【DP】hdu——4105

f[i][j][2]表示前i個數以j個數結尾，0是低穀，1是高峰

轉移方程：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 105int 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;}