2014.7.14 CodeForces C题 字符串上的动态规划

/*ChasingWaves 	446A - DZY Loves Sequences 	GNU C++ 	Accepted 	31 ms 	1200 KB */
#include <cstdio>
#include <iostream>
#include <cstring>
using namespace std;

const int maxn = 100000;
int a[maxn+5];
int l[maxn+5], r[maxn+5]; //l[i]表示以a[i]结尾的最长严格单增序列长度，r[i]表示以a[i]开头的最长严格单增序列长度

inline int max(int a, int b){
return a > b ? a : b;
}

int main(){
int n;
while(scanf("%d", &n) != EOF){
for(int i=0; i<n; i++)
scanf("%d", &a[i]);
if(n == 1){
printf("1\n");
continue;
}
else if(n == 2){
printf("2\n");
continue;
}
memset(l, 0, sizeof(l));
memset(r, 0, sizeof(r));
l[0] = 1;
for(int i=1; i<n; i++){
if(a[i] > a[i-1]) l[i] = l[i-1] + 1;
else l[i] = 1;
}
r[n-1] = 1;
for(int i=n-2; i>=0; i--){
if(a[i] < a[i+1]) r[i] = r[i+1] + 1;
else r[i] = 1;
}
int res = max(1+r[1], l[n-2]+1); //只有左半区间或只有右半区间
for(int i=1; i<n-1; i++){
if(a[i+1] >= a[i-1]+2) res = max(res, l[i-1]+1+r[i+1]); //两个区间可以连接
else res = max(res, max(l[i-1]+1, 1+r[i+1]));
}
printf("%d\n", res);
}
return 0;
}