HDU 3638 Go , SuSu

HDU_3638

    如果除去怪物的話這個題和普通的BFS走迷宮沒太大區別，因此不妨將怪物看得相對分離一點，每到一秒先將怪物的視野處理出來，然後這一秒既不走到怪物的視野內又不走到*的位置即可。

    在處理怪物的視野時，可以先用一個結構體儲存怪物的方位，每過一秒就自動更新怪物的方位，再打一個包括4個方向且每個方向上有9個相對位置的表，這樣就比較好處理出怪物的視野了。

#include<stdio.h>#include<string.h>#include<algorithm>#include<queue>#define MAXN 60int trans[] = {0, 2, 1, 4, 3};int dx[] = {0, -1, 1, 0, 0}, dy[] = {0, 0, 0, -1, 1};int ex[][9] ={    {0, 0, 0, 0, 0, 0, 0, 0, 0},    {0, -1, -1, -1, -2, -2, -2, -2, -2},    {0, 1, 1, 1, 2, 2, 2, 2, 2},    {0, -1, 0, 1, -2, -1, 0, 1, 2},    {0, -1, 0, 1, -2, -1, 0, 1, 2}};int ey[][9] ={    {0, 0, 0, 0, 0, 0, 0, 0, 0},    {0, -1, 0, 1, -2, -1, 0, 1, 2},    {0, -1, 0, 1, -2, -1, 0, 1, 2},    {0, -1, -1, -1, -2, -2, -2, -2, -2},    {0, 1, 1, 1, 2, 2, 2, 2, 2}};struct Monster{    int x, y, d;}mon[MAXN];char b[MAXN];int N, M, K, sx, sy, tx, ty, g[MAXN][MAXN], del[MAXN][MAXN], vis[MAXN][MAXN];struct Point{    int x, y, t;    Point(){}    Point(int _x, int _y, int _t) : x(_x), y(_y), t(_t){}};inline int inside(int x, int y){    return x >= 1 && x <= N && y >= 1 && y <= M;}void moveall(){    int i, x, y;    for(i = 0; i < K; i ++)    {        x = mon[i].x + dx[mon[i].d], y = mon[i].y + dy[mon[i].d];        if(inside(x, y) && g[x][y]) mon[i].x = x, mon[i].y = y;        else mon[i].d = trans[mon[i].d];    }}void delblock(){    int i, j, x, y;    for(i = 1; i <= N; i ++)        for(j = 1; j <= M; j ++) del[i][j] = 1 - g[i][j];    for(i = 0; i < K; i ++)        for(j = 0; j < 9; j ++)        {            x = mon[i].x + ex[mon[i].d][j], y = mon[i].y + ey[mon[i].d][j];            if(inside(x, y)) del[x][y] = 1;        }}void init(){    int i, j;    scanf("%d%d", &N, &M);    for(i = 1; i <= N; i ++)    {        scanf("%s", b + 1);        for(j = 1; j <= M; j ++)        {            g[i][j] = b[j] != '*';            if(b[j]== 'A') sx = i, sy = j;            else if(b[j] == 'B') tx = i, ty = j;        }    }    scanf("%d", &K);    for(i = 0; i < K; i ++) scanf("%d%d%d", &mon[i].x, &mon[i].y, &mon[i].d);}void solve(){    int i, j, x, y, cur, ans = -1;    cur = -1;    delblock();    if(del[sx][sy])    {        printf("勝敗兵家事不期 捲土重來是大俠\n");        return ;    }    std::queue <Point> q;    q.push(Point(sx, sy, 0));    while(!q.empty())    {        Point p = q.front();        q.pop();        if(p.t > 1000) break;        if(p.x == tx && p.y == ty)        {            ans = p.t;            break;        }        if(p.t > cur)        {            cur = p.t;            for(i = 1; i <= N; i ++)                for(j = 1; j <= M; j ++) vis[i][j] = 0;            moveall(), delblock();        }        for(i = 0; i < 5; i ++)        {            x = p.x + dx[i], y = p.y + dy[i];            if(inside(x, y) && !del[x][y] && !vis[x][y])                vis[x][y] = 1, q.push(Point(x, y, p.t + 1));        }    }    if(ans == -1) printf("勝敗兵家事不期 捲土重來是大俠\n");    else printf("%d\n", ans);}int main(){    int t, tt;    scanf("%d", &t);    for(tt = 1; tt <= t; tt ++)    {        init();        printf("Case %d: ", tt);        solve();    }    return 0;}