Topic 1: Discounted tickets (Hihocoder 20 Challenge)

Time limit:10000msSingle Point time limit:1000msMemory Limit:256MB

Describe

Because Miss Shinjuku's "little Sister", the island Niang plans to go to Tokyo in June, but this time it seems she needs to pay her own pocket. After a few days of battle taking, the island Niang finally exhausted, with Python grabbed all June, Shanghai to Tokyo, all n tickets. Now, please help the debt-laden island Niang to screen out the most expensive tickets that meet the requirements of the time range.

Input

The first line of the input data contains two integers n . m (1?≤ n ,? m ? ≤?105), indicating the total number of tickets, and the total number of inquiries. The next n line, two integers per line t ,? v (1?≤ t ,? v ? ≤?105), indicating the time and price per ticket departure. The next m line, two integers per line a ,? b (1?≤?a?≤?b?≤?105), which indicates the time interval required for each inquiry.

Output

For each set of queries, the output line represents the most expensive price. If there is no ticket that meets the requirements, the output line is "None".

Sample input

7 61 12 14 34 44 56 97 91 71 26 73 34 45 5

Sample output

919none5none

Analysis: It is obvious that the segment tree interval to find the most value, the difference: there is discontinuity point special treatment

/*7-5919none5none*/#include <cstdio> #include <cstring> #include <vect---- or> #include <iostream> #include <algorithm> #define INF-0XFCONST int maxn=1000005;using namespace STD;    struct node{int id; int price;};    int segtree[maxn];int a[maxn];void Build (int node, int b, int e) {if (b = = e) Segtree[node] = a[b];        else {Build (2*node, B, (b+e)/2);        Build (2*node+1, (b+e)/2+1, E);        if (SEGTREE[2 * node] >= segtree[2 * node + 1]) Segtree[node] = segtree[2 * node];    else Segtree[node] = segtree[2 * node + 1];    }}int Query (int node,int b,int e,int left,int right) {if (B>=left && e<=right) return Segtree[node];    int m= (b+e)/2;    int sum=0;    if (left<=m) Sum=max (Sum,query (node<<1,b,m,left,right));    if (right>m) Sum=max (Sum,query (node<<1|1,m+1,e,left,right)); return sum;}    int main () {int n,m; scanf ("%d%d", &n,&m);   int a1,a2;    for (int i=1;i<=n;i++) A[i]=inf;        for (int i=1;i<=n;i++) {scanf ("%d%d", &AMP;A1,&AMP;A2);        if (a[a1] = = INF) a[a1]=a2;    else A[a1]=max (A[A1],A2);    } Build (1,1,n);        for (int i=1;i<=m;i++) {scanf ("%d%d", &AMP;A1,&AMP;A2);        int Ans=query (1,1,N,A1,A2);    if (ans = = 0) printf ("none\n");//Special for Else printf ("%d\n", ans); } return 0;}

Topic 1: Discounted tickets (Hihocoder 20 Challenge)