codeforces 1107E Vasya and Binary String DP

本文解析了一道经典的动态规划题目,通过定义状态f(i)表示i个相同元素消去的最大得分,采用完全背包算法解决。将序列转换为黑白交错的点,使用g(l,r,k)表示消去[l,r]段及后续k个相同元素的最大得分,通过记忆化搜索实现高效求解。

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题目分析

听说是个DP经典套路?

f(i)f(i)f(i)表示iii个一样的连在一起的元素,被消完的最大分数,一个完全背包可以搞定。

然后将连续的一段相同元素合成一个点,原序列变成了若干黑白交错的点,记点iii中的元素个数为sz(i)sz(i)sz(i)

g(l,r,k)g(l,r,k)g(l,r,k)表示现在要消完[l,r][l,r][l,r]这一段点(和后面的kkk个元素),点rrr后面有kkk个与rrr点颜色相同的元素。那么rrr可以选择跟前面的元素一起消,也可以不一起消。

不一起消:g(l,r−1,0)+f(sz(r)+k)g(l,r-1,0)+f(sz(r)+k)g(l,r1,0)+f(sz(r)+k)

一起消:(iii[l,r−1][l,r-1][l,r1]中一个和rrr颜色相同的点)g(i+1,r−1,0)+g(l,i,k+sz(r))g(i+1,r-1,0)+g(l,i,k+sz(r))g(i+1,r1,0)+g(l,i,k+sz(r))

记忆化搜索,跑不满,加上CF评测姬快如闪电,可以过。

代码

#include<bits/stdc++.h>
using namespace std;
#define RI register int
typedef long long LL;
const int N=105;
int n,cnt;LL a[N],f[N],g[N][N][N];char S[N];
struct node{int sz,col;}b[N];

void prework() {
    for(RI i=1;i<=n;++i)
        for(RI j=i;j<=n;++j) f[j]=max(f[j],f[j-i]+a[i]);
    int now=0;
    for(RI i=1;i<=n;++i) {
        if(i==1||S[i]==S[i-1]) ++now;
        else ++cnt,b[cnt].sz=now,b[cnt].col=S[i-1]-'0',now=1;
    }
    ++cnt,b[cnt].sz=now,b[cnt].col=S[n]-'0';
}
LL DP(int l,int r,int k) {
    if(l==r) {return g[l][r][k]=f[b[l].sz+k];}
    if(g[l][r][k]!=-1) return g[l][r][k];
    LL re=DP(l,r-1,0)+f[b[r].sz+k];
    for(RI i=l;i<r-1;++i)
        if(b[i].col==b[r].col)
            re=max(re,DP(i+1,r-1,0)+DP(l,i,b[r].sz+k));
    return g[l][r][k]=re;
}
int main()
{
    scanf("%d",&n);
    scanf("%s",S+1);
    for(RI i=1;i<=n;++i) scanf("%lld",&a[i]);
    prework();
    memset(g,-1,sizeof(g));
    printf("%lld\n",DP(1,cnt,0));
    return 0;
}
### Codeforces 887E Problem Solution and Discussion The problem **887E - The Great Game** on Codeforces involves a strategic game between two players who take turns to perform operations under specific rules. To tackle this challenge effectively, understanding both dynamic programming (DP) techniques and bitwise manipulation is crucial. #### Dynamic Programming Approach One effective method to approach this problem utilizes DP with memoization. By defining `dp[i][j]` as the optimal result when starting from state `(i,j)` where `i` represents current position and `j` indicates some status flag related to previous moves: ```cpp #include <bits/stdc++.h> using namespace std; const int MAXN = ...; // Define based on constraints int dp[MAXN][2]; // Function to calculate minimum steps using top-down DP int minSteps(int pos, bool prevMoveType) { if (pos >= N) return 0; if (dp[pos][prevMoveType] != -1) return dp[pos][prevMoveType]; int res = INT_MAX; // Try all possible next positions and update 'res' for (...) { /* Logic here */ } dp[pos][prevMoveType] = res; return res; } ``` This code snippet outlines how one might structure a solution involving recursive calls combined with caching results through an array named `dp`. #### Bitwise Operations Insight Another critical aspect lies within efficiently handling large integers via bitwise operators instead of arithmetic ones whenever applicable. This optimization can significantly reduce computation time especially given tight limits often found in competitive coding challenges like those hosted by platforms such as Codeforces[^1]. For detailed discussions about similar problems or more insights into solving strategies specifically tailored towards contest preparation, visiting forums dedicated to algorithmic contests would be beneficial. Websites associated directly with Codeforces offer rich resources including editorials written after each round which provide comprehensive explanations alongside alternative approaches taken by successful contestants during live events. --related questions-- 1. What are common pitfalls encountered while implementing dynamic programming solutions? 2. How does bit manipulation improve performance in algorithms dealing with integer values? 3. Can you recommend any online communities focused on discussing competitive programming tactics? 4. Are there particular patterns that frequently appear across different levels of difficulty within Codeforces contests?
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