[NOIP模拟] 区间

---

2017.11.03 T1

Solution

    很好我已经遇见两个作为 T1 比 T2 难的NOIP题，估计是自己太弱了。
    对于每一个询问我们必须做到小于 O(n) 的复杂度才行，我们想想怎么做到，我们考虑打差分，比如对于询问 (x, y) ， 我们对一个 x， 打+1，一个y，打-1，于是我们可以想到前面相同的 sum，可以组成一个平衡区间，于是我们就把x，y的位置提出来，x 和 y 的位置的个数 ≤n 所以我们就得到小于 O(n) 的算法。

Code :

#include <cstdio>#include <cstdlib>#include <cstring>#include <string>#include <algorithm>#include <iostream>#include <cmath>#include <ctime>#include <map>#include <vector>#define mp make_pairusing namespace std;inline int read() {    int i = 0, f = 1;    char ch = getchar();    while(!isdigit(ch)) {        if(ch == '-') f = -1; ch = getchar();    }    while(isdigit(ch)) {        i = (i << 3) + (i << 1) + ch - '0'; ch = getchar();    }    return i * f;}const int MAXN = 8000 + 5;int a[MAXN], b[MAXN], n, k, m, cnt[MAXN * 2], check[MAXN * 2];map<int, int> num;vector<int> p[MAXN];map<pair<int, int>, int> ans;inline void disc_init() {    sort(b + 1, b + n + 1);    m = unique(b + 1, b + n + 1) - b - 1;    for(int i = 1; i <= n; ++i)        a[i] = lower_bound(b + 1, b + m + 1, a[i]) - b, num[b[a[i]]] = a[i];}inline int solve(int x, int y, int l) {     int sum = n; check[sum] = l, cnt[sum] = 0; int pos = 0, tmp = 0;    int i = 0, j = 0, now = 0, mn = 0x3f3f3f3f, mx = (int)0xC0000000;    while(p[x][i] != n + 1 || p[y][j] != n + 1) {        if(p[x][i + 1] <= p[y][j + 1]) {            tmp += (cnt[sum] + cnt[sum] + p[x][i + 1] - pos - 1) * (p[x][i + 1] - pos) / 2;            cnt[sum] += p[x][i + 1] - pos;            pos = p[x][++i]; ++sum;            if(check[sum] != l) check[sum] = l, cnt[sum] = 0;            //printf("%d\n", cnt[sum - 1]);        }        else {            tmp += (cnt[sum] + cnt[sum] + p[y][j + 1] - pos - 1) * (p[y][j + 1] - pos) / 2;            cnt[sum] += p[y][j + 1] - pos;            pos = p[y][++j]; --sum;             if(check[sum] != l) check[sum] = l, cnt[sum] = 0;            //printf("%d\n", cnt[sum + 1]);        }        //printf("%d %d %d %d\n", pos, sum, p[x][i], p[y][j]);    }    //printf("mx = %d mn = %d\n", mx, mn);    ans[mp(x, y)] = tmp;    return tmp;}int main() {    //freopen("1.in", "r", stdin);    n = read(), k = read();    for(int i = 1; i <= n; ++i) a[i] = read(), b[i] = a[i];    disc_init();    for(int i = 0; i <= m; ++i) p[i].push_back(0);    for(int i = 1; i <= n; ++i) p[a[i]].push_back(i);    for(int i = 0; i <= m; ++i) p[i].push_back(n + 1), p[i].push_back(n + 2);    for(int i = 1; i <= k; ++i) {        int x = read(), y = read();        x = num[x], y = num[y];        if(x > y) swap(x, y);        int now = ans[mp(x, y)];        if(!now) cout<<solve(x, y, i)<<'\n';        else cout<<now<<'\n';    }}