POJ 1952 BUY LOW, BUY LOWER DP记录数据
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最长递减子序列,加记录有多少个最长递减子序列,然后需要去重。
最麻烦的就是去重了。
基本的思路就是:全面出现重复的值,然后还是相同长度的子序列,这里的DP记录的子序列是以当前值为结尾的时候,并且一定选择这个值的最长递减子序列, 那么就需要减去前面已经出现过了的子序列。
有点绕口。
举例就是9 8 9 8 2 和 10 5 12 5 3;这些例子去重。
本类型的题目如果不用记录数据是可以使用O(nlgn)的算法的,不过暂时不知道如何记录数据。故此这里只使用DP了。
#include <stdio.h>#include <vector>#include <string.h>#include <algorithm>#include <iostream>#include <string>#include <limits.h>#include <stack>#include <queue>#include <set>#include <map>using namespace std;const int MAX_N = 5001;int arr[MAX_N], N, tbl[MAX_N], C[MAX_N];void getLongest(int &len, int &n){memset(tbl, 0, sizeof(int) * (N+1));memset(C, 0, sizeof(int) * (N+1));tbl[0] = 1; C[0] = 1;for (int i = 1; i < N; i++){tbl[i] = 1;for (int j = 0; j < i; j++){if (tbl[j] == -1) continue;if (arr[j] > arr[i] && tbl[i] < tbl[j]+1){tbl[i] = tbl[j]+1;}}for (int j = 0; j < i; j++){if (arr[j] > arr[i] && tbl[i] == tbl[j]+1){C[i] += C[j];}}if (C[i] == 0) C[i] = 1;//递增的时候/*可以不用下面这段代码for (int j = 0; j < i; j++){if (arr[i] == arr[j] && tbl[i] == tbl[j] && C[i] == C[j]){tbl[i] = -1;break;}//去掉相同的数据 9 8 9 8}if (tbl[i] == -1) continue;*/for (int j = 0; j < i; j++){if (arr[j] == arr[i] && tbl[j] == tbl[i]) C[i] -= C[j];}//特例:6 5 7 5 3 需要去掉前后5重复的地方}len = INT_MIN;for (int i = 0; i < N; i++){len = max(len, tbl[i]);}n = 0;for (int i = 0; i < N; i++){if (tbl[i] == len) n += C[i];}}int main(){while (scanf("%d", &N) != EOF){for (int i = 0; i < N; i++){scanf("%d", arr + i);}int len, n;getLongest(len, n);printf("%d %d\n", len, n);}return 0;}
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