2018 ACM School Competition J Willful

Problem J: Willful

Time Limit: 2 Sec Memory Limit: 128 MB

Topic description

The willful boss's favorite game is Fighting the Landlord, but today he wants to try a tank battle. In the game, he can use his tank to launch a laser to attack the enemy, and the laser can be reflected by a mirror.

The wayward boss now wants to calculate how many enemies the laser can kill in total.

Note: The laser passes through the enemy after killing the enemy; the laser passes directly when it encounters its own tank; the laser can choose to fire horizontally or vertically.

enter

Multiple sets of input and output

Enter two integers n, m (1 ≤ n, m ≤ 1000) in the first line, which represent the number of rows and columns of the map, respectively, and the following n lines describe each line of the map.

. Indicates open space

* Represents an obstacle (the light cannot pass through, and the propagation ends when the light hits the obstacle)

T The initial position of the wayward boss's tank

E for enemy

/ Indicates mirrors placed from bottom left to top right

\ Indicates mirrors placed from top left to bottom right

output

An integer representing the maximum number of enemies that can be killed

sample input

5 5
.*/E\
E*.*.
E*TEE
\.../
. * \ EE

Sample output

4 

Analysis: Enumerate the four directions of up, down, left, and right, and take the one that can kill the most enemies. Use vis[i][j][k] to indicate whether to pass the point (i, j) in the k direction to avoid falling into an infinite loop. 
The XOR operation can be used to realize the replacement of numbers within a certain range, reducing the judgment of if.

#include<bits/stdc++.h>
using namespace std;
char a[1100][1100];
int n,m;
const int dx[]= {-1,0,1,0};
const int dy[]= {0,1,0,-1};
int vis[1100][1100][4];
bool ok(int x,int y)
{
    for(int i=0; i<4; ++i)
        if(vis[x][y][i])return false;
    return true;
}
intmain ()
{
    //freopen("i.txt","r",stdin);
    while(scanf("%d%d",&m,&n)==2)
    {
        int xx,yy;
        for(int i=0; i<m; ++i)
            for(int j=0; j<n; ++j)
            {
                scanf(" %c",&a[i][j]);
                if(a[i][j]=='T')
                    xx=i,yy=j;
            }
        int maxn=0;
        for(int i=0; i<4; ++i)
        {
            memset(vis,0,sizeof vis);
            int p=i;
            int x=xx,y=yy;
            int cnt=0;
            while(x>=0&&x<m&&y>=0&&y<n&&!vis[x][y][p]&&a[x][y]!='*')
            {
                if(a[x][y]=='E'&&ok(x,y))
                        ++cnt;
                else if(a[x][y]=='/')
                    p^=1;
                else if(a[x][y]=='\\')
                    p^=3;
                force[x][y][p ] = 1
                x+=dx[p],y+=dy[p];
            }
            maxn=max(maxn,cnt);
        }
        printf("%d\n",maxn);
    }
}