HDU 4166 Robot Navigation

来源：互联网发布：php final static 编辑：程序博客网时间：2024/05/19 09:50
题意：
一个机器人走迷宫每一秒要么转向要么前进问最少时间的情况下有几种方案
思路：
记忆化搜索即可简单bfs
代码：
#include<cstdio>#include<iostream>#include<cstring>#include<string>#include<algorithm>#include<map>#include<set>#include<vector>#include<queue>#include<cstdlib>#include<ctime>#include<cmath>#include<bitset>using namespace std;#define N 1010int dir[4][2] = { { -1, 0 }, { 0, 1 }, { 1, 0 }, { 0, -1 } };int cas = 1, n, m, mod, ans;int dp[N][N][4][2];char maz[N][N];struct position {    int x, y, to;} u, v, S, T, qu[N * N * 4];int main() {    char face[10];    while (~scanf("%d%d%d", &n, &m, &mod)) {        if (!mod)            break;        memset(maz, '*', sizeof(maz));        for (int i = 1; i <= n; i++)            scanf("%s", maz[i] + 1);        scanf("%d%d%d%d%s", &S.x, &S.y, &T.x, &T.y, face);        memset(dp, -1, sizeof(dp));        S.x++;        S.y++;        T.x++;        T.y++;        u.x = S.x;        u.y = S.y;        if (face[0] == 'N')            u.to = 0;        else if (face[0] == 'E')            u.to = 1;        else if (face[0] == 'S')            u.to = 2;        else            u.to = 3;        dp[u.x][u.y][u.to][0] = 0;        dp[u.x][u.y][u.to][1] = 1;        int l = 0, r = 1;        qu[0] = u;        while (l < r) {            u = qu[l++];            int nowtime = dp[u.x][u.y][u.to][0] + 1;            int nowways = dp[u.x][u.y][u.to][1];//            printf("from %d %d %d %d %d\n", u.x, u.y, u.to,//                    dp[u.x][u.y][u.to][0], dp[u.x][u.y][u.to][1]);            v = u;            v.x += dir[v.to][0];            v.y += dir[v.to][1];//            printf("to %d %d\n", v.x, v.y);            if (maz[v.x][v.y] == '.') {                if (dp[v.x][v.y][v.to][0] == -1) {                    dp[v.x][v.y][v.to][0] = nowtime;                    qu[r++] = v;                    dp[v.x][v.y][v.to][1] = nowways % mod;                } else if (dp[v.x][v.y][v.to][0] == nowtime) {                    dp[v.x][v.y][v.to][1] += nowways;                    dp[v.x][v.y][v.to][1] %= mod;                }            }            v = u;            v.to = (u.to + 1) % 4;            if (maz[v.x][v.y] == '.') {                if (dp[v.x][v.y][v.to][0] == -1) {                    dp[v.x][v.y][v.to][0] = nowtime;                    qu[r++] = v;                    dp[v.x][v.y][v.to][1] = nowways % mod;                } else if (dp[v.x][v.y][v.to][0] == nowtime) {                    dp[v.x][v.y][v.to][1] += nowways;                    dp[v.x][v.y][v.to][1] %= mod;                }            }            v = u;            v.to = (u.to + 3) % 4;            if (maz[v.x][v.y] == '.') {                if (dp[v.x][v.y][v.to][0] == -1) {                    dp[v.x][v.y][v.to][0] = nowtime;                    qu[r++] = v;                    dp[v.x][v.y][v.to][1] = nowways % mod;                } else if (dp[v.x][v.y][v.to][0] == nowtime) {                    dp[v.x][v.y][v.to][1] += nowways;                    dp[v.x][v.y][v.to][1] %= mod;                }            }        }        int best = -1;        for (int i = 0; i < 4; i++)            if (best == -1 || best > dp[T.x][T.y][i][0])                best = dp[T.x][T.y][i][0];        if (best == -1) {            printf("Case %d: %d -1\n", cas++, mod);            continue;        }        ans = 0;        for (int i = 0; i < 4; i++)            if (dp[T.x][T.y][i][0] == best)                ans = (ans + dp[T.x][T.y][i][1]) % mod;        printf("Case %d: %d %d\n", cas++, mod, ans);//        for (int i = 1; i <= n; i++) {//            for (int j = 1; j <= m; j++) {//                for (int k = 0; k < 4; k++) {//                    printf("(%d,%d) %d %d %d\n", i, j, k, dp[i][j][k][0],//                            dp[i][j][k][1]);//                }//            }//        }    }    return 0;}
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