小游戏 OpenJ_Bailian - 2802

一个不算太难的Dfs问题，可以理解为找到一条到终点的最短路径同时记下变了几次方向。

题解都写在注释里了。

//小游戏
#include <iostream>
#include<cstdio>
#include<cstring>
using namespace std;

const int maxw = 80;
char board[maxw][maxw];
bool mark[maxw][maxw]; //标记数组
int minStep, w, h, to[4][2] = {{0,1}, {1,0}, {0,-1}, {-1,0}};

//定义的深搜状态为从x1,y1到x2,y2,当前分为step个线段，当前方向是f
void Dfs(int x1, int y1, int x2, int y2, int step, int f)
{
    if(step > minStep) return; //最优化剪枝

    if(x1==x2 && y1==y2){ //终点
        if(minStep > step) minStep = step;
        return;
    }
    for(int i = 0; i < 4; i++){ //对每个可以走的位置
        int x = x1+to[i][0], y = y1+to[i][1];
        //越界最多一格且是没有走过的空格或终点
        if((x>=0) && (x<=w+1) && (y>=0) && (y<=h+1) && (((board[y][x])==' ' && (mark[y][x]==false))
            || ((x==x2) && (y==y2) && (board[y][x]=='X')))){
            mark[y][x] = true; //标记为走过
            //上一步搜索方向和当前方向相同step不变，否则step+1
            if(f==i) Dfs(x, y, x2, y2, step, i);
            else Dfs(x, y, x2, y2, step+1, i);

            mark[y][x] = false; //回溯
        }
    }
}

int main()
{
    int index1 = 0;
    while(scanf("%d%d",&w,&h) != EOF){
        if(w==0 && h==0) break;
        index1++;
        printf("Board #%d:\n",index1);
        for(int i = 0; i < 75+2; i++)
            board[0][i] = board[i][0] = ' ';
        for(int i = 1; i <= h; i++){
            getchar();
            for(int j = 1; j <= w; j++)
                board[i][j] = getchar();
        }

        //在矩形最外层加一圈空格
        for(int i = 0; i <= w; i++)
            board[h+1][i+1] = ' ';
        for(int i = 0; i <= h; i++)
            board[i+1][w+1] = ' ';
        int x1, y1, x2, y2, index2 = 0;
        while(scanf("%d%d%d%d",&x1, &y1, &x2, &y2)!=EOF && x1>0){
            index2++;
            minStep = 100000;
            memset(mark, false, sizeof(mark));
            Dfs(x1, y1, x2, y2, 0, -1);

            if(minStep < 100000) printf("Pair %d: %d segments.\n",index2,minStep);
            else printf("Pair %d: impossible.\n",index2);
        }
        printf("\n");
    }
    return 0;
}