HDU 5196 DZY Loves Inversions(树状数组,二分)

这题之前CC做过类似的,思路如官方题解。

代码:

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;

const int N = 100005;
typedef long long ll;
int n, m;
ll q;
struct Hash {
    int v, id;
    void read(int id) {
        scanf("%d", &v);
        this->id = id;
    }
} h[N];

bool cmp(Hash a, Hash b) {
    return a.v < b.v;
}

int a[N];

int bit[N], l[N];

inline int lowbit(int x) {return (x&(-x));}

void add(int x, int v) {
    while (x <= n) {
        bit[x] += v;
        x += lowbit(x);
    }
}

int get(int x) {
    int ans = 0;
    while (x) {
        ans += bit[x];
        x -= lowbit(x);
    }
    return ans;
}

ll sum[N], sum2[N];
ll ans[N];
int X[N], Y[N];

void gao(ll q, int bo) {
    if (q < 0) return;
    memset(bit, 0, sizeof(bit));
    int r = 0;
    ll tot = 0;
    for (int i = 1; i <= n; i++) {
        while (r <= n && tot <= q) {
            r++;
            if (r > n) break;
            tot += (r - i) - get(a[r]);
            add(a[r], 1);
        }
        l[i] = r - 1;
        add(a[i], -1);
        tot -= get(a[i] - 1);
    }
    sum[n + 1] = 0;
    for (int i = n; i >= 1; i--)
        sum[i] = sum[i + 1] + (l[i] - i + 1);
    int x, y;
    for (int i = 0; i < m; i++) {
        int x = X[i], y = Y[i];
        int tmp = lower_bound(l + x, l + y + 1, y) - l - 1;
        ans[i] += bo * (sum[x] - sum[tmp + 1] + sum2[y - tmp]);
    }
}

int main() {
    for (int i = 1; i < N; i++)
        sum2[i] = sum2[i - 1] + i;
    while (~scanf("%d%d%I64d", &n, &m, &q)) {
        for (int i = 1; i <= n; i++)
            h[i].read(i);
        sort(h + 1, h + n + 1, cmp);
        int tmp = 1;
        a[h[1].id] = 1;
        for (int i = 1; i <= n; i++) {
            if (h[i].v != h[i - 1].v) tmp++;
            a[h[i].id] = tmp;
        }
        for (int i = 0; i < m; i++) scanf("%d%d", &X[i], &Y[i]);
        memset(ans, 0, sizeof(ans));
        gao(q, 1); gao(q - 1, -1);
        for (int i = 0; i < m; i++)
            printf("%I64d\n", ans[i]);
    }
    return 0;
}


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