UVA439 POJ2243 HDU1372 ZOJ1091 Knight Moves

问题链接：UVA439 POJ2243 HDU1372 ZOJ1091
Knight Moves。基础级练习题，用C++语言编写程序。

题意简述：给出国际象棋棋盘中的两个点，求马从一个点跳到另一个点的最少步数。

问题分析：典型的BFS问题。在BFS搜索过程中，马跳过的点就不必再跳了，因为这次再跳下去不可能比上次步数少。

程序中，使用了一个队列来存放中间节点，但是每次用完需要清空。

这里给出两个版本的程序，有一个是设置了边界的。

AC的C++语言程序如下：

/* UVA439 POJ2243 HDU1372 ZOJ1091 Knight Moves */

#include <cstdio>
#include <cstring>
#include <queue>

using namespace std;

#define MAXN 8

#define DIRECTSIZE 8

struct direct {
    int drow;
    int dcol;
} direct[DIRECTSIZE] =
    {{-2, 1}, {-1, 2}, {1, 2}, {2, 1}, {2, -1}, {1, -2}, {-1, -2}, {-2, -1}};

char grid[MAXN][MAXN];

int startcol, startrow, endcol, endrow;
int ans;

struct node {
    int row;
    int col;
    int level;
};

queue<node> q;

void bfs()
{
    memset(grid, ' ', sizeof(grid));

    ans = 0;
    node start;
    start.row = startrow;
    start.col = startcol;
    start.level = 0;
    q.push(start);

    while(!q.empty()) {
        node front = q.front();
        q.pop();

        if(front.row == endrow && front.col == endcol) {
            ans = front.level;
            break;
        }

        for(int i=0; i<DIRECTSIZE; i++) {
            int nextrow = front.row + direct[i].drow;
            int nextcol = front.col + direct[i].dcol;

            if(0 <= nextrow && nextrow < MAXN && 0 <= nextcol && nextcol < MAXN)
                if(grid[nextrow][nextcol] == ' ') {
                    node v;
                    v.row = nextrow;
                    v.col = nextcol;
                    v.level = front.level + 1;
                    q.push(v);
                }
        }

        grid[front.row][front.col] = '*';
    }
}

int main(void)
{
    char startc, endc;

    while(scanf("%c%d %c%d", &startc, &startrow, &endc, &endrow) != EOF) {
        getchar();

        while(!q.empty())
            q.pop();

        startcol = startc - 'a';
        startrow -= 1;
        endcol = endc - 'a';
        endrow -= 1;

        bfs();

        printf("To get from %c%d to %c%d takes %d knight moves.\n", startc, startrow+1, endc, endrow+1, ans);
    }

    return 0;
}

另外一版AC的C++语言程序如下：

/* UVA439 POJ2243 HDU1372 ZOJ1091 Knight Moves */

#include <cstdio>
#include <cstring>
#include <queue>

using namespace std;

#define MAXN 8

#define DIRECTSIZE 8

struct direct {
    int drow;
    int dcol;
} direct[DIRECTSIZE] =
    {{-2, 1}, {-1, 2}, {1, 2}, {2, 1}, {2, -1}, {1, -2}, {-1, -2}, {-2, -1}};

char grid[MAXN+4][MAXN+4];

int startcol, startrow, endcol, endrow;
int ans;

struct node {
    int row;
    int col;
    int level;
};

queue<node> q;

void bfs()
{
    ans = 0;
    node start;
    start.row = startrow;
    start.col = startcol;
    start.level = 0;
    q.push(start);

    while(!q.empty()) {
        node front = q.front();
        q.pop();

        if(front.row == endrow && front.col == endcol) {
            ans = front.level;
            break;
        }

        for(int i=0; i<DIRECTSIZE; i++) {
            int nextrow = front.row + direct[i].drow;
            int nextcol = front.col + direct[i].dcol;
            if(grid[nextrow][nextcol] == ' ') {
                node v;
                v.row = nextrow;
                v.col = nextcol;
                v.level = front.level + 1;
                q.push(v);
            }
        }

        grid[front.row][front.col] = '*';
    }
}

int main(void)
{
    int i;
    char startc, endc;

    while(scanf("%c%d %c%d", &startc, &startrow, &endc, &endrow) != EOF) {
        getchar();

        while(!q.empty())
            q.pop();

        startcol = startc - 'a' + 2;
        startrow += 1;
        endcol = endc - 'a' + 2;
        endrow += 1;

        memset(grid, ' ', sizeof(grid));
        for(i=0; i<MAXN+4; i++) {
            grid[0][i] = '*';
            grid[1][i] = '*';
            grid[MAXN+2][i] = '*';
            grid[MAXN+3][i] = '*';

            grid[i][0] = '*';
            grid[i][1] = '*';
            grid[i][MAXN+2] = '*';
            grid[i][MAXN+3] = '*';
        }

        bfs();

        printf("To get from %c%d to %c%d takes %d knight moves.\n", startc, startrow-1, endc, endrow-1, ans);
    }

    return 0;
}