CodeForces 599 C. Day at the Beach(贪心)
2017-02-28 19:15
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Description
给出n个数,把其分成最多的连续子段,使得每个子段升序排完之后整个序列还是升序的
Input
第一行一整数n表示序列长度,之后n个整数a[i]表示该序列(1<=n<=100000,1<=a[i]<=1e9)
Output
输出合法划分中子段最大数量
Sample Input
3
1 2 3
Sample Output
3
Solution
一个子段能够被划分出来需要满足该子段中最大值不大于后面所有数中的最小值,所以预处理出后缀最小值,之后从前往后扫一遍,更新当前段的最大值,如果该最大值不大于后面段的最小值则可以单独分出来一段,否则继续往后扫
Code
给出n个数,把其分成最多的连续子段,使得每个子段升序排完之后整个序列还是升序的
Input
第一行一整数n表示序列长度,之后n个整数a[i]表示该序列(1<=n<=100000,1<=a[i]<=1e9)
Output
输出合法划分中子段最大数量
Sample Input
3
1 2 3
Sample Output
3
Solution
一个子段能够被划分出来需要满足该子段中最大值不大于后面所有数中的最小值,所以预处理出后缀最小值,之后从前往后扫一遍,更新当前段的最大值,如果该最大值不大于后面段的最小值则可以单独分出来一段,否则继续往后扫
Code
#include<cstdio> #include<iostream> #include<cstring> #include<algorithm> #include<cmath> #include<vector> #include<queue> #include<map> #include<set> #include<ctime> using namespace std; typedef long long ll; #define maxn 111111 int n,a[maxn],b[maxn]; int main() { while(~scanf("%d",&n)) { for(int i=1;i<=n;i++)scanf("%d",&a[i]); b =a ; for(int i=n-1;i>=1;i--)b[i]=min(b[i+1],a[i]); int ans=0,c=a[1]; for(int i=2;i<=n;i++) { if(c<=b[i])ans++,c=a[i]; else { while(c>b[i]&&i<=n)c=max(c,a[i]),i++; i--; } } printf("%d\n",ans+1); } return 0; }
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