文章分类 - DFS
摘要:ZipperTime Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 5593Accepted Submission(s): 2039Problem DescriptionGiven three strings, you are to determine whether the third string can be formed by combining the characters in the first two strings. The firs
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摘要:B-numberTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1654 Accepted Submission(s): 902Problem DescriptionA wqb-number, or B-number for short, is a non-negative integer whose decimal form contains the sub- string "13" and can be divided
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摘要:Image PerimetersTime Limit: 1000MS Memory Limit: 10000KTotal Submissions: 7194 Accepted: 4299DescriptionTechnicians in a pathology lab analyze digitized images of slides. Objects on a slide are selected for analysis by a mouse click on the object. The perimeter of the boundary of an object is one u.
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摘要:滑雪Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 68987 Accepted: 25450DescriptionMichael喜欢滑雪百这并不奇怪, 因为滑雪的确很刺激。可是为了获得速度,滑的区域必须向下倾斜,而且当你滑到坡底,你不得不再次走上坡或者等待升降机来载你。Michael想知道载一个区域中最长底滑坡。区域由一个二维数组给出。数组的每个数字代表点的高度。下面是一个例子 1 2 3 4 516 17 18 19 615 24 25 20 714 23 22 21 813 12 11 10 9一个人可以从某个...
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摘要://搜索练习题,深搜广搜也行//437MS 240K 1511 B G++#include#include#include#includechar map[10][10];int mov[4][2]={0,1,1,0,0,-1,-1,0};int n,m,t;int flag;int xb,yb,xe,ye;void dfs(int x,int y,int k){ if(flag) return; if(x==xe && y==ye && k==t){ flag=1;return; } int temp=(t-k)-abs(x-xe)-a...
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