B.寻找最大值
算法:
1.正向思维如果枚举区间求最值肯定TLE,数据量很大。
2.反向思维,枚举每个点的左右区间,虽然两个for循环,我感觉是平均是O(N)的时间复杂度,动态规划的思想,可以求出。
3.这题相当诡异,不能给dt数组清空赋值,坑了我一晚上加上午,还不知道为什么。
4.我的L[I]记录的是该点能往左到达的边界,R【i]是往右到达的边界。旭他们记录的是以该点向左能扩张的长度,R【i]类似。
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#include<stdio.h> #include<stdlib.h> #include<string.h> #include<iostream> #include<vector> #include<string> #include<math.h> #include<map> #include<set> #include<algorithm> using namespace std; const long long inf = -(1LL<<60); int N; long long dt[201000]; //输入数据 long long sum[201000]; //静态,不需要修改,数组维护最长前缀和 long long L[201000]; long long R[201000]; int main( ) { while( scanf("%d",&N) != EOF) { memset(sum, 0, sizeof(sum)); for(int i = 1; i <= N; i++) { scanf("%I64d",&dt[i]); sum[i] = sum[i-1] + dt[i]; } memset(L,0,sizeof(L)); memset(R,0,sizeof(R)); for(int i = 1; i <= N; i++) { int j = i-1, len = 0; while(j >= 1){ if( dt[i] > dt[j] ) break; else{ len += 1+L[j]; j = j-L[j]-1; } } L[i] = len; } for(int i = N; i >= 1; i--) { int j = i+1, len = 0; while(j <= N ){ if( dt[i] > dt[j] ) break; else{ len += 1+R[j]; j = j+R[j]+1; } } R[i] = len; } long long maxn = inf; long long ans; for( int i = 1; i <= N; i++) { int l = i-L[i]; int r = i+R[i]; ans = (sum[r] - sum[l - 1]) * dt[i]; if( ans > maxn ) maxn = ans; } printf("%I64d\n",maxn); } return 0; }
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#include<stdio.h> #include<stdlib.h> #include<string.h> #include<iostream> #include<vector> #include<string> #include<math.h> #include<map> #include<set> #include<algorithm> using namespace std; const long long inf = (1LL<<60); int N; long long dt[301000]; //输入数据 long long sum[301000]; //静态,不需要修改,数组维护最长前缀和 int L[301000]; int R[301000]; int main( ) { while( scanf("%d",&N) != EOF) { for(int i = 1; i <= N; i++) { scanf("%I64d",&dt[i]); sum[i] = sum[i-1] + dt[i]; } memset(L,0,sizeof(L)); memset(R,0,sizeof(R)); for(int i = 1; i <= N; i++) { if( dt[i] == dt[i-1] && i != 1) { L[i] = L[i-1]; continue; } if( dt[i] > dt[i-1] || i == 1) { L[i] = i; } else { int j = i; while( --j ) { int x = L[j]; if( dt[x - 1] < dt[i] ) { L[i] = x; break; } } } } for( int i = N; i >= 1; i--) { if( dt[i] == dt[i+1] && i != N) { R[i] = R[i+1]; continue; } if( dt[i] > dt[i+1] || i == N) { R[i] = i; } else { int j = i; while( ++j <= N ) { int x = R[j]; if( dt[x + 1] < dt[i] ) { R[i] = x; break; } } } } long long maxn = -inf; for( int i = 1; i <= N; i++) { long long ans = (sum[R[i]] - sum[L[i] - 1]) * dt[i]; if( ans > maxn ) maxn = ans; } printf("%I64d\n",maxn); } return 0; }
posted on 2012-08-18 09:38 more think, more gains 阅读(174) 评论(0) 编辑 收藏 举报