【DP】hdu——4105

最新推荐文章于 2024-07-30 10:30:00 发布

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本文介绍了一种使用动态规划解决序列中峰值问题的方法，包括转移方程、边界条件和算法实现细节。

f[i][j][2]表示前i个数以j个数结尾，0是低谷，1是高峰

转移方程：f[i][j][0] = max(f[i][j][0], f[i-j][k][1] + 1);

f[i][j][1] = max(f[i][j][1], f[i-j][k][0] + 1);

#include <map>
#include <set>
#include <list>
#include <queue>
#include <deque>
#include <stack>
#include <string>
#include <time.h>
#include <cstdio>
#include <math.h>
#include <iomanip>
#include <cstdlib>
#include <limits.h>
#include <string.h>
#include <iostream>
#include <fstream>
#include <algorithm>
using namespace std;

#define LL long long
#define MIN INT_MIN
#define MAX INT_MAX
#define PI acos(-1.0)
#define FRE freopen("input.txt","r",stdin)
#define FF freopen("output.txt","w",stdout)
#define N 105
int max(int x,int y){return x > y ? x : y;}
char str[N];
int a[N];
int n;
int f[N][N][2];
int chk (int l1,int r1,int l2,int r2) {
    while (l1 <= r1 && a[l1] == 0) l1++;
    while (l2 <= r2 && a[l2] == 0) l2++;
    int l = r1 - l1 + 1;
    int r = r2 - l2 + 1;
    if (l < r) return -1;
    if (l > r) return 1;
    for (int i = l1 ,j = l2; i <= r1; i++, j++) {
        if (a[i] > a[j]) return 1;
        if (a[i] < a[j]) return -1;
    }
    return 0;
}
int main () {
    while (scanf("%d%s",&n,str) != EOF) {
        int i,j,k;
        for (i = 0; i < n ; i++) a[i+1] = str[i] - '0';
        memset(f,-1,sizeof(f));
        f[1][1][0] = 0;
        for (i = 2 ; i <= n ; i++) {
            f[i][i][0] = 1;
            for (j = 1 ; j < i; j++) {
                for (k = 1; k <= i - j; k++) {
                    if (chk(i-j-k+1, i-j, i-j+1, i) > 0) {
                        if (f[i-j][k][1] == -1) continue;
                        f[i][j][0] = max(f[i][j][0], f[i-j][k][1] + 1);
                    }
                    if (chk(i-j-k+1, i-j, i-j+1, i) < 0) {
                        if (f[i-j][k][0] == -1) continue;
                        f[i][j][1] = max(f[i][j][1], f[i-j][k][0] + 1);
                    }
                }
            }
        }
        int ans = 0;
        for (i = 1; i <= n; i++) {
            //cout<<f[n][i][0]<<" "<<f[n][i][1]<<endl;
            ans = max(ans, f[n][i][0]);
            ans = max(ans, f[n][i][1]);
        }
        printf("%d\n",ans);
    }
    return 0;
}