POJ2104Kth Number

整体二分模板题, 有些细节需要注意

#include<cstdio>
#include<cctype>
#include<climits>
#include<algorithm>
#include<cstring>
using namespace std;
inline int read()
{
    int x = 0, flag = 1;
    char c;
    while(! isgraph(c = getchar()))
        if(c == '-')
            flag *= - 1;
    while(isgraph(c))
        x = x * 10 + c - '0', c = getchar();
    return x * flag;
}
void println(int x)
{
    if(x < 0)
        putchar('-'), x *= - 1;
    if(x == 0)
        putchar('0');
    int ans[10 + (1 << 4)], top = 0;
    while(x)
        ans[top ++] = x % 10, x /= 10;
    for(; top; top --)
        putchar(ans[top - 1] + '0');
    putchar('\n');
}
const int MAXN = (int)1e5 + (1 << 5), MAXM = (int)5e4 + (1 << 4);
const int oo = INT_MAX;
int a[MAXN];
struct query
{
    int L, R, k, ID;
    query(){}
    query(int L, int R, int k, int ID): L(L), R(R), k(k), ID(ID){}
}Q[MAXN];
int res[MAXN], sum[MAXN];
int nxt[MAXN][2], pre[MAXN][2];
int _a[MAXN], map[MAXN];
query _Q[MAXM];
int ans[MAXM];
void solve(int mn, int mx, int L, int R, int QL, int QR)
{
    if(mn >= mx)
    {
        for(int i = QL; i <=QR; i ++)
            ans[Q[i].ID] = mn;
        return;
    }
    int cur = (mn + mx) >> 1;
    for(int i = L; i <= R; i ++)
        res[i] = (a[i] <= cur);
    sum[L] = res[L];
    for(int i = L + 1; i <= R; i ++)
        sum[i] = sum[i - 1] + res[i];
    pre[L][res[L]] = L, pre[L][res[L] ^ 1] = - oo;
    for(int i = L + 1; i <= R; i ++)
        pre[i][res[i]] = i, pre[i][res[i] ^ 1] = pre[i - 1][res[i] ^ 1];
    nxt[R][res[R]] = R, nxt[R][res[R] ^ 1] = oo;
    for(int i = R - 1; i >= L; i --)
        nxt[i][res[i]] = i, nxt[i][res[i] ^ 1] = nxt[i + 1][res[i] ^ 1];
    int Ltop = L, Rtop = R;
    for(int i = L; i <= R; i ++)
    {
        if(res[i])
            _a[Rtop] = a[i], map[i] = Rtop --;
        else
            _a[Ltop] = a[i], map[i] = Ltop ++;
    }
    int mid = Rtop;
    for(int i = L; i <= R; i ++)
        a[i] = _a[i];
    Ltop = QL, Rtop = QR;
    for(int i = QL; i <= QR; i ++)
    {
        int cnt = sum[Q[i].R] - sum[Q[i].L - 1];
        if(cnt < Q[i].k)
        {
            Q[i].L = map[nxt[Q[i].L][0]];
            Q[i].R = map[pre[Q[i].R][0]];
            Q[i].k -= cnt;  //当 cnt < k 时要在k中减去cnt 
            if(Q[i].L > Q[i].R)
                swap(Q[i].L, Q[i].R); //记得要判断是否需要调换顺序 
            _Q[Ltop ++] = Q[i];
        }
        else if (cnt >= Q[i].k) //当 cnt >= k时则不需要减去cnt 
        {
            Q[i].L = map[nxt[Q[i].L][1]];
            Q[i].R = map[pre[Q[i].R][1]];
            if(Q[i].L > Q[i].R)
                swap(Q[i].L, Q[i].R);
            _Q[Rtop --] = Q[i];
        }
    }
    for(int i = QL; i <= QR; i ++)
        Q[i] = _Q[i];
    for(int i = L;  i <= R; i ++)
        sum[i] = 0;
    solve(cur + 1, mx, L, mid, QL, Rtop);
    solve(mn, cur, mid + 1, R, Ltop, QR);
}
int main()
{
    #ifndef ONLINE_JUDGE
    freopen("POJ2104.in", "r", stdin);
    freopen("POJ2104.out", "w", stdout);
    #endif
    int n = read();
    int m = read();
    int mx = - oo, mn = oo;
    for(int i = 1; i <= n; i ++)
        a[i] = read(), mx = max(mx, a[i]), mn = min(mn, a[i]);
    for(int i = 1; i <= m; i ++)
    {
        int L = read(), R = read(), k = read();
        Q[i] = query(L, R, k, i);
    }
    memset(sum, 0, sizeof(sum));
    solve(mn, mx, 1, n, 1, m);
    for(int i = 1; i <= m; i ++)
        println(ans[i]);
}