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 = 3 * 1e6 + 10, INF = 1e9 + 7, mod = 998244353;
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}
int N, M;
int l[MAXN], r[MAXN], a[MAXN], date[MAXN], cnt = 0, sum[MAXN], num[MAXN];
vector<int> v[MAXN];
int fp(int a, int p) {
    int base = 1;
    while(p) {
        if(p & 1) base = (base * a) % mod;
        a = (a * a) % mod; p >>= 1;
    }
    return base % mod;
}
main() {
    N = read(); M = read();
    for(int i = 1; i <= N; i++) l[i] = read(), r[i] = read(), date[++cnt] = l[i], date[++cnt] = r[i];
    for(int i = 1; i <= M; i++) a[i] = read(), date[++cnt] = a[i];
    sort(a + 1, a + M + 1);
    sort(date + 1, date + cnt + 1);
    cnt = unique(date + 1, date + cnt + 1) - date - 1;
    for(int i = 1; i <= N; i++) {
        l[i] = lower_bound(date + 1, date + cnt + 1, l[i]) - date;
        r[i] = lower_bound(date + 1, date + cnt + 1, r[i]) - date;
        sum[l[i]]++; sum[r[i] + 1]--; 
        v[l[i]].push_back(r[i]);
    }
    for(int i = 1; i <= M; i++)  a[i] = lower_bound(date + 1, date + cnt + 1, a[i]) - date;
    for(int i = 1; i <= cnt; i++) 
        sum[i] += sum[i - 1];
    
   // for(int i = 1; i <= cnt; i++)
    //    printf("%d ", num[i]); puts("");
    int ans = 0;
    for(int i = 1; i <= M; i++) {
           int base = 0;
        for(int j = a[i - 1] + 1; j <= a[i]; j++) {
            for(int k = 0; k < v[j].size(); k++) {
                if(v[j][k] >= a[i]) base++;
            }
        }
        ans = (ans + fp(2, sum[a[i]]) - fp(2, sum[a[i]] - base) + mod) % mod;
    }
    printf("%lld
", (ans + mod) % mod);
    return 0;
}
/*
3 2
7 11
1 5
3 8
4
7
*/