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2016 ACM/ICPC Asia Regional Shenyang Online 1009 QSC and Master 区间dp

来源：程序员人生发布时间：2016-12-03 10:22:56 阅读次数：6751次

QSC and Master

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 773 Accepted Submission(s): 291

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

Sample Output

0
2
0

Source

2016 ACM/ICPC Asia Regional Shenyang Online

My Solution

状态定义： dp[i][j][u] u == 1 时表示当端点 i， j 进行合并时(取出 val[i] 、 val[j] 时)

或 i < k < k+1 < j， key[i] 和 key[k] 可以合并且key[k+1] 和 key[j]合并的，区间 [i, j]内的最大价值；

u == 0 时表示当端点 i， j 不进行合并时(不取出 val[i] 、 val[j] 时) 的区间 [i, j]内的最大价值；

初始化：全部初始化为 0；

边界：当 j - i == 1 时如果可以合并则 dp[i][j][1] = val[i] + val[j]；

状态转移方程：如果 key[i], key[j] 可以组合，则如果 dp[i + 1][j - 1][1] > 0 则 dp[i][j][1] = max(dp[i][j][1], dp[i+1][j⑴][1] + val[i] + val[j]);

然后对区间划分 k = i ~ j

dp[i][j][0] = max(dp[i][j][0], max(dp[i][k][0], dp[i][k][1]) + max(dp[k+1][j][1] , dp[k+1][j][0]));
if(dp[i][k][1] > 0 && dp[k+1][j][1] > 0) dp[i][j][1] = max(dp[i][j][1], dp[i][k][1] + dp[k+1][j][1]);

接着是对u == 0时的转移(即keyi keyj 能合并时可以选择不合并)

dp[i][j][0] = max(dp[i][j][0], max(dp[i+1][j⑴][1], dp[i+1][j⑴][0]));

如果 key[i], key[j] 不能合并，则
if(dp[i+1][j⑴][1] > 0){
dp[i][j][0] = max(dp[i][j][0], max(dp[i+1][j⑴][1], dp[i+1][j⑴][0]));
for(k = i; k < j; k++){
dp[i][j][0] = max(dp[i][j][0], max(dp[i][k][0], dp[i][k][1]) + max(dp[k+1][j][1] , dp[k+1][j][0]));
if(dp[i][k][1] > 0 && dp[k+1][j][1] > 0) dp[i][j][1] = max(dp[i][j][1], dp[i][k][1] + dp[k+1][j][1]);
}

}
else{
dp[i][j][0] = max(dp[i][j][0], dp[i+1][j⑴][0]);
for(k = i; k < j; k++){
dp[i][j][0] = max(dp[i][j][0], max(dp[i][k][0], dp[i][k][1]) + max(dp[k+1][j][1] , dp[k+1][j][0]));
if(dp[i][k][1] > 0 && dp[k+1][j][1] > 0) dp[i][j][1] = max(dp[i][j][1], dp[i][k][1] + dp[k+1][j][1]);
}
}

复杂度 O(n^3)

#include <iostream> #include <cstdio> #include <cmath> #include <cstring> using namespace std; typedef long long LL; const int maxn = 3e2 + 8; LL key[maxn], val[maxn], dp[maxn][maxn][2]; inline LL gcd(LL a, LL b) { return b == 0 ? a : gcd(b, a % b); } inline bool check(LL a, LL b) { return gcd(a, b) != 1; } int main() { #ifdef LOCAL freopen("1009.txt", "r", stdin); //freopen("1009.out", "w", stdout); #endif // LOCAL // ios::sync_with_stdio(false); cin.tie(0); LL T, n; cin >> T; while(T--){ memset(dp, 0, sizeof dp); cin >> n; for(int i = 0; i < n; i++){ cin >> key[i]; } for(int i = 0; i < n; i++){ cin >> val[i]; } LL i, j, k, ans = 0, x, y; for(i = n - 1; i >= 0; i--){ for(j = i + 1; j < n; j++){ if(check(key[i], key[j])){ if(j - i > 2){ if(dp[i+1][j⑴][1] > 0){ dp[i][j][1] = max(dp[i][j][1], dp[i+1][j⑴][1] + val[i] + val[j]); } for(k = i; k < j; k++){ dp[i][j][0] = max(dp[i][j][0], max(dp[i][k][0], dp[i][k][1]) + max(dp[k+1][j][1] , dp[k+1][j][0])); if(dp[i][k][1] > 0 && dp[k+1][j][1] > 0) dp[i][j][1] = max(dp[i][j][1], dp[i][k][1] + dp[k+1][j][1]); } dp[i][j][0] = max(dp[i][j][0], max(dp[i+1][j⑴][1], dp[i+1][j⑴][0])); } else if(j - i == 1) dp[i][j][1] = val[i] + val[j]; } if(j - i > 2){ if(dp[i+1][j⑴][1] > 0){ dp[i][j][0] = max(dp[i][j][0], max(dp[i+1][j⑴][1], dp[i+1][j⑴][0])); for(k = i; k < j; k++){ dp[i][j][0] = max(dp[i][j][0], max(dp[i][k][0], dp[i][k][1]) + max(dp[k+1][j][1] , dp[k+1][j][0])); if(dp[i][k][1] > 0 && dp[k+1][j][1] > 0) dp[i][j][1] = max(dp[i][j][1], dp[i][k][1] + dp[k+1][j][1]); } } else{ dp[i][j][0] = max(dp[i][j][0], dp[i+1][j⑴][0]); for(k = i; k < j; k++){ dp[i][j][0] = max(dp[i][j][0], max(dp[i][k][0], dp[i][k][1]) + max(dp[k+1][j][1] , dp[k+1][j][0])); if(dp[i][k][1] > 0 && dp[k+1][j][1] > 0) dp[i][j][1] = max(dp[i][j][1], dp[i][k][1] + dp[k+1][j][1]); } } } else if(j - i == 1) dp[i][j][0] = 0; //这行可以疏忽 ^_^ //*/ ans = max(ans, max(dp[i][j][0], dp[i][j][1])); } } //cout << dp[0][n⑴][0] << " " << dp[0][n⑴][1]<< endl; if(ans > 0) cout << ans << "\n"; else cout << 0 << "\n"; } return 0; }

Thank you!

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