题目链接 http://acm.hdu.edu.cn/showproblem.php?pid=1176

题意不解释了

简单的记忆化搜索可以拿来练练手，注意要从pos = 5 开始搜索

#include <iostream>

#include <cstring>

#include <algorithm>

#include <cstdio>

#include <cmath>

using namespace std;

const int M = 1e5 + 10;

int dp[15][M] , t2[M][12];

int dfs(int pos , int count) {

    if(count == 0) {

        if(pos == 6) {

            return 0;

        }

        else {

            return -1;

        }

    }

    if(dp[pos][count] != -1)

        return dp[pos][count];

    int sum = 0;

    if(pos == 1) {

        sum += max(dfs(pos + 1 , count - 1) , dfs(pos , count - 1));

    }

    else if(pos == 11) {

        sum += max(dfs(pos - 1 , count - 1) , dfs(pos , count - 1));

    }

    else {

        sum += max(max(dfs(pos - 1 , count - 1) , dfs(pos + 1 , count - 1)) , dfs(pos , count - 1));

    }

    if(sum != -1) {

        sum += t2[count][pos];

        dp[pos][count] = sum;

    }

    return sum;

}

int main() {

    int n;

    while(scanf("%d" , &n) != EOF) {

        if(n == 0)

            break;

        memset(t2 , 0 , sizeof(t2));

        int T = 0;

        for(int i = 0 ; i < n ; i++) {

            int x , y;

            scanf("%d%d" , &x , &y);

            t2[y][x + 1]++;

            T = max(T , y);

        }

        memset(dp , -1 , sizeof(dp));

        int MAX = 0;

        for(int i = 1 ; i <= 11 ; i++) {

            MAX = max(MAX , dfs(i , T));

        }

        printf("%d\n" , MAX);

    }

    return 0;

}