2015 多校联赛 ——HDU5372（树状数组）

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Sample Input

30 00 30 150 10 01 10 10 0

Sample Output

Case #1:000Case #2:0102

有0,1两个操作，0 x代表添加从x 到 x + i（带表第i次添加）的线段，每次添加时问被其覆盖的线段有多少。

1 x代表删除第i次添加的。

思路：每一次添加后，求出小于x的左节点个数x1，小于等于y的右节点个数x2。 x2- x1即可

改变单个节点，所以树状数组更加合适

#include<iostream>#include<cstdio>#include<cstring>#include<string>#include<algorithm>using namespace std;#define N 200050#define mod 258280327int le[N],re[N],lr[N],rr[N];int oper[N],Left[N],Right[N],id[N];int c[N];int tot;int lowbit(int x){    return x&-x;}void update(int *c,int n,int k,int v){    while(k <= n)    {        c[k] += v;        k += lowbit(k);    }}int query(int *c ,int k){    int ans = 0;    while(k > 0)    {        ans += c[k];        k -= lowbit(k);    }    return ans;}int main(){    //freopen("4.txt","r",stdin);    int cas=1,L=1,l,ans,nul,nur,n,Q;    while(scanf("%d",&n)!=EOF)    {        Q = nul = nur = 0;        for(int i = 1; i <= n; i++)        {            scanf("%d%d",&oper[i],&l);            if(oper[i] == 0)            {                id[++Q] = i;                le[i] = l;                re[i] = l+L;                Left[nul++] = le[i];                Right[nur++] = re[i];                L++;            }            else            {                le[i] = l;            }        }        printf("Case #%d:\n",cas++);        memset(lr,0,sizeof(lr));        memset(rr,0,sizeof(rr));        sort(Left, Left + nul);        nul = unique(Left, Left + nul) - Left;        sort(Right, Right + nur);        nur = unique(Right, Right + nur) - Right;        for(int i = 1; i <= n; i++)        {            if(oper[i] == 0)            {                int x1 = lower_bound(Left,Left+nul,le[i]) - Left + 1;                int x2 = lower_bound(Right,Right+nur,re[i]) - Right + 1;                ans = query(rr,x2)-query(lr,x1-1);                update(rr,nur,x2,1);                update(lr,nul,x1,1);                printf("%d\n",ans);            }            else            {                int v = id[le[i]];                int x1 = lower_bound(Left,Left+nul,le[v]) - Left + 1;                int x2 = lower_bound(Right,Right+nur,re[v]) - Right + 1;                update(lr,nul,x1,-1);                update(rr,nur,x2,-1);            }        }    }    return 0;}

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