hdu 1728 逃离迷宫(搜索:BFS+优先队列)

来源：互联网发布：日本经济数据怎么查编辑：程序博客网时间：2024/06/05 17:25

问在给定转弯次数内能不能到达终点

因为限制了转弯次数,所以要想到每次取转弯次数最小的情况

看了网上好多人都是BFS+DFS,找到某一方向后沿着方向一直走

但我是用优先队列解决,每次取出转弯次数最小的情况,当然这样会慢一些

解决第一次转弯不计次数的方法是判断四个方向是否能加入队列,能的话加入即可

代码如下:

#include <queue>#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#define MAXN 100#define LL long longusing namespace std;int m, n;struct Position {    int x, y, k, dir;    bool operator < (const Position &rhs) const {        return k > rhs.k;    }}start, end, cur, tmp;int g[MAXN][MAXN];bool vis[MAXN][MAXN][12][5];priority_queue<Position> q;//定义在外面不要忘了清空bool BFS(int k, int x1, int y1, int x2, int y2) {        while(!q.empty())        q.pop();    memset(vis, 0, sizeof(vis));    start.x = x1;    start.y = y1;    start.k = 0;    end.x = x2;    end.y = y2;    end.k = k;        cur.k = 0;        if(x1-1>=1 && g[x1-1][y1]==0) {//向上        cur.x = x1-1;        cur.y = y1;        cur.dir = 1;        q.push(cur);    }    if(y1-1>=1 && g[x1][y1-1]==0) {//向左        cur.x = x1;        cur.y = y1-1;        cur.dir = 2;        q.push(cur);    }    if(x1+1<=m && g[x1+1][y1]==0) {//向下        cur.x = x1+1;        cur.y = y1;        cur.dir = 4;        q.push(cur);    }    if(y1+1<=n && g[x1][y1+1]==0) {//向右        cur.x = x1;        cur.y = y1+1;        cur.dir = 3;        q.push(cur);    }    while(!q.empty()) {        tmp = q.top();        q.pop();        vis[tmp.x][tmp.y][tmp.k][tmp.dir] = true;        if(tmp.x==end.x && tmp.y==end.y) {            return true;        }        if(tmp.x-1>=1 && tmp.dir!=4 && g[tmp.x-1][tmp.y]==0) {//向上            cur.x = tmp.x-1;            cur.y = tmp.y;            cur.dir = 1;            cur.k = tmp.k;            if(tmp.dir != 1)                cur.k++;            if(cur.k <= end.k && !vis[cur.x][cur.y][cur.k][cur.dir])                q.push(cur);        }        if(tmp.y-1>=1 && tmp.dir!=3 && g[tmp.x][tmp.y-1]==0) {//向左            cur.x = tmp.x;            cur.y = tmp.y-1;            cur.dir = 2;            cur.k = tmp.k;            if(tmp.dir != 2)                cur.k++;            if(cur.k <= end.k && !vis[cur.x][cur.y][cur.k][cur.dir])                q.push(cur);        }        if(tmp.x+1<=m && tmp.dir!=1 && g[tmp.x+1][tmp.y]==0) {//向下            cur.x = tmp.x+1;            cur.y = tmp.y;            cur.dir = 4;            cur.k = tmp.k;            if(tmp.dir != 4)                cur.k++;            if(cur.k <= end.k && !vis[cur.x][cur.y][cur.k][cur.dir])                q.push(cur);        }        if(tmp.y+1<=n && tmp.dir!=2 && g[tmp.x][tmp.y+1]==0) {//向右            cur.x = tmp.x;            cur.y = tmp.y+1;            cur.dir = 3;            cur.k = tmp.k;            if(tmp.dir != 3)                cur.k++;            if(cur.k <= end.k && !vis[cur.x][cur.y][cur.k][cur.dir])                q.push(cur);        }    }    return false;}int main(void) {    int T, k, x1, y1, x2, y2;    char ch;    scanf("%d", &T);    while(T--) {        scanf("%d%d", &m, &n);        memset(g, 0, sizeof(g));        for(int i=1; i<=m; ++i) {            getchar();            for(int j=1; j<=n; ++j) {                ch = getchar();                if(ch == '.')                    g[i][j] = 0;                else g[i][j] = -1;            }        }        scanf("%d%d%d%d%d", &k, &y1, &x1, &y2, &x2);        if(BFS(k, x1, y1, x2, y2))            cout << "yes" << endl;        else cout << "no" << endl;    }    return 0;}

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