题目链接 http://poj.org/problem?id=3984

中文题题意不解释了

反正就是简单的结构体套结构体存一下路径就行了

#include <iostream>#include <cstring>#include <deque>#include <queue>using namespace std;int map[6][6];struct ss {    int x , y;};struct TnT {    deque<ss>mm;};int vis[6][6] , dr[4][2] = {1 , 0 , -1 , 0 , 0 , 1 , 0 , -1};TnT bfs(int x , int y) {    memset(vis , 0 , sizeof(vis));    TnT gg;    ss gl;    gl.x = x , gl.y = y;    vis[x][y] = 1;    gg.mm.push_back(gl);    queue<TnT>q;    q.push(gg);    while(!q.empty()) {        TnT ml = q.front();        ss xg = ml.mm.back();        if(xg.x == 4 && xg.y == 4) {            return ml;        }        for(int i = 0 ; i < 4 ; i++) {            TnT sg = ml;            ss xl = ml.mm.back();            xl.x += dr[i][0];            xl.y += dr[i][1];            if(xl.x >= 0 && xl.x < 5 && xl.y >= 0 && xl.y < 5 && !map[xl.x][xl.y] && !vis[xl.x][xl.y]) {                vis[xl.x][xl.y] = 1;                sg.mm.push_back(xl);                q.push(sg);            }        }        q.pop();    }    return q.front();}int main(){    for(int i = 0 ; i < 5 ; i++) {        for(int j = 0 ; j < 5 ; j++) {            cin >> map[i][j];        }    }    TnT q = bfs(0 , 0);    while(!q.mm.empty()) {        ss sl = q.mm.front();        q.mm.pop_front();        cout << "(" << sl.x << ", " << sl.y << ")" << endl;    }    return 0;}

poj3984 迷宫问题（简单的输出路径的bfs）