hdu 5372 Segment Game 【 树状数组 】

给出一些操作，

0是将第i次增加的线段放在b位置，第i次放的线段的长度为i

1是将第b次增加操作放的线段删除

每次增加操作完之后，询问这条线段上面的完整的线段的条数

每次询问统计比这条线段左端点大的线段的条数 L，比这条线段右端点大的线段的条数 R,两个相减就是完整的线段的条数

另外因为给的b很大，所以需要离散化一下，而且b可能会相同，所以相同大小的应该占据一个编号

然后就像求逆序对那样的算

#include<iostream>
#include<cstdio>
#include<cstring>
#include <cmath>
#include<stack>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<algorithm>
using namespace std;

typedef long long LL;
const int INF = (1<<30)-1;
const int mod=1000000007;
const int maxn=500005;

int n;
int a[maxn],c[maxn],b[maxn];

int lowbit(int x){ return x &(-x);}

int sum1(int x){
int ret =0;
while(x>0){
ret+=c[x];x-=lowbit(x);
}
return ret;
}

void add1(int x,int d){
while(x<=maxn){
c[x]+=d;x+=lowbit(x);
}
}

int sum2(int x){
int ret =0;
while(x>0){
ret+=b[x];x-=lowbit(x);
}
return ret;
}

void add2(int x,int d){
while(x<=maxn){
b[x]+=d;x+=lowbit(x);
}
}

struct node{
int x,y;
int l,r;
int lb,ub;
int id;
int idx;
};

int cmp1(node n1,node n2){
return n1.l < n2.l;
}

int cmp2(node n1,node n2){
return n1.r < n2.r;
}

node p[maxn],cmd[maxn],add[maxn];

int main(){
int q;
int kase = 0;
while(scanf("%d",&q) != EOF){
memset(a,0,sizeof(a));
memset(b,0,sizeof(b));
memset(c,0,sizeof(c));

int cnt = 1;
for(int i = 1;i <= q;i++){
scanf("%d %d",&cmd[i].x,&cmd[i].y);
if(cmd[i].x == 0){
cmd[i].id = cnt;
add[cnt].l = cmd[i].y; p[cnt].l = cmd[i].y;
add[cnt].r = add[cnt].l + cnt;p[cnt].r = add[cnt].l + cnt;
add[cnt].idx = cnt;p[cnt].idx = cnt;
cnt++;
}
}
printf("Case #%d:\n",++kase);

sort(p+1,p+cnt,cmp1);
for(int i=1,j=0;i< cnt;i++){
if(i==1||p[i].l != p[i-1].l) j++;
add[p[i].idx].lb=j;
}

sort(p+1,p+cnt,cmp2);
for(int i=1,j=0;i< cnt;i++){
if(i==1||p[i].r != p[i-1].r) j++;
add[p[i].idx].ub=j;
}

for(int i = 1;i <= q;i++){
if(cmd[i].x == 0){
int k = cmd[i].id;
int c1 = add[k].lb;
int c2 = add[k].ub;
int l = (k-1) - sum1(c1 -1);
int r = (k-1)- sum2(c2 );
int ans = abs(l-r);
printf("%d\n",ans);
add1(c1,1);add2(c2,1);
}
if(cmd[i].x == 1){
int k = cmd[i].y;
int c1 = add[k].lb;
int c2 = add[k].ub;
int d1 = sum1(c1) - sum1(c1-1);
if(d1 != 0) add1(c1,-1);

int d2 = sum2(c2) - sum2(c2-1);
if(d2 != 0) add2(c2,-1);
}
}
}
return 0;
}

View Code