题意：N个数，M个询问，每个询问为一个区间，求区间最长连续子序列，要求每个数都不同（perfect sequence，简称PS）。
题解：很容易求出以每个数为结尾的ps，也就是求区间的最大值。有一个不同就是长度可能会超出询问范围，所以先对PS的首位置二分，然后RMQ。注意一点，序列有可能出现负数，所以先加最大值变为正数。其实也不算变形，挺裸的……
这题卡线段树，然而我只会线段树，心塞……
代码（树状数组）：
#include <iostream> #include <cstring> #include <cstdio> #include <algorithm> using namespace std; const int N = 2000005; int a[N], pos[N], len[N], nt[N]; #define lowbit(x) ((x)&(-x)) int idx[N]; void init(int n) { for(int i=1;i<=n;i++) { idx[i] = len[i]; for(int j=1;j<lowbit(i);j<<=1){ idx[i]=max(idx[i],idx[i-j]); } } } int query(int l, int r) { int ans=len[r]; while(true) { ans=max(ans,len[r]); if(r==l) break; for(r-=1;r-l>=lowbit(r);r-=lowbit(r)){ ans=max(ans,idx[r]); } } return ans; } int main() { int n, m; while (~scanf("%d%d", &n, &m)) { for (int i = 1; i <= n; ++i) { scanf("%d", &a[i]); a[i] += 1000000; } memset(pos, 0, sizeof pos); nt[0] = 1; for (int i = 1; i <= n; ++i) { nt[i] = max(nt[i-1], pos[a[i]]+1);//nt[i]以i为结尾的最长子序列的首端 len[i] = i-nt[i]+1; pos[a[i]] = i; } init(n); while (m--) { int l, r; scanf("%d%d", &l, &r); l++, r++; int p = upper_bound(nt+1, nt+1+n, l) - nt; int ans = 0; if (p > r) { printf("%d ", r-l+1); continue; } if (p > l) ans = p-l; ans = max(ans, query(p, r)); printf("%d ", ans); } } return 0; }
代码（ST算法）：
#include <iostream> #include <cstring> #include <cstdio> #include <algorithm> using namespace std; const int N = 2000005; int a[N], pos[N], len[N], nt[N]; int f[N][20]; void init(int n) { // f[i,j]表示[i,i+2^j-1]区间最大值 // f[i,j]=max(d[i,j-1], d[i+2^(j-1),j-1]) for (int i = 1; i <= n; ++i) f[i][0] = len[i]; for (int j = 1; (1<<j) <= n; ++j) for (int i = 1; i+j-1 <= n; ++i) f[i][j] = max(f[i][j-1], f[i+(1<<j-1)][j-1]); } int query(int l, int r) { int k = 0; while ((1<<k+1 <= r-l+1)) ++k; return max(f[l][k], f[r-(1<<k)+1][k]); } int main() { int n, m; while (~scanf("%d%d", &n, &m)) { for (int i = 1; i <= n; ++i) { scanf("%d", &a[i]); a[i] += 1000000; } memset(pos, 0, sizeof pos); nt[0] = 1; for (int i = 1; i <= n; ++i) { nt[i] = max(nt[i-1], pos[a[i]]+1);//nt[i]以i为结尾的最长子序列的首端 len[i] = i-nt[i]+1; pos[a[i]] = i; } init(n); while (m--) { int l, r; scanf("%d%d", &l, &r); l++, r++; int p = upper_bound(nt+1, nt+1+n, l) - nt; int ans = 0; if (p > r) { printf("%d ", r-l+1); continue; } if (p > l) ans = p-l; ans = max(ans, query(p, r)); printf("%d ", ans); } } return 0; }
还有一种网上流传的算法，类似kmp。想法很巧妙，但是最坏复杂度貌似是O(n^2)，不过也能A这道题。而且速度不必上面慢。数据水吧~
nt[i]记录想要字串中含有a[(i+1)-pos[i+1]]的最大位置。
#include <iostream> #include <cstring> #include <cstdio> using namespace std; const int N = 2000005; int a[N], pos[N], len[N], nt[N]; int main() { int n, m; while (~scanf("%d%d", &n, &m)) { for (int i = 1; i <= n; ++i) { scanf("%d", a+i); a[i] += 1000000; } memset(pos, 0, sizeof pos); len[0] = 0; nt[0] = 0; for (int i = 1; i <= n; ++i) { if (len[i-1]+1 <= i-pos[a[i]]) { len[i] = len[i-1]+1; nt[i] = nt[i-1]; } else { len[i] = i-pos[a[i]]; nt[i] = i-1; } pos[ a[i] ] = i; //printf("%d %d %d ", len[i], nt[i], pos[a[i]]); } while (m--) { int l, r; scanf("%d%d", &l, &r); l++, r++; int res = len[r]; if (res >= r-l+1) { printf("%d ", r-l+1); continue; } while (nt[r] > 0) { r = nt[r]; res = max(res, min(len[r], r-l+1)); if (res >= r-l+1) { break; } } printf("%d ", res); } } return 0; }