POJ 3264 Balanced Lineup (simple line segment tree or RMQ)

Language:DefaultBalanced Lineup Time Limit: 5000MS Memory Limit: 65536K Total Submissions: 36833 Accepted: 17252 Case Time Limit: 2000MS Description For the daily milking, Farmer John's n cows (1≤ n ≤50,000) always line up in the same order. One day Farmer John decides to organize a game of Ultimate frisbee with some of the cows. To keep things simple, he'll take a contiguous range of cows from the milking lineup to play the game. However, for all the cows to has fun they should not differ too much in height. Farmer John has made a list of Q (1≤ q ≤200,000) Potential groups of cows and their heights (1≤ H Eight ≤1,000,000). For each group, he wants your help to determine the difference in height between the shortest and the tallest cow in the G Roup. Input Line 1:two space-separated integers,NandQ. Lines 2..N+1:lineI+1 contains a single integer which is the height of cowI LinesN+2..N+Q+1:two integersAandB(1≤A≤B≤N), representing the range of cows fromAToBInclusive. Output Lines 1..Q: Each line contains a single integer so is a response to a reply and indicates the difference in height between the Tal Lest and shortest cow in the range. Sample Input 6 31734251 54) 62 2 Sample Output 630 Source Usaco January Silver

To find the maximum interval minimization

Line segment Tree Code:

#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include < cmath> #include <queue> #include <stack> #include <vector> #include <set> #include <map > #define L (x) (x<<1) #define R (x) (x<<1|1) #define MID (x, y) ((x+y) >>1) #define EPS 1e-8typedef __ Int64 ll;using namespace std; #define N 50005int n,m,a[n];struct stud{int le,ri; int mi,ma;} f[n*4];void Build (int pos,int le,int ri) {f[pos].le=le;f[pos].ri=ri;if (Le==ri) {F[pos].mi=f[pos].ma=a[le];return;} int Mid=mid (LE,RI), Build (L (POS), Le,mid), Build (R (POS), Mid+1,ri), F[pos].ma=max (F[l (POS)].ma,f[r (POS)].ma); F[pos]. Mi=min (F[l (POS)].mi,f[r (POS)].MI);} int querymin (int pos,int le,int ri) {if (F[pos].le>=le&&f[pos].ri<=ri) return f[pos].mi; int Mid=mid (F[pos ].le,f[pos].ri); if (Mid>=ri) return Querymin (L (POS), Le,ri); if (Mid<le) return Querymin (R (POS), Le,ri); return min (Querymin (L (POS), Le,mid), Querymin (R (POS), Mid+1,ri));} int Querymax (int pos,int le,inT ri) {if (F[pos].le>=le&&f[pos].ri<=ri) return f[pos].ma; int Mid=mid (F[POS].LE,F[POS].RI); if (mid>= RI) return Querymax (L (POS), Le,ri); if (Mid<le) return Querymax (R (POS), Le,ri); Return Max (Querymax (L (POS), Le,mid), Querymax (R (POS), Mid+1,ri));}    int main () {#ifndef Online_judge freopen ("In.txt", "R", stdin); #endif//Online_judgeint I,j;while (~scanf ("%d%d", &n,&m)) {for (i=1;i<=n;i++) scanf ("%d", &a[i]); build        (1,1,n); int Le,ri;while (m--) {scanf ("%d%d", &le,&ri);p rintf ("%d\n", Querymax (1,le,ri)-querymin (1,le,ri));}} return 0;}

RMQ Code:

#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include < cmath> #include <queue> #include <stack> #include <vector> #include <set> #include <map > #define L (x) (x<<1) #define R (x) (x<<1|1) #define MID (x, y) ((x+y) >>1) #define EPS 1e-8typedef __ Int64 ll;using namespace std; #define N 50005int dpmin[n][20];int dpmax[n][20];int n,m,a[n];void inint () {int i,j;for (i=1;  i<=n;i++) dpmin[i][0]=dpmax[i][0]=a[i];for (j=1; (1<<j) <=n+1;j++) for (i=1;i+ (1<<j) -1<=n;i++) {  Dpmin[i][j]=min (dpmin[i][j-1],dpmin[i+ (1<< (j-1))][j-1]); Dpmax[i][j]=max (dpmax[i][j-1],dpmax[i+ (1<< (j-1))][j-1]);}} inline int getmax (int le,int ri) {int k= (int) (log (ri-le+1+0.0)/log (2.0)); return Max (dpmax[le][k],dpmax[ri-(1<<k ) +1][k]);} inline int getmin (int le,int ri) {int k= (int) (log (ri-le+1+0.0)/log (2.0)), Return min (dpmin[le][k],dpmin[ri-(1<<k ) +1][k]);} int main () {//#ifndef online_judge//freopen ("In.txt", "R", sTdin);//#endif//Online_judge int i,j;while (~scanf ("%d%d", &n,&m)) {for (i=1;i<=n;i++) scanf ("%d", &a[i ]); Inint (); int Le,ri;while (m--) {scanf ("%d%d", &le,&ri);p rintf ("%d\n", Getmax (Le,ri)-getmin (Le,ri));}} return 0;}

POJ 3264 Balanced Lineup (simple line segment tree or RMQ)