Codeforces 327A Flipping Game(区间DP，暴力枚举)

Flipping Game

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Iahub got bored, so he invented a game to be played on paper. He writes n integers a1, a2, …, an. Each of those integers can be either 0 or 1. He’s allowed to do exactly one move: he chooses two indices i and j (1 ≤ i ≤ j ≤ n) and flips all values ak for which their positions are in range [i, j] (that is i ≤ k ≤ j). Flip the value of x means to apply operation x = 1 - x. The goal of the game is that after exactly one move to obtain the maximum number of ones. Write a program to solve the little game of Iahub. Input The first line of the input contains an integer n (1 ≤ n ≤ 100). In the second line of the input there are n integers: a1, a2, …, an. It is guaranteed that each of those n values is either 0 or 1. Output Print an integer — the maximal number of 1s that can be obtained after exactly one move.

Example Input 5 1 0 0 1 0 Output 4

Input 4 1 0 0 1 Output 4

Note In the first case, flip the segment from 2 to 5 (i = 2, j = 5). That flip changes the sequence, it becomes: [1 1 1 0 1]. So, it contains four ones. There is no way to make the whole sequence equal to [1 1 1 1 1]. In the second case, flipping only the second and the third element (i = 2, j = 3) will turn all numbers into 1

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题意：

给定一串01序列，选择其中一段翻转，也就是0变1，1变0.求一次翻转后的最多1的数目。

思路：

暴力枚举： 用个数组储存前缀和，表示i之前1的数目，在枚举每个区间，数据比较小，也能过。 区间DP： 预处理差不多，DP[i][j]表示I到j之间一次翻转后一最多的数量。 对于每个区间而言，在中间找个断点k，一次翻转可能发生在前面I到K，也可能发生在K到j之间， 那我们就得到了状态转移方程 dp[I][j]=max(dp[i][k]+sum[j]-sum[k],dp[k+1][j]+sum[k]-sum[i-1])); (i<=K < j)

#include<iostream> #include<cstdio> #include<algorithm> #include<math.h> #include<string.h> #define mem(a,b) memset((a),(b),sizeof(b)); using namespace std; int main() { ios::sync_with_stdio(false); // freopen("text.txt", "r", stdin); int a[105]; int sum[105]; int dp[105][105]; int n; cin>>n; sum[0]=0; memset(dp,0,sizeof(dp)); for(int i=1;i<=n;i++) { cin>>a[i]; sum[i]=a[i]+sum[i-1]; dp[i][i]=1-a[i]; } for(int l=1;l<n;l++) for(int i=1;i+l<=n;i++) { int j=i+l; dp[i][j]=j-i+1-sum[j]+sum[i-1]; for(int k=i;k<j;k++) dp[i][j]=max(dp[i][j], max(dp[i][k]+sum[j]-sum[k], dp[k+1][j]+sum[k]-sum[i-1])); //cout<<i<<' '<<j<<' '<<dp[i][j]<<endl; } cout<<dp[1][n]<<endl; return 0; }

所谓水题就是拿来练习，只有对某种方法足够熟练，有一定理解才能用起来得心应手，虽然我现在菜的不行，但坚持下去，我想会有收获的。