ZR#330. 【18 提高 3】矿石(容斥)

题意

题目链接

Sol

挺显然的，首先对每个矿排序

那么答案就是$2^x - 2^y$

$x$表示能覆盖到它的区间，$y$表示的是能覆盖到它且覆盖到上一个的区间

第一个可以差分维护

第二个直接vector暴力插入扫就行，

时间复杂度：$O(nlogn)$

#include<cstdio>
#include<algorithm>
#include<bitset>
#include<vector>
#define Pair pair<int, ull>
#define MP(x, y) make_pair(x, y)
#define fi first
#define se second
#define ull unsigned long long
#define LL long long
#define int long long
using namespace std;
const int MAXN =  * 1e6 + , INF = 1e9 + , mod = ;
inline int read() {
    char c = getchar(); int x = , f = ;
    while(c < '' || c > '') {if(c == '-') f = -; c = getchar();}
    while(c >= '' && c <= '') x = x *  + c - '', c = getchar();
    return x * f;
}
int N, M;
int l[MAXN], r[MAXN], a[MAXN], date[MAXN], cnt = , sum[MAXN], num[MAXN];
vector<int> v[MAXN];
int fp(int a, int p) {
    int base = ;
    while(p) {
        if(p & ) base = (base * a) % mod;
        a = (a * a) % mod; p >>= ;
    }
    return base % mod;
}
main() {
    N = read(); M = read();
    for(int i = ; i <= N; i++) l[i] = read(), r[i] = read(), date[++cnt] = l[i], date[++cnt] = r[i];
    for(int i = ; i <= M; i++) a[i] = read(), date[++cnt] = a[i];
    sort(a + , a + M + );
    sort(date + , date + cnt + );
    cnt = unique(date + , date + cnt + ) - date - ;
    for(int i = ; i <= N; i++) {
        l[i] = lower_bound(date + , date + cnt + , l[i]) - date;
        r[i] = lower_bound(date + , date + cnt + , r[i]) - date;
        sum[l[i]]++; sum[r[i] + ]--;
        v[l[i]].push_back(r[i]);
    }
    for(int i = ; i <= M; i++)  a[i] = lower_bound(date + , date + cnt + , a[i]) - date;
    for(int i = ; i <= cnt; i++)
        sum[i] += sum[i - ];

   // for(int i = 1; i <= cnt; i++)
    //    printf("%d ", num[i]); puts("");
    int ans = ;
    for(int i = ; i <= M; i++) {
           int base = ;
        for(int j = a[i - ] + ; j <= a[i]; j++) {
            for(int k = ; k < v[j].size(); k++) {
                if(v[j][k] >= a[i]) base++;
            }
        }
        ans = (ans + fp(, sum[a[i]]) - fp(, sum[a[i]] - base) + mod) % mod;
    }
    printf("%lld\n", (ans + mod) % mod);
    return ;
}
/*
3 2
7 11
1 5
3 8
4
7
*/