poj 1069 The Bermuda Triangle

poj 1069 The Bermuda Triangle,第1张

poj 1069 The Bermuda Triangle
#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;int hexagon[110][110];int ori_size,n,cut_size[11];bool can_cut(int x,int y,int size){    int i,j;    if (y+2*size-2>ori_size*4 || x+size-1>ori_size*2)        return false;    if (y%2==1)    {        for (i=0;i<size;i++) for (j=0;j<2*i+1;j++)     if (hexagon[x+i][y+j]==false)         return false;    }    else    {        for (i=0;i<size;i++) for (j=2*i;j<2*size-1;j++)     if (hexagon[x+i][y+j]==false)         return false;    }    return true;}void cut(int x,int y,int size){    int i,j;    if (y%2==1)        for (i=0;i<size;i++) for (j=0;j<2*i+1;j++)     hexagon[x+i][y+j]=false;    else        for (i=0;i<size;i++) for(j=2*i;j<2*size-1;j++)     hexagon[x+i][y+j]=false;}void decut(int x,int y,int size){    int i,j;    if (y%2==1)        for (i=0;i<size;i++) for (j=0;j<2*i+1;j++)     hexagon[x+i][y+j]=true;    else        for (i=0;i<size;i++) for(j=2*i;j<2*size-1;j++)     hexagon[x+i][y+j]=true;}bool dfs(int x,int y){    int i,j;    if (x>ori_size*2)        return true;    if (y>4*ori_size)        return dfs(x+1,1);    if (hexagon[x][y]==false)    {        for (j=y+1;j<=4*ori_size;j++) if (hexagon[x][j])     break;        return dfs(x,j);    }    for (i=0;i<n;i++)    {        if (can_cut(x,y,cut_size[i]))        { cut(x,y,cut_size[i]); if (dfs(x,y+1))     return true; decut(x,y,cut_size[i]);        }        else break;    }    return false;}int main(int argc, char** argv){    int test_cases,i,j,k;    bool flag;    scanf("%d",&test_cases);    while (test_cases--)    {        scanf("%d%d",&ori_size,&n);        for (i=0,flag=false;i<n;i++)        { scanf("%d",&cut_size[i]); if (ori_size%cut_size[i]==0)     flag=true;        }        if (flag)        { printf("YESn"); continue;        }        memset(hexagon,false,sizeof(hexagon));        for (i=1;i<=ori_size;i++) for (j=1;j<=ori_size*2-1+2*i;j++)     hexagon[i][j]=true;        for (i=ori_size+1;i<=ori_size*2;i++) for (j=(i-ori_size)*2;j<=ori_size*4;j++)     hexagon[i][j]=true;        sort(cut_size,cut_size+n);        for (i=0;i<n;i++) if (cut_size[i]>ori_size) {     n=i-1;     break; }        for (i=0;i<n;i++) for (j=0;j<i;j++)     if (cut_size[i]%cut_size[j]==0)     {         for (k=i;k<n-1;k++)  cut_size[k]=cut_size[k+1];         i--;         n--;         break;     }        if (dfs(1,1)) printf("YESn");        else printf("NOn");    }    return 0;}

欢迎分享,转载请注明来源:内存溢出

原文地址: https://outofmemory.cn/zaji/4896561.html

(0)
打赏 微信扫一扫 微信扫一扫 支付宝扫一扫 支付宝扫一扫
上一篇 2022-11-12
下一篇 2022-11-12

发表评论

登录后才能评论

评论列表(0条)

保存