解救小Q

http://acm.uestc.edu.cn/#/problem/show/149
其实没有想象中那么难,就是bfs。
要注意在传送门中往返,走一个门封一个门。

#include<iostream>
#include <cstring>
#include  <cstdio>
#include   <queue>
#include     <set>
#include     <map>
using namespace std;
const int INT = 2147483647;
int dir[4][2] = { { 1,0 },{ -1,0 },{ 0,1 },{ 0,-1 } };
int lx, ly, qx, qy, n, m, ans;
char mapn[55][55];
int vis[55][55];
struct Note {
    int x;
    int y;
    int spet;
}note;
map<char, Note> mm;
void BFS(Note nn) {
    queue<Note> q;
    q.push(nn);
    while (!q.empty()) {
        for (int i = 0; i < 4; i++) {
            note.x = q.front().x + dir[i][0];
            note.y = q.front().y + dir[i][1];
            note.spet = q.front().spet + 1;
            if (note.x >= 0 && note.x < n && note.y >= 0 && note.y<m && mapn[note.x][note.y] != '#'&& vis[note.x][note.y]>note.spet) {
                if (note.x == qx && note.y == qy) {
                    ans = note.spet;
                    return;
                }
                else if ('a' <= mapn[note.x][note.y] && mapn[note.x][note.y] <= 'z') {
                    int x = note.x, y = note.y;
                    vis[x][y] = note.spet;
                    note.x = mm[mapn[x][y] - 'a' + 'A'].x;
                    note.y = mm[mapn[x][y] - 'a' + 'A'].y;
                    q.push(note);//允许往返一次
                    continue;
                }
                else if ('A' <= mapn[note.x][note.y] && mapn[note.x][note.y] <= 'Z') {
                    int x = note.x, y = note.y;
                    vis[x][y] = note.spet;
                    note.x = mm[mapn[x][y] - 'A' + 'a'].x;
                    note.y = mm[mapn[x][y] - 'A' + 'a'].y;
                    q.push(note);
                    continue;
                }
                q.push(note);
                vis[note.x][note.y] = note.spet;
            }
        }
        q.pop();
    }
}
int main() {
    int T;
    scanf("%d", &T);
    while (T--) {
        ans = -1;
        mm.clear();
        scanf("%d%d", &n, &m);
        getchar();
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                vis[i][j] = INT;
            }
        }
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                scanf("%c", &mapn[i][j]);
                if (mapn[i][j] == 'L') {
                    lx = i;
                    ly = j;
                }
                if (mapn[i][j] == 'Q') {
                    qx = i;
                    qy = j;
                }
                if ('a' <= mapn[i][j] && mapn[i][j] <= 'z') {
                    note.x = i;
                    note.y = j;
                    map<char, Note>::iterator it;;
                    it = mm.find(mapn[i][j]);
                    if (it == mm.end()) {
                        mm.insert(pair <char, Note>(mapn[i][j], note));
                    }
                    else {
                        mapn[i][j] = mapn[i][j] - 'a' + 'A';
                        mm.insert(pair <char, Note>(mapn[i][j], note));
                    }
                }
            }
            getchar();
        }
        note.x = lx;
        note.y = ly;
        note.spet = 0;
        vis[lx][ly] = 0;
        BFS(note);
        printf("%d\n", ans);
    }
    return 0;
}

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转载自blog.csdn.net/lxworld123/article/details/51900639