【POJ3083】Children of the Candy Corn

本题传送门

本题知识点：深度优先搜索 + 宽度优先搜索

本题题意是求三个路径长度，第一个是一直往左手边走的距离，第二个是一直往右手边走的距离，第三个是最短距离。

第三个很好办，就是一个简单的bfs的模板，问题出在第一二个。

但第一二个只是方向的丝丝不同，所以会其中一个也就解决了。

读懂题意后很简单，问题是怎么简单高效地实现。

推荐这篇博客

一些感慨：该题做了我差不多4个小时吧，前前后后de了bug又发现有新bug，包括调试的代码，旧代码起码达到了300多行。已经有东西南北方向的想法了，可就是差那么一点点。还是刷题不够吧。总之这4个小时里感到兴奋的心情比难过迷茫的心情要多很多。还是要以学习算法为前提去A题才是快乐A题呀！

// 3083
#include<iostream>
#include<cstdio>
#include<queue>
#include<cstring>
using namespace std;

char maze[50][50];
bool take[50][50];
int len[50][50];
int T, W, H, now;
int bh, bw, eh, ew;
int left_len, right_len, min_len;
bool ok;
struct node{
    int h, w;
};
queue<node> que;
int rw[] = { -1, 0, 1, 0  };
int rh[] = { 0, 1,  0, -1 };

bool check(int h, int w){
    if(0 <= h && h < H && 0 <= w && w < W && maze[h][w] != '#' && !take[h][w]) return true;
    return false;
}

void build(int W, int H){
    memset(take, false, sizeof(take));
    left_len = right_len = min_len = 0;
    for(int i = 0; i < H; i++){
        scanf("%s", maze[i]);
        for(int j = 0; j < W; j++){
            if(maze[i][j] == 'S'){
                bh = i; bw = j;
//                maze[i][j] = '#';
            }
            if(maze[i][j] == 'E'){
                eh = i; ew = j;
            }
        }
    }
    if(bh == 0) now = 2;
    else if(bh == H - 1) now = 0;
    else if(bw == 0) now = 3;
    else if(bw == W - 1) now = 1;
}

void l_dfs(int h, int w, int go){
    left_len++;
//    take[h][w] = true;
    if(h == eh && w == ew){
        ok = true;
        return ;
    }

    switch(go)
    {
    case 0:
        {
            if(check(h, w - 1)) l_dfs(h, w - 1, 1);
            else if(check(h - 1, w)) l_dfs(h - 1, w, 0);
            else if(check(h, w + 1)) l_dfs(h, w + 1, 3);
            else if(check(h + 1, w)) l_dfs(h + 1, w, 2);
            break;
        }
    case 1:
        {
            if(check(h + 1, w)) l_dfs(h + 1, w, 2);
            else if(check(h, w - 1)) l_dfs(h, w - 1, 1);
            else if(check(h - 1, w)) l_dfs(h - 1, w, 0);
            else if(check(h, w + 1)) l_dfs(h, w + 1, 3);
            break;
        }
    case 2:
        {
            if(check(h, w + 1)) l_dfs(h, w + 1, 3);
            else if(check(h + 1, w)) l_dfs(h + 1, w, 2);
            else if(check(h, w - 1)) l_dfs(h, w - 1, 1);
            else if(check(h - 1, w)) l_dfs(h - 1, w, 0);
            break;
        }
    case 3:
        {
            if(check(h - 1, w)) l_dfs(h - 1, w, 0);
            else if(check(h, w + 1)) l_dfs(h, w + 1, 3);
            else if(check(h + 1, w)) l_dfs(h + 1, w, 2);
            else if(check(h, w - 1)) l_dfs(h, w - 1, 1);
            break;
        }
    }

}

void r_dfs(int h, int w, int go){
    right_len++;
    if(h == eh && w == ew){
        ok = true;
        return ;
    }

    switch(go)
    {
    case 0:
        {
            if(check(h, w + 1)) r_dfs(h, w + 1, 3);
            else if(check(h - 1, w)) r_dfs(h - 1, w, 0);
            else if(check(h, w - 1)) r_dfs(h, w - 1, 1);
            else if(check(h + 1, w)) r_dfs(h + 1, w, 2);
            break;
        }
    case 1:
        {
            if(check(h - 1, w)) r_dfs(h - 1, w, 0);
            else if(check(h, w - 1)) r_dfs(h, w - 1, 1);
            else if(check(h + 1, w)) r_dfs(h + 1, w, 2);
            else if(check(h, w + 1)) r_dfs(h, w + 1, 3);
            break;
        }
    case 2:
        {
            if(check(h, w - 1)) r_dfs(h, w - 1, 1);
            else if(check(h + 1, w)) r_dfs(h + 1, w, 2);
            else if(check(h, w + 1)) r_dfs(h, w + 1, 3);
            else if(check(h - 1, w)) r_dfs(h - 1, w, 0);
            break;
        }
    case 3:
        {
            if(check(h + 1, w)) r_dfs(h + 1, w, 2);
            else if(check(h, w + 1)) r_dfs(h, w + 1, 3);
            else if(check(h - 1, w)) r_dfs(h - 1, w, 0);
            else if(check(h, w - 1)) r_dfs(h, w - 1, 1);
            break;
        }
    }
}

void bfs(){
    len[bh][bw] = 1;
    while(!que.empty()){
        node now = que.front(), next; que.pop();
        int d = len[now.h][now.w];
//        printf("d:%d\n", d);
        if(now.h == eh && now.w == ew) break;
        for(int i = 0; i < 4; i++){
            next.h = now.h + rh[i]; next.w = now.w + rw[i];
            if(check(next.h, next.w)) {
                take[next.h][next.w] = true;
                len[next.h][next.w] = d + 1;
                que.push(next);
            }
        }
    }
}

int main()
{
    scanf("%d", &T);
    while(T--){
        scanf("%d %d", &W, &H);
        build(W, H);

        // right
        ok = false;
        memset(take, false, sizeof(take));
        l_dfs(bh, bw, now);
//        cout << endl;

        ok = false;
        memset(take, false, sizeof(take));
        r_dfs(bh, bw, now);

        while(!que.empty()) que.pop();
        memset(take, false, sizeof(take));
        memset(len, 0, sizeof(len));
        node a; a.h = bh; a.w = bw;
        que.push(a);
        bfs();

        printf("%d %d %d\n", left_len, right_len, len[eh][ew]);
    }

    return 0;
}

//6
//8 8
//########
//#......#
//#.####.#
//#.####.#
//#.####.#
//#.####.#
//#...#..#
//#S#E####
//9 5
//#########
//#.#.#.#.#
//S.......E
//#.#.#.#.#
//#########
//8 8
//########
//#......#
//#.####.#
//#.####.#
//#.####.#
//#.####.#
//#..#...#
//####E#S#
//9 5
//#########
//#.#.#.#.#
//E.......S
//#.#.#.#.#
//#########
//9 9
//####E####
//#.......#
//#..#.#..#
//#..###..#
//#.......#
//#..#.#..#
//#..#.#..#
//#..#.#..#
//####S####
//9 9
//####S####
//#.......#
//#..#.#..#
//#.#####.#
//#.......#
//#..#.#..#
//#..#.#..#
//#..#.#..#
//####E####

标签：int,Corn,50,Candy,bool,len,include,Children,本题
来源： https://www.cnblogs.com/Ayanowww/p/11555066.html