Poj3061 subsequence & poj3320 Jessica's reading problem (ruler acquisition)

Both questions are based on the ruler. The ruler acquisition method is a common technique in the challenge Programming Competition.

It is O (n) scanning the array, and the answer will come out after scanning, this process requires the problem to have this nature: After the head pointer goes forward (s ++, the tail pointer (t) either does not move or moves forward. If this feature is met, the ruler acquisition method can be considered.

Poj3061 is relatively simple. It can also be done in binary mode. time complexity O (N * logn ). You can use the ruler acquisition method O (n) to solve the problem.

# Include <iostream> # include <cstdio> # include <cstdlib> # include <cstring> # include <cmath> # include <map> # include <set> # include <vector> # include <algorithm> # include <stack> # include <queue> # include <cctype> # include <sstream> using namespace STD; # define INF 1000000000 # define EPS 1e-8 # define PII pair <int, int> # define ll long intint T, N, S, A [100010]; int main () {// freopen ("in7.txt", "r", stdin); // freopen ("out.txt", "W", stdout); scanf ("% d ", & T); While (t --) {scanf ("% d", & N, & S); int ans = inf; For (INT I = 0; I <n; I ++) {scanf ("% d", & A [I]);} int S = 0, T = 0, sum = 0; while (1) {While (T <n & sum <s) {sum + = A [T]; t ++;} If (sum <s) break; ans = min (ANS, t-s); // note that the distance here is not (t-s + 1), because the last t ++ in my previous while, so // The range from S to T is left closed right open sum-= A [s]; s ++;} If (ANS <inf) cout <ans <Endl; else cout <'0' <Endl;} // fclose (stdin); // fclose (stdout); Return 0 ;}

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Poj3320 is very convenient to use set and map for data processing, and the core part is the ruler acquisition method.

# Include <iostream> # include <cstdio> # include <cstdlib> # include <cstring> # include <cmath> # include <map> # include <set> # include <vector> # include <algorithm> # include <stack> # include <queue> # include <cctype> # include <sstream> using namespace STD; # define INF 1000000000 # define EPS 1e-8 # define PII pair <int, int> # define ll long intint P, A [1000010]; set <int> st; Map <int, int> MP; int main () {// freopen ("in8.txt", "r", stdin); // freopen ("out.txt", "W", stdout ); scanf ("% d", & P); For (INT I = 0; I <p; I ++) {scanf ("% d ", & A [I]); ST. insert (a [I]);} int Tol = ST. size (); int S = 0, T = 0; int ans = inf; For (;) {While (T <P & MP. size () <TOL) {If (MP. count (A [T]) MP [A [t ++] ++;/* use the count function in map, and 1 is returned, 0 */else MP [A [t ++] = 1;} If (MP. size () <TOL) break; ans = min (ANS, t-s); MP [A [s] --; if (MP [A [s] = 0) MP. erase (A [s]); s ++;} cout <ans <Endl; // fclose (stdin); // fclose (stdout); Return 0 ;}

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Poj3061 subsequence & poj3320 Jessica's reading problem (ruler acquisition)