zoj3211 按增长量排序然后dp

最新推荐文章于 2019-09-01 10:29:51 发布

greentea_acmer

最新推荐文章于 2019-09-01 10:29:51 发布

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分类专栏： acm 文章标签： zoj dp zju 3211 排序

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本文介绍ZOJ3211问题的解决方法，采用动态规划算法求解最大值问题。通过按增长量排序并进行状态转移，实现了O(mn)的时间复杂度。

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zoj3211 DP题

按增长量排序，然后dp，dp[i][j]=max(dp[i-1][k]+a[i]+(i-1)*b[j]),其实有O(mn)的dp

//1882141 2009-05-23 22:11:52 Accepted  3211 C++ 810 448 green tea 
#include <cstdlib>
#include <cctype>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <vector>
#include <string>
#include <iostream>
#include <sstream>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <fstream>
#include <numeric>
#include <iomanip>
#include <bitset>
#include <list>
#include <ctime>
#include <utility>

using namespace std;

#define inf (1<<30)
#define PB push_back
#define mset(x,a) memset(x,(a),sizeof(x))
#define SIZE(X) ((int)X.size())
typedef vector<int> VI;
typedef vector<char> VC;
typedef vector<string> VS;
typedef long long LL;
typedef unsigned long long uLL;
#define twoL(X) (((LL)(1))<<(X))
const double PI=acos(-1.0);
const double eps=1e-11;
template <class T> T sqr(T x) {return x*x;}
template <class T> T gcd(T a, T b) {if(a<0) return (gcd(-a,b)); if(b<0) return (gcd(a,-b)); return (b==0)?a:gcd(b,a%b);}
template <class T> T lcm(T a, T b) {return a*b/gcd(a,b);}
LL toLL(string s) { istringstream sin(s); LL t; sin>>t; return t;}
int toInt(string s) {istringstream sin(s); int t; sin>>t; return t;}
string toString(LL v) {ostringstream sout; sout<<v; return sout.str();}
string toString(int v) {ostringstream sout; sout<<v; return sout.str();}
#define FOREACH(it, a) for(typeof((a).begin()) it = (a).begin(); it!=(a).end(); ++it)
#define ALL(x) ((x).begin, (x).end())
#define cross(a, b, c)  ((c).x-(a).x)*((b).y-(a).y)-((b).x-(a).x)*((c).y-(a).y)
#define sq_dist(p, q) ((p).x-(q).x)*((p).x-(q).x)+((p).y-(q).y)*((p).y-(q).y)

int T;

struct point {
    int w, add;
};
point a[260];

bool cmp(point p1, point p2) {
    return p1.add < p2.add;
}

int dp[260][260];

//dp[i][j]´ú±íÇ°iÌìÒÔj½áÎ²µÄ×î´óÖµ

int main()
{
    int n, m;
    scanf("%d", &T);
    while ( T-- ) {
        scanf("%d%d", &n, &m);
        for ( int i = 1; i <= n; ++i )
            scanf("%d", &a[i].w);
        for ( int i = 1; i <= n; ++i )
            scanf("%d", &a[i].add);
        sort(a+1, a+n+1, cmp);
        mset(dp, 0);
        for ( int i = 1; i <= n; ++i )
            dp[1][i] = a[i].w;
        for ( int i = 2; i <= m; ++i )
            for ( int j = i; j <= n; ++j )
                for ( int k = i-1; k < j; ++k )
                    dp[i][j] = max(dp[i][j], dp[i-1][k]+a[j].w+(i-1)*a[j].add);
        int ans = 0;
        for ( int i = 1; i <= n; ++i )
            ans = max(ans, dp[m][i]);
        printf("%d\n", ans);
    }
    return 0;
}