[四分树] Spatial Structures (ACM/ICPC World Finals 1998, UVa806)

题目描述

图像常用一棵四分树表示。黑白图像有两种表示法：点阵表示和路径表示。你的任务是在这两种表示方法中转换。

题解

题不难，理解题意按照要求做就行。再注意一些细节就行。

AC代码

#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>
using namespace std;

const int dx[4]={0,0,1,1};
const int dy[4]={0,1,0,1};
const int powf[7]={1,5,25,125,625,3125,15625};
int N;
char  mp [65][65];
vector<int> ans,num;


bool check(int x, int y, int len){
    for(int i = x; i<x+len; i++)
        for(int j = y; j<y+len; j++)
            if(mp[i][j] == '0')return false;
    return true;
}

bool checkw(int x, int y, int len){
    for(int i = x; i<x+len; i++)
        for(int j = y; j<y+len; j++)
            if(mp[i][j] == '1')return false;
    return true;
}

void dfs(int x, int y, int len, int pt, int ly){
    for(int i = 0; i<4; ++i){
        int blen = len/2;
        int bx = x+dx[i]*blen, by = y+dy[i]*blen;
        int bpt = (i+1)*powf[ly]+pt;
        if(check(bx, by, blen)) {ans.push_back(bpt);continue;}
        if(checkw(bx,by, blen)) continue;
        dfs(bx, by, blen, bpt, ly+1);
    }
}

void pton(){
    if(checkw(0, 0, N)) return;
    if(check(0,0,N)){ans.push_back(0);return;}
    dfs(0,0, N, 0, 0);
}

void draw(int x, int y, int len){
    for(int i = x; i<x+len; i++)
        for(int j = y; j<y+len; j++)
            mp[i][j] = '*';
}

void trans(int x){
    while(x){
        ans.push_back(x%5);
        x/=5;
    }
}

void ntop(){
    if(!num.size())return;
    if(num[0] == 0){draw(0, 0, N); return;}
    for(int i = 0, k = num.size(); i<k; ++i){
        ans.clear();
        trans(num[i]);
        int x = 0, y = 0, len = N;
        for(int j=0, kj=ans.size(); j<kj; ++j){
            len/=2;
            x = x+dx[ans[j]-1]*len;
            y = y+dy[ans[j]-1]*len;
        }
        draw(x, y, len);
    }
}

int main(){
    freopen("in.txt", "r", stdin);
    freopen("out.txt", "w", stdout);
    int kase = 0;
    while(scanf("%d", &N)!=EOF && N){
        if(kase) printf("\n");
        if(N>0){
            memset(mp, 0, sizeof(mp));
            ans.clear();

            for(int i = 0; i<N; ++i)
                scanf("%s", mp[i]);



            pton();

            sort(ans.begin(), ans.end());

            printf("Image %d", ++kase);
            if(ans.size()){
            for(int i = 0 , k = ans.size(); i<k; i++){
                if(i%12 == 0)printf("\n%d", ans[i]);
                else  printf(" %d", ans[i]);
            }
            printf("\n");
            }
            else printf("\n");
            printf("Total number of black nodes = %d\n", ans.size());
        }
        else{
            N = -N;
            memset(mp, '.', sizeof(mp));
            num.clear();

            int buf;
            while(scanf("%d", &buf) && buf != -1)
                num.push_back(buf);

            ntop();

            printf("Image %d\n", ++kase);
            for(int i = 0; i<N; i++){
                for(int j = 0; j<N; j++)
                    printf("%c", mp[i][j]);
                printf("\n");
            }
        }
    }
    return 0;

}