HDU5040- Instrusive(BFS+记忆化搜索)

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题目链接

题意：给定一张图，Matt要从M移动到T取偷东西，N，W，S，E表示这些位置有监控器，字母表示这些监控器的初始方向，并且每一秒顺时针旋转90度。现在Matt每移动一格需要花一秒，但是如果当前他所在的位置或者他要去的位置被监控器监视，那么如果他要移动，就必须躲在箱子里移动，时间需要花费3秒。或者也可以选择不移动，躲在箱子里1秒，问说Matt最少花费多少时间移动到T。

思路：借鉴的是小伙伴的想法，先预处理整张图，用二进制表示在一个周期(4S)内每个格子是否会被监控器监控到，之后状态要多开一维，表示4S一个周期，然后进行状态转移。

代码：

#include <iostream>#include <cstdio>#include <cstring>#include <queue>#include <algorithm>using namespace std;const int MAXN = 505;const int INF = 0x3f3f3f3f;const int dx[] = {-1, 0, 1, 0};const int dy[] = {0, 1, 0, -1}; struct node{    node(int xx, int yy, int tt) {        x = xx;         y = yy;        t = tt;    }    int x, y, t;};char map[MAXN][MAXN];int g[MAXN][MAXN], d[MAXN][MAXN][5];int n, sx, sy, ex, ey;int change(char c) {    if (c == 'N') return 0;    if (c == 'E') return 3;    if (c == 'S') return 2;    if (c == 'W') return 1;    return -1;}int bfs(int x, int y, int t) {    queue<node> q;    node s(x, y, 0);    q.push(s);     memset(d, INF, sizeof(d));    d[s.x][s.y][s.t] = 0;    int ans = INF;     while (!q.empty()) {        node u = q.front();         q.pop();        node v = u;                               if (v.x == ex && v.y == ey)            ans = min(ans, d[v.x][v.y][v.t]);        v.t = (v.t + 1) % 4;        if (d[v.x][v.y][v.t] > d[u.x][u.y][u.t] + 1) {            d[v.x][v.y][v.t] = d[u.x][u.y][u.t] + 1;            q.push(v);        }        for (int i = 0; i < 4; i++) {            node v = u;             v.x = u.x + dx[i];            v.y = u.y + dy[i];            int step = 0;            if (v.x < 0 || v.x >= n || v.y < 0 || v.y >= n || g[v.x][v.y] == -1) continue;            if ((g[u.x][u.y] & (1<<u.t)) || (g[v.x][v.y] & (1<<u.t))) {                v.t = (v.t + 3) % 4;                 step = 3;            }            else {                v.t = (v.t + 1) % 4;                 step = 1;            }            if (d[v.x][v.y][v.t] > d[u.x][u.y][u.t] + step) {                d[v.x][v.y][v.t] = d[u.x][u.y][u.t] + step;                q.push(v);            }        }    }    return ans;}int main() {    int cas, t = 1;        scanf("%d", &cas);    while (cas--) {        scanf("%d", &n);          memset(g, 0, sizeof(g));        memset(map, 0, sizeof(map));        for (int i = 0; i < n; i++) {            scanf("%s", map[i]);            for (int j = 0; j < n; j++) {                if (map[i][j] == 'M') {                    sx = i;                     sy = j;                }                else if (map[i][j] == 'T') {                    ex = i;                    ey = j;                }                else if (map[i][j] == '#')                    g[i][j] = -1;                else if (change(map[i][j]) != -1) {                    g[i][j] = 15;                     for (int k = 0; k < 4; k++) {                        int tx = i + dx[k];                         int ty = j + dy[k];                        if (tx < 0 || tx >= n || ty < 0 || ty >= n || map[tx][ty] == '#') continue;                        g[tx][ty] |= (1<<((change(map[i][j]) + k) % 4));                    }                 }            }        }        printf("Case #%d: ", t++);        int ans = bfs(sx, sy, 0);        if (ans >= INF)            printf("-1\n");        else             printf("%d\n", ans);    }    return 0;}

用记忆化搜索差点就超时了，所以用BFS+优先队列会更好，思想其实是差不多的。

#include <iostream>#include <cstdio>#include <cstring>#include <queue>#include <algorithm>using namespace std;const int N = 505;const int dx[] = {-1, 0, 1, 0, 0};const int dy[] = {0, -1, 0, 1, 0};int vis[N][N][5], b[N][N][5];char g[N][N];int n, sx, sy, ex, ey;struct node{    node(int xx, int yy, int dd) {        x = xx;         y = yy;        d = dd;    }    int x, y, d;    friend bool operator < (node a, node b) {        return a.d > b.d;     }};int bfs(int x,int y) {    priority_queue <node> q ;    node s(x, y, 0);    q.push(s) ;    memset(vis,0,sizeof(vis)) ;    vis[x][y][0] = 1 ;    while(!q.empty()) {        node u = q.top() ;        q.pop();        if(u.x == ex && u.y == ey)             return u.d ;         for(int i = 0 ;i < 5 ;i++) {            int tx = u.x + dx[i] ;            int ty = u.y + dy[i] ;            if(tx < 0 || tx >= n || ty < 0 || ty >= n || g[tx][ty] == '#')continue ;            node p = u;            if(b[tx][ty][u.d%4] || b[u.x][u.y][u.d%4]) {                if(tx == u.x && ty == u.y && !vis[tx][ty][(u.d+1)%4]) {                    p.d++ ;                    vis[tx][ty][p.d%4]=1 ;                    q.push(p) ;                }                else if(!vis[tx][ty][(u.d+3)%4]) {                    p.x=tx;                    p.y=ty ;                    p.d += 3 ;                    vis[tx][ty][p.d%4]=1 ;                    q.push(p) ;                }            }            else if(!vis[tx][ty][(u.d+1)%4]) {                p.x=tx;                p.y=ty;                p.d++ ;                vis[tx][ty][p.d%4]=1 ;                q.push(p) ;            }        }    }    return -1 ;}int main() {    int cas, t = 1;    scanf("%d", &cas);    while (cas--) {        scanf("%d", &n);        for (int i = 0; i < n; i++)               scanf("%s", g[i]);        memset(b, 0, sizeof(b));        for (int i = 0; i < n; i++)            for (int j = 0; j < n; j++) {                if (g[i][j] == 'M') {                    sx = i;                     sy = j;                 }                 else if (g[i][j] == 'T') {                    ex = i;                     ey = j;                 }                else if (g[i][j] == 'N') {                    b[i][j][0] = b[i][j][1] = b[i][j][2] = b[i][j][3] = 1;                    if (i - 1 >= 0) b[i - 1][j][0] = 1;                    if (j + 1 < n) b[i][j + 1][1] = 1;                    if (i + 1 < n) b[i + 1][j][2] = 1;                    if (j - 1 >= 0) b[i][j - 1][3] = 1;                }                else if (g[i][j] == 'W') {                    b[i][j][0] = b[i][j][1] = b[i][j][2] = b[i][j][3] = 1;                    if (j - 1 >= 0) b[i][j - 1][0] = 1;                    if (i - 1 >= 0) b[i - 1][j][1] = 1;                    if (j + 1 < n) b[i][j + 1][2] = 1;                     if (i + 1 < n) b[i + 1][j][3] = 1;                }                else if (g[i][j] == 'S') {                    b[i][j][0] = b[i][j][1] = b[i][j][2] = b[i][j][3] = 1;                    if (i + 1 < n) b[i + 1][j][0] = 1;                    if (j - 1 >= 0) b[i][j - 1][1] = 1;                    if (i - 1 >= 0) b[i - 1][j][2] = 1;                    if (j + 1 < n) b[i][j + 1][3] = 1;                 }                else if (g[i][j] == 'E') {                    b[i][j][0] = b[i][j][1] = b[i][j][2] = b[i][j][3] = 1;                    if (j + 1 < n) b[i][j + 1][0] = 1;                     if (i + 1 < n) b[i + 1][j][1] = 1;                    if (j - 1 >= 0) b[i][j - 1][2] = 1;                    if (i - 1 >= 0) b[i - 1][j][3] = 1;                }            }        printf("Case #%d: %d\n", t++, bfs(sx, sy));    }    return 0;}

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