Balanced Lineup POJ - 3264
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 will take a contiguous range of cows from the milking lineup to play the game. However, for all the cows to have 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 ≤ height ≤ 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 group.
Input
Line 1: Two space-separated integers, N and Q.
Lines 2.. N+1: Line i+1 contains a single integer that is the height of cow i
Lines N+2.. N+ Q+1: Two integers A and B (1 ≤ A ≤ B ≤ N), representing the range of cows from A to Binclusive.
Lines 2.. N+1: Line i+1 contains a single integer that is the height of cow i
Lines N+2.. N+ Q+1: Two integers A and B (1 ≤ A ≤ B ≤ N), representing the range of cows from A to Binclusive.
Output
Lines 1.. Q: Each line contains a single integer that is a response to a reply and indicates the difference in height between the tallest and shortest cow in the range.
Sample Input
6 3 1 7 3 4 2 5 1 5 4 6 2 2
Sample Output
6 3 0
解一：线段树
#include <iostream> #include <algorithm> #include <cstring> #include <string> #include <vector> #include <map> #include <set> #include <list> #include <deque> #include <queue> #include <stack> #include <cstdlib> #include <cstdio> #include <cmath> #include <iomanip> #define ull unsigned long long #define ll long long #define pb push_back #define mem(sum,x) memset(sum,x,sizeof(sum)) #define rep(i,start,end) for(int i=start;i<=end;i++) #define per(i,end,start) for(int i=end;i>=start;i--) #define tle ios::sync_with_stdio(false); cin.tie(0); cout.tie(0); using namespace std; const int mod = 998244353; const int mxn = 1e5 +7; int _,n,m,t,u,v,ans,cnt,ok,lim; ll w[mxn] , cost[mxn] , dp[mxn] , far[mxn] , siz[mxn]; double num[mxn]; #define lc now<<1 #define rc now<<1|1 string str ; struct node {int l,r,val,lazy,mx,mn;}p[mxn<<4]; void pushup(int now){ /// p[now].val = p[lc].val+p[rc].val; p[now].mx = max(p[lc].mx , p[rc].mx); p[now].mn = min(p[lc].mn , p[rc].mn); } void pushdown(int now) { if(p[now].lazy) { p[lc].lazy = p[rc].lazy = p[now].lazy; p[lc].val = p[now].lazy*(p[lc].r-p[lc].l+1); p[rc].val = p[now].lazy*(p[rc].r-p[rc].l+1); p[now].lazy = 0 ; } } void build(int l,int r,int now) { p[now].l = l,p[now].r = r ;/// p[now].lazy = 0 ; if(l==r){ /// cin>>p[now].val; scanf("%d",&p[now].val); p[now].mx = p[now].mn = p[now].val ; return ; } int mid = (l+r)>>1; build(l,mid,lc);build(mid+1,r,rc);pushup(now); } void ask(int l,int r,int now) { if(l<=p[now].l && r>=p[now].r){ u = max(u,p[now].mx); v = min(v,p[now].mn); return ; } int mid = (p[now].l+p[now].r)>>1; if(l<=mid) ask(l,r,lc); if(r>mid) ask(l,r,rc); } int main() { tle; ///while(cin>>n>>m) while(~scanf("%d %d",&n,&m)) { build(1,n,1); int l,r; while(m--) { /// cin>>l>>r; scanf("%d %d",&l,&r); u = -1 , v = mod ;ask(l,r,1); printf("%d ",u-v); /// cout<<u-v<<endl; } } }
解二：RMQ
1 #include <iostream> 2 #include <algorithm> 3 #include <cstring> 4 #include <string> 5 #include <vector> 6 #include <map> 7 #include <set> 8 #include <list> 9 #include <deque> 10 #include <queue> 11 #include <stack> 12 #include <cstdlib> 13 #include <cstdio> 14 #include <cmath> 15 #include <iomanip> 16 #define ull unsigned long long 17 #define ll long long 18 #define pb push_back 19 #define mem(sum,x) memset(sum,x,sizeof(sum)) 20 #define rep(i,start,end) for(int i=start;i<=end;i++) 21 #define per(i,end,start) for(int i=end;i>=start;i--) 22 #define tle ios::sync_with_stdio(false); cin.tie(0); cout.tie(0); 23 using namespace std; 24 const int mod = 998244353; 25 const int mxn = 1e5 +7; 26 int _,n,m,t,u,v,ans,cnt,ok,lim; 27 ll w[mxn] , cost[mxn] , far[mxn] , siz[mxn]; 28 double num[mxn]; 29 #define lc now<<1 30 #define rc now<<1|1 31 string str ; 32 struct node {int l,r,val,lazy,mx,mn;}p[mxn<<4]; 33 void pushup(int now){ 34 /// p[now].val = p[lc].val+p[rc].val; 35 p[now].mx = max(p[lc].mx , p[rc].mx); 36 p[now].mn = min(p[lc].mn , p[rc].mn); 37 } 38 void pushdown(int now) 39 { 40 if(p[now].lazy) 41 { 42 p[lc].lazy = p[rc].lazy = p[now].lazy; 43 p[lc].val = p[now].lazy*(p[lc].r-p[lc].l+1); 44 p[rc].val = p[now].lazy*(p[rc].r-p[rc].l+1); 45 p[now].lazy = 0 ; 46 } 47 } 48 void build(int l,int r,int now) 49 { 50 p[now].l = l,p[now].r = r ;/// p[now].lazy = 0 ; 51 if(l==r){ 52 /// cin>>p[now].val; 53 scanf("%d",&p[now].val); 54 return ; 55 } 56 int mid = (l+r)>>1; 57 build(l,mid,lc);build(mid+1,r,rc);pushup(now); 58 } 59 int dp[mxn][20][2]; 60 void rmq() 61 { 62 for(int j=1;j<=log((double)n)/log(2.0)+1;j++) 63 { 64 for(int i=1;i+(1<<j)-1<=n;i++) 65 { 66 dp[i][j][0] = max(dp[i][j-1][0] , dp[i+(1<<(j-1))][j-1][0] ); 67 dp[i][j][1] = min(dp[i][j-1][1] , dp[i+(1<<(j-1))][j-1][1] ); 68 } 69 } 70 } 71 int main() 72 { 73 tle; 74 while(~scanf("%d %d",&n,&m)) 75 { 76 for(int i=1,k;i<=n;i++) 77 scanf("%d",&k) ,dp[i][0][0] = dp[i][0][1] = k ; 78 rmq(); 79 int l,r; 80 while(m--) 81 { 82 /// cin>>l>>r; 83 scanf("%d %d",&l,&r); 84 int k = log((double)(r-l+1))/log(2.0); 85 printf("%d ", max(dp[l][k][0],dp[r-(1<<k)+1][k][0]) - min(dp[l][k][1],dp[r-(1<<k)+1][k][1]) ); 86 /// cout<<u-v<<endl; 87 } 88 } 89 }