HDU 5900 QSC and Master 区间DP

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原文链接：http://www.cnblogs.com/zxhl/p/5882761.html

本文介绍了一个关于从一组数对中选出键值互为非互质的连续数对以获得最大价值总和的问题，并提供了完整的算法实现，包括如何通过动态规划解决此问题。

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QSC and Master

Problem Description

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

3 3 1 2 3 1 1 1 3 1 2 4 1 1 1 4 1 3 4 3 1 1 1 1

Sample Output

0 2 0

#include<iostream>
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<cstring>
using namespace std;
#pragma comment(linker, "/STACK:102400000,102400000")
#define ls i<<1
#define rs ls | 1
#define mid ((ll+rr)>>1)
#define pii pair<int,int>
#define MP make_pair
typedef long long LL;
const long long INF = 1e18;
const double Pi = acos(-1.0);
const int N = 600+10, M = 1e2+11, mod = 1e9+7, inf = 0x3fffffff;

LL sum[N],dp[N][N];
int f[N][N],n,m,value[N],key[N];
int gcd(int a,int b) { return b == 0 ? a : gcd(b, a%b);}
void DP() {
        for(int i = 1; i < n; ++i) f[i][i+1] = gcd(key[i],key[i+1]) == 1? 0 : 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(key[i],key[r])!=1);
                for(int k=i;k<r;++k)
                    f[i][r] += (f[i][k]&&f[k+1][r]);
            }
        }
}
void solve() {
        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]) {
                    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() {
        int T;
        scanf("%d",&T);
        while(T--) {
            sum[0] = 0;
            scanf("%d",&n);
            memset(dp,0,sizeof(dp));
             memset(f,0,sizeof(f));
            for(int i = 1; i <= n; ++i) scanf("%d",&key[i]);
            for(int i = 1; i <= n; ++i) scanf("%d",&value[i]),sum[i] = sum[i-1] + value[i];
            DP();
            solve();
            printf("%I64d\n",dp[1][n]);
        }
        return 0;
}