http://poj.org/problem?id=1260 Pearls Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 8474   Accepted: 4236 Description In Pearlania everybody is fond of pearls. One company, called The Royal Pearl, produces a lot of jewelry with pearls…

http://poj.org/problem?id=2192 Zipper Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 17585   Accepted: 6253 Description Given three strings, you are to determine whether the third string can be formed by combining the characters in the…

Pearls Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 7947   Accepted: 3949 Description In Pearlania everybody is fond of pearls. One company, called The Royal Pearl, produces a lot of jewelry with pearls in it. The Royal Pearl has its…

Tickets Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 7925    Accepted Submission(s): 4032 Problem Description Jesus, what a great movie! Thousands of people are rushing to the cinema. Howeve…

[POJ 3071] Football(DP) Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 4350   Accepted: 2222 Description Consider a single-elimination football tournament involving 2n teams, denoted 1, 2, -, 2n. In each round of the tournament, all tea…

Description A plotter is a vector graphics printing device that connects to a computer to print graphical plots. There are two types of plotters: pen plotters and electrostatic plotters. Pen plotters print by moving a pen across the surface of a piec…

http://codeforces.com/gym/101341/problem/K 题意:给出n个区间,每个区间有一个l, r, w,代表区间左端点右端点和区间的权值,现在可以选取一些区间,要求选择的区间不相交,问最大的权和可以是多少,如果权和相同,则选区间长度最短的.要要求输出区间个数和选了哪些区间. 思路:把区间按照右端点排序后,就可以维护从左往右,到p[i].r这个点的时候,已经选择的最大权和是多少,最小长度是多少,区间个数是多少. 因为可以二分找右端点小于等于当前区间的左端点的某个区间…

http://acm.hdu.edu.cn/showproblem.php?pid=5791 Two Problem Description   Alice gets two sequences A and B. A easy problem comes. How many pair of sequence A' and sequence B' are same. For example, {1,2} and {1,2} are same. {1,2,4} and {1,4,2} are not…

题意:有两棵树,标号为1和2,在Tmin内,每分钟都会有一个苹果从其中一棵树上落下,问最多移动M次的情况下(该人可瞬间移动),最多能吃到多少苹果.假设该人一开始在标号为1的树下. 分析: 1.dp[x][y][z]---第x分钟移动了y次的情况下,现在位于标号为z的树下最多吃到的苹果数. 2.枚举所有的时间.次数和可能位于的树下. 若第i分钟该人位于标号为k的树下,第i + 1分钟苹果从标号为a[i+1]的树上落下. (1)k==a[i], 此人站在标号为k的树下不移动,即可再吃到一个苹果:dp…

题目链接 题意 : 中文题不详述. 思路 : 黑书上116页讲的很详细.不过你需要在之前预处理一下面积,那样的话之后列式子比较方便一些. 先把均方差那个公式变形, 另X表示x的平均值,两边平方得 平均值是一定的,所以只要让每个矩形的总分的平方和尽量小即可.左上角坐标为(x1,y1)右下角坐标为(x2,y2)的棋盘,设总和为s[][][][],切割k次以后得到k+1块矩形的总分平方和是d[k][][][][],则可以沿着横线切也可以沿着竖线切,然后选一块接着切,递归下去,状态转移方程 d[k,x1…