去年多校的時候用最暴力的辦法水過了,忽然看到,覺得還是得用正統的方式過了.............
#include <iostream>#include <algorithm>#include <cmath>#include<functional>#include <cstdio>#include <cstdlib>#include <cstring>#include <string>#include <vector>#include <set>#include <queue>#include <stack>#include <climits>//形如INT_MAX一類的#define MAX 100050#define INF 0x7FFFFFFF#define L(x) x<<1#define R(x) x<<1|1using namespace std;inline void RD(int &ret) { char c; do { c = getchar(); } while(c < '0' || c > '9') ; ret = c - '0'; while((c=getchar()) >= '0' && c <= '9') ret = ret * 10 + ( c - '0' );}void OT(int a) { if(a >= 10)OT(a / 10); putchar(a % 10 + '0');}int n,m,cnt,size;int x[MAX * 3],xx[MAX * 3],le[MAX],ri[MAX],qes[MAX];struct Node { int l,r,mid; int add,v;} tr[MAX * 4];void up(int x) { tr[x].v = tr[L(x)].v + tr[R(x)].v;}void down(int x) { if(tr[x].add != 0) { int l,r; tr[L(x)].add += tr[x].add; tr[R(x)].add += tr[x].add; l = tr[L(x)].l; r = tr[L(x)].r; tr[L(x)].v += (r - l + 1) * tr[x].add; //tr[L(x)].v = (r - l + 1) * tr[L(x)].add; l = tr[R(x)].l; r = tr[R(x)].r; tr[R(x)].v += (r - l + 1) * tr[x].add; //tr[R(x)].v = (r - l + 1) * tr[R(x)].add; tr[x].add = 0; }}void build(int l,int r,int x) { tr[x].l = l; tr[x].r = r; tr[x].mid = (l + r) >> 1; tr[x].v = 0; tr[x].add = 0; if(l == r) return ; build(l,tr[x].mid,L(x)); build(tr[x].mid + 1,r,R(x));}void update(int l,int r,int x,int va) { if(l <= tr[x].l && r >= tr[x].r) { tr[x].add += va; tr[x].v += (tr[x].r - tr[x].l + 1) * va; //tr[x].v = (tr[x].r - tr[x].l + 1) * tr[x].add; return ; } down(x); int mid = tr[x].mid; if(r > mid) update(l,r,R(x),va); if(l <= mid) update(l,r,L(x),va); up(x);}int ans = 0;void query(int l,int x) { if(tr[x].l == tr[x].r) { ans += tr[x].v; return ; } down(x); int mid = tr[x].mid; if(l > mid) query(l,R(x)); if(l <= mid) query(l,L(x));}int main() { int T; cin >> T; int ca = 1; while(T--) { RD(n); RD(m); cnt = 0; for(int i=0; i<n; i++) { RD(le[i]); RD(ri[i]); x[cnt++] = le[i]; x[cnt++] = ri[i]; } for(int i=0; i<m; i++) { RD(qes[i]); x[cnt++] = qes[i]; } sort(x,x+cnt); size = unique(x,x+cnt) - x; build(1,size,1); for(int i=0; i<n; i++) { int ll = lower_bound(x,x+size,le[i]) - x + 1; int rr = lower_bound(x,x+size,ri[i]) - x + 1; update(ll,rr,1,1); } printf("Case #%d:\n",ca++); for(int i=0; i<m; i++) { int q = lower_bound(x,x+size,qes[i]) - x + 1; ans = 0; query(q,1); OT(ans); puts(""); } } return 0;}