HDU 5040 instrusive (BFS)

HDU 5040 instrusive

Given a picture, Matt will move from m to t to steal things. N, W, S, e indicates that these locations have monitors, and letters indicate the initial direction of these monitors, and Rotate 90 degrees clockwise every second. Now it takes one second for Matt to move one cell. However, if the current location or the location where he is going is monitored by the monitor, if he wants to move it, he must move it in the box, it takes 3 seconds. You can also choose not to move, hide in the box for one second, and ask how much time Matt will spend moving to T.

Solution: There is nothing to do at the beginning, and it is estimated that it is a pitfall question. It is good to pre-process the graph. Because the monitor has a cycle of 4 seconds, you only need to compute the time in each position, 0, 1, 2, and 3 seconds to be monitored, use a V [x] [Y] to represent a binary number record. Then DP [x] [y] [T] indicates the optimal time (t =, 2, 3) at the position X and Y ).

#include <cstdio>#include <cstring>#include <queue>#include <algorithm>using namespace std;const int maxn = 505;const int dir[4][2] = {{-1, 0}, {0, 1}, {1, 0}, {0, -1}};const int INF = 0x3f3f3f3f;struct state {    int x, y, t;    state (int x = 0, int y = 0, int t = 0) {        this->x = x;        this->y = y;        this->t = t;    }};int N, sx, sy, ex, ey, v[maxn][maxn];int dp[maxn][maxn][4];char g[maxn][maxn];inline int cal (char ch) {    if (ch == ‘N‘) return 0;    else if (ch == ‘E‘) return 1;    else if (ch == ‘S‘) return 2;    else return 3;}void look (int x, int y) {    v[x][y] = 15;    int b = cal(g[x][y]);    for (int i = 0; i < 4; i++) {        int idx = (i + b) % 4;        int p = x + dir[idx][0];        int q = y + dir[idx][1];        if (p < 0 || p >= N || q < 0 || q >= N || g[p][q] == ‘#‘)            continue;        v[p][q] |= (1<<i);    }}void init () {    scanf("%d", &N);    memset(v, 0, sizeof(v));    for (int i = 0; i < N; i++)        scanf("%s", g[i]);    for (int i = 0; i < N; i++) {        for (int j = 0; j < N; j++) {            if (g[i][j] == ‘#‘)                v[i][j] = -1;            else if (g[i][j] == ‘M‘) {                sx = i, sy = j;            } else if (g[i][j] == ‘T‘) {                ex = i, ey = j;            } else if (g[i][j] != ‘.‘)                look(i, j);        }    }}int bfs () {    queue<state> que;    que.push(state(sx, sy, 0));    memset(dp, INF, sizeof(dp));    dp[sx][sy][0] = 0;    while (!que.empty()) {        state u = que.front();        que.pop();        int x = u.x, y = u.y, t = u.t;        for (int i = 0; i < 4; i++) {            int p = x + dir[i][0];            int q = y + dir[i][1];            if (p < 0 || p >= N || q < 0 || q >= N || v[p][q] == -1)                continue;            int add = 1;            if ((v[x][y] & (1<<t)) || (v[p][q] & (1<<t))) {                int ti = (t + add) % 4;                if (dp[x][y][ti] > dp[x][y][t] + add) {                    dp[x][y][ti] = dp[x][y][t] + add;                    que.push(state(x, y, ti));                }                add = 3;            }            int ti = (t + add) % 4;            if (dp[p][q][ti] > dp[x][y][t] + add) {                dp[p][q][ti] = dp[x][y][t] + add;                que.push(state(p, q, ti));            }        }    }    int ans = INF;    for (int i = 0; i < 4; i++)        ans = min(ans, dp[ex][ey][i]);    if (ans == INF)        return -1;    else        return ans;}int main () {    int cas;    scanf("%d", &cas);    for (int kcas = 1; kcas <= cas; kcas++) {        init();        printf("Case #%d: %d\n", kcas, bfs());    }    return 0;}

HDU 5040 instrusive (BFS)