UVA 12663 (线段树区间更新+二分查找,Q次区间更新后,求满足结果大于K的个数)_线段树 区间大于k的个数

看见这种区间更新的题目，首先想到的就是线段树的区间更新。本体套用线段树的区间更新模板，更新的区间要二分去找。

#include<stdio.h>
#include<string.h>
#include<algorithm>
#define MAX 200009
#define LL long long int
using namespace std;
LL a[MAX],cnt;
struct Node {
    LL L;
    LL R;
    Node *pL;
    Node *pR;
    LL sum;
    LL add;
} tree[4*MAX];
LL n,m,k;
LL ans;
LL mid(struct Node *root) {
    return (root->L+root->R)/2;
}
LL Search(LL low,LL high,LL goal) {
    LL mid;
    while(low<=high){
        mid=(low+high)/2;
        if(goal<=a[mid]){
            high=mid-1;
        }else{
            low=mid+1;
        }
    }
    return low;
}
void build(struct Node *root,LL L,LL R) {
    root->L=L;
    root->R=R;
    root->add=0;
    if(L==R) return;
    LL midd=(L+R)/2;
    cnt++;
    root->pL=tree+cnt;
    build(root->pL,L,midd);
    cnt++;
    root->pR=tree+cnt;
    build(root->pR,midd+1,R);
}
void add(struct Node *root, LL L, LL R, LL val) {
    if(root->L==L&&root->R==R) {
        root->add+=val;
        return;
    }
    LL midd=mid(root);
    if(R<=midd)add(root->pL, L, R, val);
    else if(L>midd) add(root->pR, L, R, val);
    else {
        add(root->pL, L, mid(root), val);
        add(root->pR, mid(root)+1, R,val);
    }
}
LL query(struct Node *root, LL L, LL R){
    if(L == R){
        if(root->add>=k) ans ++;
        return 0;
    }
    LL midd=mid(root);
    add(root->pL, root->L, midd, root->add);
    add(root->pR, midd+1, root->R, root->add);
    root->add=0;
    if(R<=midd) return query(root->pL, L, R);
    else if(L>midd) return query(root->pR, L, R);
    else return query(root->pL, L, midd)+query(root->pR, midd+1, R);
}
int main() {
    int t=1;
    while(~scanf("%lld %lld %lld",&n,&m,&k)) {
        cnt=0;
        ans=0;
        for(LL i=1; i<=n; i++) {
            scanf("%lld",&a[i]);
        }
        sort(a+1,a+n+1);
        build(tree,1,n);
        LL first=2,second;
        for(LL i=0; i<m; i++) {
            scanf("%lld",&second);
            LL firsttmp=lower_bound(a + 1, a + n + 1, first) - a;
            LL secondtmp = upper_bound(a + 1, a + n + 1, second) - a - 1;
            if(firsttmp>secondtmp) continue;
//            if(secondtmp>n) secondtmp=n;
            scanf("%lld",&first);
            if(secondtmp < firsttmp) continue;
            add(tree,firsttmp,secondtmp,1);
            first++;
        }
        query(tree,1,n);
        printf("Case %d: %lld\n",t++, ans);
    }
    return 0;
}
/*
10 10 2
2 4 6 9 6 2 4 3 19 39
5 2
6 3
9 2
8 3
4 3
9 6
8 2
Case 1: 6
9 8 2
3 7 2 9 8 4 3 2 1
5 2
6 3
9 2
8 4
9 2
10 6
46 5
10 3
Case 2: 6
4 2 2
2 2 2 2
5 2
7 2
Case 3: 0
*/

还有一种方法，叫做差分，这适用于Q次更新，但是只有最后一次查询的状况，每次查询的时间复杂度是O(n).

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
#define MAX 100009
#define inf 100000009
int a[MAX],num[MAX],ans[MAX];
int main(){
    int n,m,k,t=1;
    while(~scanf("%d %d %d",&n,&m,&k)){
        memset(num,0,sizeof(num));
        for(int i=1;i<=n;i++){
            scanf("%d",&a[i]);
        }
        sort(a+1,a+n+1);
        int first=2,second;
        for(int i=0;i<m;i++){
            scanf("%d",&second);
            int firsttmp=lower_bound(a+1,a+n+1,first)-a;
            int secondtmp=upper_bound(a+1,a+1+n,second)-a-1;
            num[firsttmp]++;
            num[secondtmp+1]--;
            scanf("%d",&first);
            first++;
        }
        int cnt=0;
        ans[0]=0;
        for(int i=1;i<=n;i++){
            ans[i]=ans[i-1]+num[i];
            if(ans[i]>=k) cnt++;
        }
        printf("Case %d: %d\n",t++, cnt);
    }
return 0;
}