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  • UVA439 POJ2243 HDU1372 ZOJ1091 Knight Moves【BFS】

    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。

    程序说明: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 <iostream>
    #include <cstring>
    #include <queue>
    
    using namespace std;
    
    const int 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}};
    
    const int MAXN = 8;
    char grid[MAXN][MAXN];
    
    struct node {
        int row;
        int col;
        int level;
    };
    
    node start, end2;
    int ans;
    
    void bfs()
    {
        queue<node> q;
    
        memset(grid, ' ', sizeof(grid));
    
        grid[start.row][start.col] = '*';
    
        ans = 0;
        q.push(start);
    
        while(!q.empty()) {
            node front = q.front();
            q.pop();
    
            if(front.row == end2.row && front.col == end2.col) {
                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] == ' ') {
                        grid[nextrow][nextcol] = '*';
                        node v;
                        v.row = nextrow;
                        v.col = nextcol;
                        v.level = front.level + 1;
                        q.push(v);
                    }
            }
        }
    }
    
    int main(void)
    {
        char startc, endc;
    
        while(cin >> startc >> start.row >> endc >> end2.row) {
            start.row--;
            start.col = startc - 'a';
            start.level = 0;
            end2.row--;
            end2.col = endc - 'a';
    
            bfs();
    
            printf("To get from %c%d to %c%d takes %d knight moves.
    ", startc, start.row+1, endc, end2.row+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.
    ", startc, startrow-1, endc, endrow-1, ans);
        }
    
        return 0;
    }


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  • 原文地址:https://www.cnblogs.com/tigerisland/p/7564459.html
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