Aizu 2325 Mysterious Maze

走迷宫 ~ 

不同的是题目给了你转向的方向序列 

dis[x][y]表示到(x,y) 使用了最少的转向次数

#include<cstdio>#include<cstring>#include<cmath>#include<algorithm>#include <queue>using namespace std;const int maxn = 1010;const int inf = (1<<30);int dx[4] = {-1,0,1,0};int dy[4] = {0,1,0,-1};struct node{    int x, y;    node(int i = 0, int j = 0)    {        x = i, y = j;    }};char s[1000010];int g[maxn][maxn], dis[maxn][maxn], cost[1000010][4], dir[1000010];int W,H,N;queue<node>Q;void init(){    memset(dis, -1, sizeof(dis));    memset(g, 0, sizeof(g));    dir[0] = 0;    for(int i = 1; i <= N; ++ i)        if(s[i] == 'L') dir[i] = (dir[i-1]+3)%4;        else dir[i] = (dir[i-1]+1)%4;    for(int i = N; i >= 0; -- i)        for(int j = 0; j < 4; ++ j)        {            if(dir[i] == j) cost[i][j] = 0;            else            {                if(i == N) cost[i][j] = inf;                else cost[i][j] = cost[i+1][j] +1;            }        }    for(int i = 0; i < 4; ++ i) cost[N+1][i] = inf;    while(!Q.empty()) Q.pop();}bool solve(int sx, int sy, int ex, int ey){    Q.push(node(sx, sy));    dis[sx][sy] = 0;    while(!Q.empty())    {        node u = Q.front();        Q.pop();        int now = dis[u.x][u.y];        for(int i = 0; i < 4; ++ i)        {            int nx = u.x+dx[i], ny = u.y+dy[i];            if(g[nx][ny] && cost[now][i] < inf)            {                if(nx == ex && ny == ey) return true;                int tem = now+cost[now][i];                if(dis[nx][ny] == -1 || tem < dis[nx][ny])                {                    dis[nx][ny] = tem;                    Q.push(node(nx, ny));                }            }        }    }    return false;}void show(){    for(int i = 0; i <= W+1; ++ i)    {        for(int j = 0; j <= H+1; ++ j)            printf("%d ", g[i][j]);        puts("");    }}int main(){    while(scanf("%d%d%d", &W, &H, &N) ==3 && W+H+N)    {        scanf("%s", s+1);        init();        int sx, sy, ex, ey;        for(int i = 1; i <= W; ++ i)        {            scanf("%s", s+1);            for(int j = 1; j <= H; ++ j)            {                if(s[j] != '#') g[i][j] = 1;                if(s[j] == 'S') sx = i, sy = j;                if(s[j] == 'G') ex = i, ey = j;            }        }       // show();        if(solve(sx, sy, ex, ey)) puts("Yes");        else puts("No");    }    return 0;}