map[i][j]);
}
dfs(1, 0);
printf("%5d\n", max);

}
return 0;
}

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2. 可以参考算法导论中的时间戳。就是结束访问时间，最后结束的顶点肯定是入度为0的顶点，因为DFS要回溯

3. for(int i=1; i<=m; i++){
for(int j=1; j<=n; j++){
dp = dp [j-1] + 1;
if(s1.charAt(i-1) == s3.charAt(i+j-1))
dp = dp[i-1] + 1;
if(s2.charAt(j-1) == s3.charAt(i+j-1))
dp = Math.max(dp [j - 1] + 1, dp );
}
}
这里的代码似乎有点问题？ dp(i)(j) = dp(i)(j-1) + 1;这个例子System.out.println(ils.isInterleave("aa","dbbca", "aadbbcb"));返回的应该是false

4. 嗯 分析得很到位，确实用模板编程能让面试官对你的印象更好。在设置辅助栈的时候可以这样：push时，比较要push的elem和辅助栈的栈顶，elem<=min.top()，则min.push(elem).否则只要push（elem）就好。在pop的时候，比较stack.top()与min.top(),if(stack.top()<=min.top()),则{stack.pop();min.pop();}，否则{stack.pop();}.