Somebody I can kiss CSU

In a maze of r rows and c columns, your task is to collect as many coins as possible.
Each square is either your start point “S”(which will become empty after you leave), an empty square “.”, a coin square “C” (which will become empty after you step on this square and thus collecting the coin), a rock square “O” or an obstacle square “X”.
At each step, you can move one square to the up, down, left or right. You cannot leave the maze or enter an obstacle square, but you can push each rock at most once (i.e. You can treat a rock as an obstacle square after you push it).
To push a rock, you must stand next to it. You can only push the rock along the direction you’re facing, into an neighboring empty square (you can’t push it outside the maze, and you can’t push it to a squarecontiaining a coin).For example, if the rock is to your immediate right, you can only push it to its right neighboring square.
Find the maximal number of coins you can collect.
Input
The first line of input contains a single integer T (T<=25), the number of test cases.
Each test case begins with two integers r and c (2<=r,c<=10), then followed by r lines, each with c columns.
There will be at most 5 rocks and at most 10 coins in each maze.
Output
For each test case, print the maximal number of coins you can collect.

Sample Input
3
3 4
S.OC
..O.
.XCX
4 6
S.X.CC
..XOCC
…O.C
….XC
4 4
.SXC
OO.C
..XX
.CCC
Sample Output
1
6
3
题意：有一个n*m的矩阵，S是人开始的位置，O是石头，C是金币，X是墙。石头最多只能推一次，现在问你最多能拿到多少个金币
思路：我们不能直接用人去做BFS一条路走到死。肯定会TLE，因为在碰到石头之前的状态其实都是一样的，而你用BFS会多出很多很多状态。
所以我们先用BFS跑出石头没动时人能走到的所有的位置，并且把吃到的金币数记下来，对应的图从C改成.
然后用DFS去枚举石头移动的所有情况（向四个方向移动），最后在所有情况里面取吃到的金币数最大的那个就可以了。注意要开三维数组保存上一层的图和标记数组。

#include<map>#include<cmath>#include<queue>#include<stack>#include<string>#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>#define inf 0x3f3f3f3f#define ll long longusing namespace std;#define pi 3.14159265358979323846264338327950288419716939937510582097494459230781640628620899char mapp[6][15][15];int v[6][15][15];int dir[4][2]= {{1,0},{0,1},{-1,0},{0,-1}};int n,m,cnt;int x2,y2;int go(int x,int y,int k){    if(0<=x&&x<n&&0<=y&&y<m&&mapp[k][x][y]!='X'&&mapp[k][x][y]!='O')        return 1;    return 0;}struct node{    int x,y;} stone[6];int ma,last_ma,num,flag[6];void  bfs(node a,int k){    node st,ed;    queue<node>q;    q.push(a);    v[k][a.x][a.y]=1;    while(!q.empty())    {        st=q.front();        q.pop();        for(int i=0; i<4; i++)        {            ed.x=st.x+dir[i][0];            ed.y=st.y+dir[i][1];            if(!go(ed.x,ed.y,k)||v[k][ed.x][ed.y]) continue;            if(mapp[k][ed.x][ed.y]=='C')            {                ma++;                last_ma=max(last_ma,ma);                mapp[k][ed.x][ed.y]='.';            }            v[k][ed.x][ed.y]=1;            q.push(ed);        }    }}void dfs(int k){    for(int i=0; i<cnt; i++)    {        if(flag[i]) continue;        for(int j=0; j<4; j++)        {            int tx=stone[i].x+dir[j][0];            int ty=stone[i].y+dir[j][1];            if(tx<0||tx>=n||ty<0||ty>=m||mapp[k-1][tx][ty]!='.') continue;            int ttx=stone[i].x-dir[j][0];            int tty=stone[i].y-dir[j][1];            if(ttx<0||ttx>=n||tty<0||tty>=m||!v[k-1][ttx][tty]) continue;            memcpy(v[k],v[k-1],sizeof(v[k]));            memcpy(mapp[k],mapp[k-1],sizeof(mapp[k]));            mapp[k][tx][ty]='X',mapp[k][stone[i].x][stone[i].y]='.';            flag[i]=1;            bfs(stone[i],k);            if(last_ma==num) return;            dfs(k+1);            if(last_ma==num) return;            for(int i=0; i<n; i++)                for(int j=0; j<m; j++)                    if(mapp[k-1][i][j]=='C'&&mapp[k][i][j]=='.')                        ma--;            flag[i]=0;        }    }}int main(){    int t;    node a;    scanf("%d",&t);    while(t--)    {        ma=last_ma=num=0;        cnt=0;        scanf("%d%d",&n,&m);        for(int i=0; i<n; i++)        {            scanf("%s",mapp[0][i]);            for(int j=0; j<m; j++)            {                if(mapp[0][i][j]=='S')                {                    a.x=i,a.y=j;                    mapp[0][i][j]='.';                }                else if(mapp[0][i][j]=='O')                    stone[cnt].x=i,stone[cnt++].y=j;                else if(mapp[0][i][j]=='C')                    num++;            }        }        memset(v[0],0,sizeof(v[0]));        memset(flag,0,sizeof(flag));        bfs(a,0);        dfs(1);        printf("%d\n",last_ma);    }}