題意：給一個長度為n（n<=200000)的序列，刪除一個連續(xù)的子序列，使得剩下的序列中有一個長度最大的連續(xù)遞增子序列。

分析：從前往后記錄每個位置的最長連續(xù)遞增序列，從后往前記錄每個位置的最長連續(xù)遞增序列。

作者代碼

#include<cstdio>#include<set>#include<cassert>using namespace std;const int maxn = 200000 + 5;int n, a[maxn], f[maxn], g[maxn];struct Candidate {int a, g;Candidate(int a, int g):a(a),g(g) {}bool operator < (const Candidate& rhs) const {return a < rhs.a;}};set<Candidate> s;int main() {int T;scanf("%d", &T);while(T--) {scanf("%d", &n);for(int i = 0; i < n; i++)scanf("%d", &a[i]);if(n == 1) { printf("1\n"); continue; }// g[i] is the length of longest increasing continuous subsequence ending at ig[0] = 1;for(int i = 1; i < n; i++)if(a[i-1] < a[i]) g[i] = g[i-1] + 1;else g[i] = 1;// f[i] is the length of longest increasing continuous subsequence starting from if[n-1] = 1;for(int i = n-2; i >= 0; i--)if(a[i] < a[i+1]) f[i] = f[i+1] + 1;else f[i] = 1;s.clear();s.insert(Candidate(a[0], g[0]));int ans = 1;for(int i = 1; i < n; i++) {Candidate c(a[i], g[i]);set<Candidate>::iterator it = s.lower_bound(c); // first one that is >= cbool keep = true;if(it != s.begin()) {Candidate last = *(--it); // (--it) points to the largest one that is < c找到比它小的可以直接接上去int len = f[i] + last.g; ans = max(ans, len);if(c.g <= last.g) keep = false;}if(keep) {s.erase(c); // if c.a is already present, the old g must be <= c.gs.insert(c);it = s.find(c); // this is a bit cumbersome and slow but it's clearit++;while(it != s.end() && it->a > c.a && it->g <= c.g) s.erase(it++);}}printf("%d\n", ans);}return 0;}

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