Codeforces Round #FF (Div. 2)C. DZY Loves Sequences
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从两个方向进行预处理,然后枚举出最大值(改子序列中的一个 或者不改两种情况)。
#include <iostream>#include <cstdio>#include <cstring>#include <vector>#include <math.h>#include <map>#include <algorithm>using namespace std;const int maxn = 100005;int dp1[maxn],dp2[maxn],num[maxn];int main(){ int n; while(~scanf("%d",&n)){ for(int i = 1;i <= n;i++) scanf("%d",num+i); memset(dp1, 0, sizeof(dp1)); memset(dp2, 0, sizeof(dp2)); for(int i = 1;i <= n;i++){//以i结尾的最长连续递增子序列 if(num[i] > num[i-1]) dp1[i] = dp1[i-1]+1; else dp1[i] = 1; } for(int i = n;i >= 1;i--){//以i开头的最长连续递增子序列 if(num[i] < num[i+1]) dp2[i] = dp2[i+1]+1; else dp2[i] = 1; } /*for(int i = 1;i <= n;i++) printf("%d ",dp1[i]); printf("\n"); for(int i = 1;i <= n;i++) printf("%d ",dp2[i]);*/ int ans_max = 0; for(int i = 1;i <= n;i++){//枚举 if(num[i+1] - num[i-1] >= 2){ int temp = dp1[i-1] + dp2[i+1] + 1; ans_max = max(ans_max,temp); } ans_max = max(ans_max,max(dp1[i-1],dp2[i+1])+1); } printf("%d\n",ans_max); } return 0;}
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