Hihocode #1299 Discount Tickets

Test instructions very simple is to give you two number n and m,n to indicate that there are n tickets, m for M query, the next n lines, two numbers per line, respectively, the time and price of the flight departure, the next M-line, each line of two number represents the most expensive flight of the two hours of the price. If no ticket is required, the output is "None". This question is a typical RMQ problem, which is the interval maximum value query problem. There are two solutions available.

1. The segment tree can be solved and is a bare topic of a line segment tree.

Segment Tree#include <iostream> #include <cstdio> #include <cstring>using namespace Std;const int N =    100005;int result;struct tree{int left;    int right; int value;}    Tree[n*4];void Build (int l, int r, int p) {tree[p].left = l;    Tree[p].right = R;    int mid = (l+r)/2;        if (L = = r) {tree[p].value = 0;    return;    } build (L, Mid, p*2);    Build (Mid+1, R, P*2+1); Tree[p].value = Max (Tree[p*2].value, tree[p*2+1].value);}    void update (int t, int value, int p) {int mid = (tree[p].left + tree[p].right)/2;        if (Tree[p].left = = tree[p].right) {tree[p].value = max (tree[p].value, value);    return;    } if (t<=mid) update (t, value, p*2);    else Update (t, value, p*2+1); Tree[p].value = Max (Tree[p*2].value, tree[p*2+1].value);}    void query (int l, int r, int p) {int mid = (tree[p].left + tree[p].right)/2; if (Tree[p].left = = L && tree[p].right = = r) {result = max (result, tree[p]. value);    return;    } if (r <= mid) query (L, R, P*2);    else if (L > Mid) query (L, R, P*2+1);        else {query (L, Mid, p*2);    Query (mid+1, R, P*2+1);    }}int Main () {int n, m;    scanf ("%d%d", &n, &m);    Build (1, N, 1);        for (int i=0; i<n; i++) {int T, V;        scanf ("%d%d", &t, &v);    Update (t, V, 1);        } for (int i=0; i<m; i++) {int L, R;        result = 0;        scanf ("%d%d", &l, &r);        Query (L, r, 1);        if (result = = 0) cout<< "None" <<endl;    else cout<<result<<endl; } return 0;}

2. The ST algorithm can also be solved.

Algorithm analysis: The preprocessing time complexity is O (N*log (n)), but the complexity of the query is O (1), preprocessing is to use a two-dimensional array data[a][b] to represent the maximum value from a to 2^b, and then use a DP to update all values, and finally query the output results.

RMQ Resolving Interval-Maximum problems # include <iostream> #include <cstring> #include <cstdio> #include <cmath>using namespace Std;const int N = 100005;int input[n];int price[n][40];void rmq (int num) {int tmp = INT (log (num*1.0)/log (2).    0));    for (int i=1; i<=num; i++) price[i][0] = Input[i];                for (int j=1; j<=tmp; j + +) for (int i=1; i<=num; i++) {if (i+ (1<<j) <= num)        PRICE[I][J] = max (price[i][j-1], price[i+ (1<< (j-1))][j-1]);    }}int Main () {int n, m;    scanf ("%d%d", &n, &m);    memset (input, 0, sizeof (input));        for (int i=0; i<n; i++) {int pos, PRI;        scanf ("%d%d", &pos, &pri);    Input[pos] = max (PRI, Input[pos]);    } RMQ (N-1);        for (int i=0; i<m; i++) {int result, l, R;        scanf ("%d%d", &l, &r);        int tmp = log ((r-l+1) *1.0)/log (2.0);        result = Max (price[l][tmp], price[r-(1<<tmp) +1][tmp]);           if (!result) cout<< "None" <<endl;    else cout<<result<<endl; } return 0;}

Hihocode #1299 Discount Tickets