uestc 1855, dfs, 模拟

C.S.I.: P15

Time Limit: 1000 ms Memory Limit: 65535 kB Solved: 23 Tried: 249

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Description

 

You have been cast as the computer genius hero-of-the-day for the season finale of the show C.S.I.: P15 (coming this fall). Somewhat unsurprisingly,there is that camera feed that needs to be analyzed. The camera in question is recording pictures in HD-9000 quality with extra regression and the stream is then internally matched by a re-inverted isomorphic bit coefficient matrix, then plasma shifted five times for good measure. You then view the feed through Netscape Navigator 4 Platinum Edition. (Note that "internally" is just fancy talk for "inside the camera".)
Unfortunately, a saboteur turned on ASCII mode on the camera and set the camera in picture burst mode. So now all you have is a bunch of still ASCII images. And now,for reasons that will be revealed later in the show, you are to design and implement a deterministic algorithm for counting the number of flowers and birds in a given still image.
The pictures always include the ground,which will show up as a contiguous row of '=' characters. The ground will always be the bottom-most row of "ASCII pixels". There will never be anything else on that row (though, on one of the pictures taken before the sabotage there is a stray electron that a someone will accidentally nd by zooming in too far, but that is for a later episode).
Air is marked in the feed as a '.' (a dot). The ground is the last line of the feed, and it looks like this: '==========='. A flower is defined as any 8-connected component which consists of characters from the set { '|', '/', '\', '-', '@'}, and which is also connected to the ground. Two cells belong to the same 8-connected component if there is a path between them that goes through elements from the forementioned set only, such that each step in the path is to an 8-connected neighbour (horizontally, vertically, or diagonally adjacent cell). A bird is an occurence of '/\/\', such that all neighbouring cells are either air or edges of the image. So if you see something that looks like a bird on the ground, it is a flower (possibly an ex-parrot, but that is also a flower for our purposes).

 

Input

 

The first line of the input consists of a single integer T, the number of test cases. Each of the following T cases then begins with a line of two integers separated by a space,the height H and width W, and ends with H lines describing the picture. Each line of the picture has exactly W characters. All lines but the last consist of only the following characters: { '.', '|', '/', '\', '-', '@'}. The last line consists of '=' characters only.

 0 < T <= 100
 0 < W <= 30
 0 < H <= 30

 

Output

 

For each test case, output two lines. If the number of flowers is F and the number of birds is B, the output should read
Flowers: F
Birds: B

 

Sample Input

 

1
12 28
............................
............................
\@/.../\/\..../\/\..........
.|..........................
.|....\@/.........../\/\....
.|.....|.............|......
.|.....|.............|......
.|.....|..\@/....\@/.|......
.|.....|....\..../...|.|-|..
.|.....|.....\../....|.|.|..
.|.....|......\/.....|.|.|..
============================

 

Sample Output

 

Flowers: 5
Birds: 2

 

Source

 

IDI Open 2013 Programming Contest

模拟+深搜

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<cmath>
#include<vector>
#include<cstdlib>
#include<algorithm>

using namespace std;

#define LL long long
#define ULL unsigned long long
#define UINT unsigned int
#define MAX_INT 0x7fffffff
#define MAX_LL 0x7fffffffffffffff
#define MAX(X,Y) ((X) > (Y) ? (X) : (Y))
#define MIN(X,Y) ((X) < (Y) ? (X) : (Y))

#define MAXN 33
char g[MAXN][MAXN];

//const char ok[]={'|', '/', '\\', '-', '@'};

const int dx[]={-1, -1, -1, 0, 0, 1, 1, 1},
          dy[]={0, -1, 1, -1, 1, -1, 0, 1};

int w, h, birds, flo;
bool vis[MAXN][MAXN];

bool cango(int x, int y){
    if(x<0 || y<0 || x>h-2
       || y>w-1 || g[x][y]=='.'
       ||   vis[x][y]) return false;
    return true;
}

void dfs(int x, int y){
    vis[x][y]=true;
    for(int i=0; i<8; i++){
        int tx=x+dx[i], ty=y+dy[i];
        if(cango(tx, ty)) dfs(tx, ty);
    }
}

bool ok_lft(int x, int y){
    return y<0 || g[x][y]=='.';
}

bool ok_rit(int x, int y){
    return y>=w || g[x][y]=='.';
}

bool ok_up(int x, int y){
    if(x<0) return true;
    for(int k=0; k<6 && y<w && y>-1; k++, y--) if(g[x][y]!='.')
        return false;
    return true;
}

bool ok_down(int x, int y){
    for(int k=0; k<6 && y<w && y>-1; k++, y--) if(g[x][y]!='.')
        return false;
    return true;
}

bool isbird(int x, int y){
    if(ok_lft(x, y-4) && ok_rit(x, y+1)
       && ok_up(x-1, y+1) && ok_down(x+1, y+1)) return true;
    return false;
}

int main(){
//    freopen("C:\\Users\\Administrator\\Desktop\\in.txt","r",stdin);
    int T;
    scanf(" %d", &T);
    while(T--){
        int i, j;
        scanf(" %d %d", &h, &w);    getchar();
        for(i=0; i<h; i++) gets(g[i]);

//        for(i=0; i<h; i++) printf("%s\n",g[i]);

        memset(vis, 0, sizeof(vis));
        birds=flo=0;

        if(h==1){
            printf("Flowers: %d\nBirds: %d\n", flo, birds);
            continue;
        }
        for(i=0; i<w; i++) if(cango(h-2, i)){
            dfs(h-2, i);
            flo++;
//            cout<<cnt<<' '<<i<<endl;
        }

        string bid;
        bid+='/';   bid+='\\';  bid+='/';   bid+='\\';

        for(i=0; i<h-2; i++)
            for(j=0; j<w; j++) if(g[i][j]=='/' && w-j>=4 && !vis[i][j]){
                string t;
                for(int q=0; q<4 && !vis[i][j]; q++, j++) t+=g[i][j];
                if(t==bid && isbird(i, j-1)) birds++;
            }
        printf("Flowers: %d\nBirds: %d\n", flo, birds);
    }
    return 0;
}