zoj 1788 Quad Trees

四分树的题目

利用dfs建立一棵四分树。

然后bfs遍历四分树。

最后把遍历结果转换为十六进制输出。

#include <iostream>#include <cstdio>#include <algorithm>#include <set>#include <list>#include <queue>#include <stack>#include <vector>#include <cstring>#include <cmath>#include <map>#include <iomanip>#define pb push_back#define mp make_pair#define fi first#define se second#define all(a) (a).begin(),(a).end()#define FOR(i,a,b) for (int i=(a);i<(b);i++)#define FORD(i,a,b) for (int i=(a); i>=(b); i--)#define REP(i,b) FOR(i,0,b)#define sf scanf#define pf printfusing namespace std;#define N 550struct quad_tree{    char value[3];    quad_tree* child[4];};int store[N][N];bool cmp(quad_tree* lhs,quad_tree *rhs){    if (lhs->value[0]=='1'||rhs->value[0]=='1') return false;    FOR(i,0,2)        if(lhs->value[i]!=rhs->value[i])            return false;    return true;}quad_tree* dfs(int r,int c,int s){    quad_tree* current = new quad_tree();    if (s==1)    {        current->value[0] = '0';        current->value[1] = '0'+store[r][c];        current->value[2] = '\0';        return current;    }    s /=2;    current->child[0] = dfs(r,c,s);    current->child[1] = dfs(r,c+s,s);    current->child[2] = dfs(r+s,c,s);    current->child[3] = dfs(r+s,c+s,s);    bool f = true;    FOR(i,1,4)    {        if (!cmp(current->child[0],current->child[i]))        {            f =false;            break;        }    }    if (f) {        current->value[0] = '0';        current->value[1] = current->child[0]->value[1];        current->value[2] = '\0';        FOR(i,0,4)        {            delete current->child[i];            current->child[i]=0;        }    }    else    {        current->value[0] = '1';        current->value[1] = '\0';    }    return current;}queue<quad_tree*> bfslist;char result[N*N];int ct;char hexout[N*N];void bfs(quad_tree* root){    bfslist.push(root);    quad_tree *tmp;    while(!bfslist.empty())    {        tmp = bfslist.front();        bfslist.pop();        if (tmp->value[0]=='1')        {            result[ct++] = '1';            FOR(i,0,4)            {                bfslist.push(tmp->child[i]);            }        }        else        {            result[ct++] = '0';            result[ct++] = tmp->value[1];        }    }}int main(){    int n,m;    char hex []= {'0','1','2','3','4','5','6','7','8','9','A','B','C','D','E','F'};    sf("%d",&n);    while(n--)    {        sf("%d",&m);        FOR(i,0,m)            FOR(j,0,m)            {                sf("%d",&store[i][j]);            }        quad_tree *root;        root=dfs(0,0,m);        while(!bfslist.empty()) bfslist.pop();        ct = 0;        bfs(root);        //四位二进制数为一个十六进制，记录能否组成一个十六进制        int chex = 0;        //记录二进制转十进制累计值        int dec = 0;        //记录十六进制输出的数组规模        int ch = 0;        for(int i=ct-1;i>=0;i--)        {            chex++;            dec = dec+ ((result[i]-'0') <<(chex-1));            if (chex==4)            {                hexout[ch++] = hex[dec];                dec = 0 ;                chex = 0 ;            }        }        if (chex>0)        {            hexout[ch++] = hex[dec];        }        for(int i=ch-1;i>=0;i--)        {            pf("%c",hexout[i]);        }        pf("\n");    }    return 0;}