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Codeforces 18D Seller Bob && 18E Flag 2 简单dp

D题很恶心的要用大数。

dp[i] 表示到第 i 条信息的最大收益。

显然,dp[1]  = 0。

对于i > 1有,若当前信息为 win x,那么显然有dp[i] = dp[i-1]。

若当前信息为sell x,那么dp[i] = max(dp[i-1] , dp[j] + 2^x),j 需满足j < i && 第j条信息为 win y && y == x。

import java.util.Scanner;
import java.math.BigInteger;
import java.io.*;

public class Main
{
	public static void main(String[] args)
	{
		Scanner cin = new Scanner(System.in);
		
		String []op = new String[5010];
		int []val = new int[5010];
		BigInteger []dp = new BigInteger[5010];
		int i,j,n;
		n = cin.nextInt();
		BigInteger TWO = BigInteger.valueOf(2);
		for(i = 1;i <= n; ++i)
		{
			op[i] = cin.next();
			val[i] = cin.nextInt();
		}
		
		dp[1] = BigInteger.ZERO;
		
		for(i = 2;i <= n; ++i)
		{
			dp[i] = dp[i-1];
			if(op[i].length() == 4)
			{	
				
				for(j = i-1;j >= 1; --j)
				{
					if(val[i] == val[j] && op[j].length() == 3)
					{
						dp[i] = dp[i].max(dp[j].add(TWO.pow(val[j])));
					}
				}
			}
		}
		System.out.println(dp[n]);
	}
}

E题让我见识了CF服务器的强大。2*10^8的算法才跑了900+。

思路:

因为相邻块颜色不同且每行有且只有两种颜色,所以每行的格式必为ABABABA。

所以我们预处理出val[i][A][B]所需花费。i为行,A,B分别表示此行用这两种颜色。o(n*(m+26*26)。

然后为o(n*26^4)的枚举,详见代码。

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <queue>
#include <cmath>
#include <stack>
#include <map>
#include <ctime>
#include <iomanip>

#pragma comment(linker, "/STACK:1024000000");
#define EPS (1e-6)
#define _LL long long
#define ULL unsigned long long
#define LL __int64
#define INF 0x3f3f3f3f
#define Mod 1000000007

/** I/O Accelerator Interface .. **/
#define g (c=getchar())
#define d isdigit(g)
#define p x=x*10+c-'0'
#define n x=x*10+'0'-c
#define pp l/=10,p
#define nn l/=10,n
template<class T> inline T& RD(T &x)
{
    char c;
    while(!d);
    x=c-'0';
    while(d)p;
    return x;
}
template<class T> inline T& RDD(T &x)
{
    char c;
    while(g,c!='-'&&!isdigit(c));
    if (c=='-')
    {
        x='0'-g;
        while(d)n;
    }
    else
    {
        x=c-'0';
        while(d)p;
    }
    return x;
}
inline double& RF(double &x)      //scanf("%lf", &x);
{
    char c;
    while(g,c!='-'&&c!='.'&&!isdigit(c));
    if(c=='-')if(g=='.')
        {
            x=0;
            double l=1;
            while(d)nn;
            x*=l;
        }
        else
        {
            x='0'-c;
            while(d)n;
            if(c=='.')
            {
                double l=1;
                while(d)nn;
                x*=l;
            }
        }
    else if(c=='.')
    {
        x=0;
        double l=1;
        while(d)pp;
        x*=l;
    }
    else
    {
        x=c-'0';
        while(d)p;
        if(c=='.')
        {
            double l=1;
            while(d)pp;
            x*=l;
        }
    }
    return x;
}
#undef nn
#undef pp
#undef n
#undef p
#undef d
#undef g
using namespace std;

char Map[510][510];

int ans[2][26];

int val[510][26][26];
int pre[510][26][26][2];

void dfs(int n,int m,int px,int py)
{
    if(n == 1)
    {
        for(int i = 1; i <= m; ++i)
            if(i&1)
                printf("%c",(char)(px+'a'));
            else
                printf("%c",(char)(py+'a'));
        puts("");
        return ;
    }

    dfs(n-1,m,pre[n][px][py][0],pre[n][px][py][1]);
    for(int i = 1; i <= m; ++i)
        if(i&1)
            printf("%c",(char)(px+'a'));
        else
            printf("%c",(char)(py+'a'));
    puts("");
}

int main()
{
    int n,m,i,j,k,l,p;

    scanf("%d %d",&n,&m);

    for(i = 1; i <= n; ++i)
        scanf("%s",Map[i]+1);

    for(i = 1; i <= n; ++i)
    {
        memset(ans,0,sizeof(ans));

        for(j = 1; j <= m; ++j)
            ans[j&1][Map[i][j]-'a']++;

        for(j = 0; j < 26; ++j)
            for(k = 0; k < 26; ++k)
                val[i][j][k] = m-ans[1][j]-ans[0][k];
    }

    int Min;

    for(i = 2; i <= n; ++i)
    {
        for(j = 0; j < 26; ++j)
            for(k = 0; k < 26; ++k)
            {
                Min = INF;
                for(l = 0; l < 26; ++l)
                {
                    if(j == l)
                        continue;
                    for(p = 0; p < 26; ++p)
                    {
                        if(k == p || l == p)
                            continue;
                        if(val[i-1][l][p] < Min)
                            Min = val[i-1][l][p],pre[i][j][k][0] = l,pre[i][j][k][1] = p;
                    }
                }
                val[i][j][k] += Min;
            }
    }

    Min = INF;
    int px,py;
    for(i = 0; i < 26; ++i)
        for(j = 0; j < 26; ++j)
            if(i != j && val[n][i][j] < Min)
                Min = val[n][i][j],px = i,py = j;
    printf("%d\n",Min);

    dfs(n,m,px,py);

    return 0;
}


Codeforces 18D Seller Bob && 18E Flag 2 简单dp