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

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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,ntemplate<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 gusing 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;}

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