浙江大学2015年校赛F题 ZOJ 3865 Superbot BFS 搜索

 不知道为什么比赛的时候一直想着用DFS 来写

一直想剪枝结果还是TLE = = 

这题数据量不大，又是问最优解，那么一般来说是用 BFS 来写

 

int commandi[4] = {1, 2, 3, 4};

我定义了一个方向数组，其实题目意思中的，指针移动还有操作版的变化本质上都是指针的移动

在此只需要 额外定义一个变量 cur 在数组 commandi 中循环移动即可

 

这道题目还是因为数据量不大吧，直接用 STL 中的 Queue 即可，优先队列肯定会更快。

总体来说，还是一道容易题。

 

Source Code :

//#pragma comment(linker, "/STACK:16777216") //for c++ Compiler

#include <stdio.h>

#include <iostream>

#include <fstream>

#include <cstring>

#include <cmath>

#include <stack>

#include <string>

#include <map>

#include <set>

#include <queue>

#include <list>

#include <vector>

#include <algorithm>

#define Max(a,b) (((a) > (b)) ? (a) : (b))

#define Min(a,b) (((a) < (b)) ? (a) : (b))

#define Abs(x) (((x) > 0) ? (x) : (-(x)))

#define MOD 1000000007

#define pi acos(-1.0)



using namespace std;



typedef long long           ll      ;

typedef unsigned long long  ull     ;

typedef unsigned int        uint    ;

typedef unsigned char       uchar   ;



template<class T> inline void checkmin(T &a,T b){if(a>b) a=b;}

template<class T> inline void checkmax(T &a,T b){if(a<b) a=b;}



const double eps = 1e-7      ;

const int N = 210            ;

const int M = 1100011*2      ;

const ll P = 10000000097ll   ;

const int MAXN = 10900000    ;

const int INF = 0x3f3f3f3f   ;



const int dir_x[5] = {0, 0, 0, -1, 1};

const int dir_y[5] = {0, -1, 1, 0, 0};



struct sc {

    int x, y;

    int time;

    int cur;

} st, ed;



int n, m, p;

char a[20][20];

bool vis[20][20][5];

int commandi[4] = {1, 2, 3, 4};

queue <sc> q;



bool check (int x, int y) {

    return x >= 1 && x <= n && y >= 1 && y <= m;

}



int left (int &cur) {

    cur = (cur - 1 + 4) % 4;

}



int right (int & cur) {

    cur = (cur + 1 + 4) % 4;

}



void solve (struct sc v) {

    ++v.time;

    if (v.time % p == 0) {

        left (v.cur);

    }

    if (vis[v.x][v.y][commandi [v.cur]] == false) {

        vis[v.x][v.y][commandi [v.cur]] = true;

        q.push (v);

    }

}



void bfs () {

    int ans = -INF;



    sc e;

    e = st;

    e.time = 0;

    e.cur = 0;



    memset (vis, 0, sizeof (vis));

    vis [e.x][e.y][commandi [e.cur]] = true;

    q.push (e);



    while (!q.empty ()) {

        sc u = q.front ();

        q.pop ();

        if ((u.x == ed.x && u.y == ed.y) || '$' == a[u.x][u.y]) {

            ans = u.time;

            break;

        }



        sc v = u;   //to left

        left (v.cur);

        solve (v);



        v = u;      //to right

        right (v.cur);

        solve (v);



        v = u;      //wait

        solve (v);



        v = u;      //press

        int dr = commandi [v.cur];

        v.x += dir_x[dr];

        v.y += dir_y[dr];

        if (!check (v.x, v.y))  continue;

        if ('*' == a[v.x][v.y]) continue;

        solve (v);

    }



    if (-INF == ans) {

        cout << "YouBadbad" << endl;

    } else {

        cout << ans << endl;

    }

}



int main () {

    int i, j, t;



    cin >> t;

    while (t--) {

        while (!q.empty ()) q.pop ();

        cin >> n >> m >> p;

        for (i = 1; i <= n; ++i) {

            for (j = 1; j <= m; ++j) {

                cin >> a[i][j];

                if ('@' == a[i][j]) {

                    st.x = i;

                    st.y = j;

                } else if ('$' == a[i][j]) {

                    ed.x = i;

                    ed.y = j;

                }

            }

        }

        bfs ();

    }



    return 0;

}





/*

10 10 3

@.........

..........

..........

..........

..........

..........

..........

..........

..........

.........$



10 5 3

@....

.....

.....

.....

.....

.....

.....

.....

.....

....$

*/