uva1587 - Box
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题目链接:
https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=4462
Sample Input
1345 2584
2584 683
2584 1345
683 1345
683 1345
2584 683
1234 4567
1234 4567
4567 4321
4322 4567
4321 1234
4321 1234
Sample Output
POSSIBLE
IMPOSSIBLE
题意:给定6个矩形的长和宽wi和hi(1<=wi,hi<=1000),判断能否构成长方体的六个面。
思路:找规律,排序判断。最开始结构体数组越界在本地未报错,提交了几次才注意到。
#include<cstdio>#include<algorithm>using namespace std;struct rectangular{ int w,h;}rectangular[6];bool cmp(struct rectangular a,struct rectangular b){ if(a.w!=b.w) return a.w<b.w; else return a.h<b.h;}/*void print(){ for(int i=0;i<6;i++){ printf("%d %d\n",rectangular[i].w,rectangular[i].h); }}*/bool is_box(){ for(int i=1;i<6;i+=2){ if((rectangular[i].h!=rectangular[i-1].h)||(rectangular[i].w!=rectangular[i-1].w)) return false; } if((rectangular[1].w!=rectangular[3].w)||(rectangular[3].h!=rectangular[5].h)||(rectangular[5].w!=rectangular[1].h)) return false; return true;}int main(){ //freopen("in.txt","r",stdin); //freopen("output.txt","w",stdout); while(scanf("%d%d",&rectangular[0].w,&rectangular[0].h)==2){ for(int i=1;i<6;i++){ scanf("%d%d",&rectangular[i].w,&rectangular[i].h); } for(int i=0;i<6;i++){ if(rectangular[i].w>rectangular[i].h){ swap(rectangular[i].w,rectangular[i].h); }//wi里存放较小的值 } sort(rectangular,rectangular+6,cmp); //print();//打印出排序后的情况,可以看到规律 if(is_box()){ printf("POSSIBLE\n"); } else{ printf("IMPOSSIBLE\n"); } } return 0;}
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