HDU 5900 QSC and Master (区间DP)

题意：给出n对数keyi,vali表示当前这对数的键值和权值，可以操作将连续的两个数合并，如果满足gcd(a[i],a[i+1])>1，得到的价值是两个数的权值和，

每次合并两个数之后，这两个数就会消失，然后旁边的数会接上.

析：区间DP，首先dp[i][j] 表示区间第 i 段到第 j 段所能得到的最大值，然后分情况讨论一下，第一种是区间内的所有的都可以合并，这个预处理一下就好，

第二种不能全合并，那么选取最大的那一个就好。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <tr1/unordered_map>
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
using namespace std :: tr1;

typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const LL LNF = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 3e2 + 5;
const int mod = 1e9 + 7;
const int N = 1e6 + 5;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
inline LL Min(LL a, LL b){ return a < b ? a : b; }
inline LL Max(LL a, LL b){ return a > b ? a : b; }
inline bool is_in(int r, int c){
    return r >= 0 && r < n && c >= 0 && c < m;
}
LL sum[maxn];
LL dp[maxn][maxn], a[maxn], val[maxn];
bool f[maxn][maxn];

int main(){
    int T;   cin >> T;
    while(T--){
        scanf("%d", &n);
        for(int i = 1; i <= n; ++i)  scanf("%I64d", &a[i]);
        for(int i = 1; i <= n; ++i){
            scanf("%I64d", &val[i]);
            sum[i] = sum[i-1] + val[i];
        }

        memset(f, false, sizeof f);
        for(int i = 1; i < n; ++i)  f[i][i+1] = __gcd(a[i], a[i+1]) != 1;

        for(int i = 2; i <= n; i += 2){
            for(int j = 1; j+i-1 <= n; ++j){
                int k = j+i-1;
                if(__gcd(a[k-1], a[k]) != 1 && f[j][k-2])  f[j][k] = true;
                else if(__gcd(a[j+1], a[j]) != 1 && f[j+2][k])  f[j][k] = true;
                else if(__gcd(a[j], a[k]) != 1 && f[j+1][k-1])  f[j][k] = true;
            }
        }

        memset(dp, 0, sizeof dp);
        for(int i = 2; i <= n; ++i){
            for(int j = 1; j+i-1 <= n; ++j){
                int l = j+i-1;
                if(f[j][l])  dp[j][l] = sum[l] - sum[j-1];
                else{
                    for(int k = j; k < l; ++k)
                        dp[j][l] = Max(dp[j][l], dp[j][k] + dp[k+1][l]);
                }
            }
        }
        printf("%I64d\n", dp[1][n]);
    }
    return 0;
}