zoukankan      html  css  js  c++  java
  • HDU 5900 QSC and Master 区间DP

    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
    ",dp[1][n]);
            }
            return 0;
    }
     
  • 相关阅读:
    Java并发编程有多难?这几个核心技术你掌握了吗?
    「mysql优化专题」主从复制面试宝典!面试官都没你懂得多!(11)
    「mysql优化专题」高可用性、负载均衡的mysql集群解决方案(12)
    「mysql优化专题」什么是慢查询?如何通过慢查询日志优化?(10)
    「mysql优化专题」详解引擎(InnoDB,MyISAM)的内存优化攻略?(9)
    vsftp虚拟用户配置
    Oracle shrink space
    linux加程序是否当掉检测脚本
    oracke创建db link
    Oracle函数日期转换成秒(时间戳)
  • 原文地址:https://www.cnblogs.com/zxhl/p/5882761.html
Copyright © 2011-2022 走看看