zoj3211 按增长量排序 然后dp

将自己历史的AC共享

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°iÌìÒÔj½áβµÄ×î´óÖµint 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;}

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