Uvalive 6531 Go up the Ultras monotone stack +rmq

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Test instructions

Given N-Mount Mountain

The following n numbers indicate the height of Mount N

If this mountain is lawful, you must satisfy:

1, if you have a higher than u on the left side of the mountain, set V is U left distance u recent and strictly higher than U mountain, between [v,u] at least one mountain X, so that the height difference between x and U >=15000

2, the right also the same.

It is lawful to satisfy 1 or 2 at the same time.

Ask:

Output all valid mountains.

Ideas:

To find a point near the nearest mountain is to maintain a monotonous stack, and then to the height of the mountain to seek a rmq.

Write and write ...

#pragma COMMENT (linker, "/stack:1024000000,1024000000") #include <bits/stdc++.h>template <class t>inline    BOOL Rd (T &ret) {char c; int sgn;    if (C=getchar (), c==eof) return 0; while (c!= '-' && (c< ' 0 ' | |    C> ' 9 ')) C=getchar ();    sgn= (c== '-')? -1:1;    ret= (c== '-')? 0: (c ' 0 ');    while (C=getchar (), c>= ' 0 ' &&c<= ' 9 ') ret=ret*10+ (c ' 0 ');    RET*=SGN; return 1;}        Template <class t>inline void pt (T x) {if (x <0) {Putchar ('-');    x =-X;    } if (X>9) pt (X/10); Putchar (x%10+ ' 0 ');}  Using namespace Std;typedef long long ll;const int n = 100050;int d[n*2][20];void rmq_init (int *a, int N) {for (int i = 1; I <= N; ++i) D[i][0] = a[i];for (int j = 1; (1 << j) <= N; ++J) for (int i = 1; i + j-1 <= N; ++i) {D[i][j] = min (d[i][j-1], d[i + (1 << (j-1))][j-1]);}} int RMQ (int L, int R) {int k = 0;while ((1 << (k + 1)) <= R-l + 1) ++k;return min (d[l][k], D[r-(1 << k) + 1][k]);} int n, H[n], StAck[n], Top;int lh[n], rh[n];void work () {for (int i = 1; I <= N; i++) Rd (H[i]);    top = 0;        for (int i = 1, i <= N; i++) {while (top && h[stack[top-1]] <= h[i]) top--;        if (top) lh[i] = stack[top-1];        else lh[i] = 0;    stack[top++] = i;    } top = 0;        for (int i = n; i; i--) {while (top && h[stack[top-1]]<= h[i]) top--;        if (top) rh[i] = stack[top-1];        else rh[i] = 0;    stack[top++] = i;        }}const int hehe = 150000;vector<int>ans;int Main () {while (cin>>n) {work ();        Rmq_init (h, N);        Ans.clear ();            for (int i =1; i <= n; i++) {if (lh[i] = = 0 && rh[i] = = 0) ans.push_back (i);                else if (lh[i] = = 0) {int v = RMQ (i, rh[i]);            if (H[i]-v >= hehe) ans.push_back (i);                } else if (rh[i] = = 0) {int v = RMQ (lh[i], i);               if (H[i]-v >= hehe)      Ans.push_back (i);                } else {int u = RMQ (Lh[i], i), V = RMQ (i, rh[i]);                int maxx = max (U, v);            if (H[i]-maxx >= hehe) ans.push_back (i);            }} for (int i = 0; i < ans.size (); i++) {pt (ans[i]);            if (I==ans.size ()-1) puts ("");        else Putchar ("); }}return 0;}

Uvalive 6531 Go up the Ultras monotone stack +rmq