CodeForces - 89C - Chip Play - （DFS，十字链表）

题目链接：http://codeforces.com/problemset/problem/89/C

C. Chip Play

time limit per test

4 seconds

The first line contains two integers n and m (1 ≤ n, m, n × m ≤ 5000). Then follow n lines containing m characters each — that is the game field description. "." means that this square is empty. "L", "R", "U", "D" mean that this square contains a chip and an arrow on it says left, right, up or down correspondingly.

It is guaranteed that a field has at least one chip.

Output

Print two numbers — the maximal number of points a player can get after a move and the number of moves that allow receiving this maximum number of points.

Examples

input

Copy

4 4
DRLD
U.UL
.UUR
RDDL

output

Copy

10 1

input

Copy

3 5
.D...
RRRLL
.U...

output

Copy

6 2

Note

In the first sample the maximum number of points is earned by the chip in the position (3, 3). You can see its progress at the following picture:

All other chips earn fewer points.

题意：给出n*m的矩阵，其中有的方格内是箭头（上、下、左、右四种），可以从一个箭头走到它所指方向最近的箭头（一个箭头只可以到一次），对于一个点(x,y)开始的答案是从(x,y)按箭头方向一直走下去所经过的箭头数量，求从某点开始的最大答案。

解析：直接dfs会超时，这里用十字链表存储每个箭头上下左右最近的箭头位置，基于此来dfs就可以

代码：

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll M = 5005;

char maze[M][M];
int n,m;
//L[i]是位于i处的箭头左边第一个箭头的位置
//R[i]是位于i处的箭头右边第一个箭头的位置
//U[i]是位于i处的箭头上边第一个箭头的位置
//D[i]是位于i处的箭头下边第一个箭头的位置
int U[M],D[M],L[M],R[M];

void getUDLR()//初始化完成之后，每个箭头存储了上下左右最近的箭头位置
{
    for(int i=0;i<n;i++)
    {
        int l=-1;
        for(int j=0;j<m;j++)
        {
            if(maze[i][j]!='.')
            {
                int index=i*m+j;
                L[index]=l;
                if(l!=-1) R[l]=index;
                l=index;
            }
        }
        R[l]=-1;
    }
    for(int j=0;j<m;j++)
    {
        int u=-1;
        for(int i=0;i<n;i++)
        {
            if(maze[i][j]!='.')
            {
                int index=i*m+j;
                U[index]=u;
                if(u!=-1) D[u]=index;
                u=index;
            }
        }
        D[u]=-1;
    }
}

void Delete(int x,int y)
{
    int index=x*m+y;
    //index上面的箭头为U[index],删除index之后,“U[index]下面的箭头”不再是“index”而是“index下面的箭头D[index]”
    if(U[index]!=-1) D[U[index]]=D[index];
    if(D[index]!=-1) U[D[index]]=U[index];
    if(L[index]!=-1) R[L[index]]=R[index];
    if(R[index]!=-1) L[R[index]]=L[index];
}

int dfs(int x,int y)
{
    Delete(x,y);//本次dfs中将用过的箭头删除
    int index=x*m+y;
    switch(maze[x][y])
    {
    case 'U':
        if(U[index]==-1) return 1;
        else return 1+dfs(U[index]/m,y);
        break;
    case 'D':
        if(D[index]==-1) return 1;
        else return 1+dfs(D[index]/m,y);
        break;
    case 'L':
        if(L[index]==-1) return 1;
        else return 1+dfs(x,L[index]%m);
        break;
    case 'R':
        if(R[index]==-1) return 1;
        else return 1+dfs(x,R[index]%m);
        break;
    }
}


int main()
{
    scanf("%d%d",&n,&m);
    for(int i=0;i<n;i++)
        scanf("%s",maze[i]);

    int maxV=-1,num=0;
    for(int i=0;i<n;i++)
    {
        for(int j=0;j<m;j++)
        {
            if(maze[i][j]!='.')
            {
                getUDLR();
                int res=dfs(i,j);
                if(res>maxV) { maxV=res; num=1;}
                else if(res==maxV){ num++;}
            }
        }
    }
    printf("%d %d\n",maxV,num);
    return 0;
}

参考：https://www.cnblogs.com/debugcool/archive/2011/06/21/CF74.html