【HDU5900】【区间DP】QSC and master 题解

最新推荐文章于 2020-04-28 22:01:05 发布

原创最新推荐文章于 2020-04-28 22:01:05 发布 · 534 阅读

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#HDU #DP #区间DP

本文通过一则东北大学的神秘传说引入了一个有趣的算法问题。故事中的主人公QSCI需要解决一道关于从数对中获取最大分数的问题。文章详细阐述了问题的输入输出格式，并提供了一段AC代码作为解答思路的实现，涉及区间DP算法的应用。

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QSC and master

Every school has some legends, Northeastern University is the same.

Enter from the north gate of Northeastern University，You are facing the main building of Northeastern University.Ninety-nine percent of the students have not been there，It is said that there is a monster in it.

QSCI am a curious NEU_ACMer，This is the story he told us.

It’s a certain period，QSCI am in a dark night, secretly sneaked into the East Building，hope to see the master.After a serious search，He finally saw the little master in a dark corner. The master said：

“You and I, we’re interfacing.please solve my little puzzle！

There are N pairs of numbers，Each pair consists of a key and a value，Now you need to move out some of the pairs to get the score.You can move out two continuous pairs，if and only if their keys are non coprime(their gcd is not one).The final score you get is the sum of all pair’s value which be moved out. May I ask how many points you can get the most？

The answer you give is directly related to your final exam results~The young man~”

QSC is very sad when he told the story，He failed his linear algebra that year because he didn’t work out the puzzle.

Could you solve this puzzle?

（Data range：1<=N<=300
1<=Ai.key<=1,000,000,000
0<=Ai.value<=1,000,000,000）

Input

First line contains a integer T，means there are T(1≤T≤10) test case。

Each test case start with one integer N . Next line contains N integers，means Ai.key.Next line contains N integers，means Ai.value.

Output

For each test case,output the max score you could get in a line.

Sample Input

1 2 3

1 1 1

1 2 4

1 1 1

1 3 4 3

1 1 1 1

Sample Output

区间DP
记录i到j的可以取得的最大值。

附AC代码

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <string>
#include <iomanip>
#include <ctime>
#include <climits>
#include <cctype>
#include <algorithm>
#define clr(x) memset(x,0,sizeof(x))
#define LL long long
#ifdef WIN32
#define AUTO "%I64d"
#else
#define AUTO "%lld"
#endif

using namespace std; 

const int maxn = 305;
int n,m,t;
LL sum[maxn],dp[maxn][maxn];
int f[maxn][maxn],num[maxn],val[maxn];

void DP() {
    for(int i = 1; i < n; i++)
        if(__gcd(num[i], num[i+1]) > 1) f[i][i+1] = 1;
    for(int l = 3; l <= n; l++)
        for(int i = 1; i+l-1 <= n; i++) {
            int r = i+l-1;
            f[i][r] = (f[i+1][r-1] && __gcd(num[i], num[r]) != 1);
            for(int k = i; k < r; k++) f[i][r] += (f[i][k] && f[k+1][r]);
        }
}

void work() {
    for(int l = 2; l <= n; l++)
        for(int i = 1; i+l-1 <= n; i++) {
            int r = i+l-1;
            if(f[i][r] == 1) {
                dp[i][r] = sum[r]-sum[i-1];
                continue;
            }
            for(int k = i; k < r; ++k) dp[i][r] = max(dp[i][r], dp[i][k]+dp[k+1][r]);
        }
}

int main() {
    scanf("%d",&t);
    while(t--) {
        sum[0] = 0;
        scanf("%d",&n);
        clr(f); clr(dp);
        for(int i = 1; i <= n; i++) scanf("%d",&num[i]);
        for(int i = 1; i <= n; i++) scanf("%d",&val[i]), sum[i] = sum[i-1]+val[i];
        DP();
        work();
        printf(AUTO"\n",dp[1][n]);
    }
    return 0;
}