POJ 1952 BUY LOW BUY LOWER 【DP】最长降序子序列及其计数问题
| Time Limit: 1000MS | Memory Limit: 30000K | |
| Total Submissions: 4981 | Accepted: 1663 |
Description
"Buy low; buy lower"
Each time you buy a stock, you must purchase it at a lower price than the previous time you bought it. The more times you buy at a lower price than before, the better! Your goal is to see how many times you can continue purchasing at ever lower prices.
You will be given the daily selling prices of a stock (positive 16-bit integers) over a period of time. You can choose to buy stock on any of the days. Each time you choose to buy, the price must be strictly lower than the previous time you bought stock. Write a program which identifies which days you should buy stock in order to maximize the number of times you buy.
Here is a list of stock prices:
Day 1 2 3 4 5 6 7 8 9 10 11 12
Price 68 69 54 64 68 64 70 67 78 62 98 87
The best investor (by this problem, anyway) can buy at most four times if each purchase is lower then the previous purchase. One four day sequence (there might be others) of acceptable buys is:
Day 2 5 6 10
Price 69 68 64 62
Input
* Lines 2..etc: A series of N space-separated integers, ten per line except the final line which might have fewer integers.
Output
* The length of the longest sequence of decreasing prices
* The number of sequences that have this length (guaranteed to fit in 31 bits)
In counting the number of solutions, two potential solutions are considered the same (and would only count as one solution) if they repeat the same string of decreasing prices, that is, if they "look the same" when the successive prices are compared. Thus, two different sequence of "buy" days could produce the same string of decreasing prices and be counted as only a single solution.
Sample Input
12 68 69 54 64 68 64 70 67 78 62 98 87
Sample Output
4 2
Source
/* (1)首先利用DP + 二分法求最大降序子序列，时间复杂度是nlogn得到以下数组： val[i]:用来存储输入元素 maxSeqLen[i]:表示以第i个元素结尾可以得到的最大降序子序列的长度; maxTailVal[i]:表示当前长度为i的所有最大降序子序列中尾元素的最大值 最后maxTailVal数组的长度maxTailLen即为最大降序子序列的长度 (2)然后利用maxSeqLen来计数，技术时需要考虑重复情况 利用countv[i]表示以第i个元素结尾可以可以得到的长度为maxSeqLen[i]的最大降序子序列的个数 利用maxLenSets[i](是一个List容器)表示maxSeqLen为i的元素的下标集合 加入一个哨兵元素n + 1,设maxSeqLen[n+1]为maxTailLen + 1, val[n + 1] = -1; 那么从1往n+1遍历，每次遍历采取以下策略来计算countv和维护maxLenSets ->维护maxLenSets:假设d = maxSeqLen[i],首先遍历容器maxLenSets[d]的所有元素如果存在一个 元素j使得val[j] = val[i],则删除j元素;最后插入i元素 ->计算countv[i]:遍历maxLenSets[d - 1]中的所有元素j,如果val[j] > val[i],则countv[i] += countv[j] 最后countv[n + 1]即为所求的计数结果,maxLenSets[i]中元素的唯一性确保了最后结果的唯一性. */ #include <iostream> #include <list> #define MAX_N 5005 using namespace std; int maxSeqLen[MAX_N + 1]; int maxTailVal[MAX_N + 1], maxTailLen; int val[MAX_N + 1], n; list<int> maxLenSets[MAX_N + 1]; int countv[MAX_N + 1]; int main() { int i; scanf("%d", &n); for(i = 1; i <= n; i++) { scanf("%d", &val[i]); //二分DP int l = 1, r = maxTailLen; while(l <= r) { int mid = (l + r) / 2; if(maxTailVal[mid] <= val[i]) r = mid - 1; else l = mid + 1; } maxTailVal[l] = val[i]; maxSeqLen[i] = l; if(l > maxTailLen) maxTailLen = l; } //第一个元素需要特殊处理 countv[1] = 1; maxLenSets[1].push_back(1); val[n + 1] = -1; maxSeqLen[n + 1] = maxTailLen + 1; //遍历后面的n个元素 for(i = 2; i <= n + 1; i++) { int d = maxSeqLen[i]; list<int>::iterator iter = maxLenSets[d].begin(); //扫描d长度List，代替相同的元素 for(; iter != maxLenSets[d].end(); ++iter) if(val[*iter] == val[i]) { maxLenSets[d].erase(iter); break; } maxLenSets[d].push_back(i); //同样d = 1的元素由于没有前驱，所以需要特殊处理 if(d == 1) { countv[i] = 1; continue; } //计算countv[i] iter = maxLenSets[d - 1].begin(); for(; iter != maxLenSets[d - 1].end(); ++iter) if(val[*iter] > val[i]) countv[i] += countv[*iter]; } printf("%d %d/n", maxTailLen, countv[n + 1]); return 0; }