题意
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 */