Segment Game

3
0 0
0 3
0 1
5
0 1
0 0
1 1
0 1
0 0

Case #1:
0
0
0
Case #2:
0
1
0
2
Hint
For the second case in the sample:

At the first add operation,Lillian adds a segment [1,2] on the line.

At the second add operation,Lillian adds a segment [0,2] on the line.

At the delete operation,Lillian deletes a segment which added at the first add operation.

At the third add operation,Lillian adds a segment [1,4] on the line.

At the fourth add operation,Lillian adds a segment [0,4] on the line
 

题意：两种操作，添加线段和删除线段，第i次添加时告诉线段起点并且要添加长度为i的线段，删除第i次添加的线段，问每次添加后有多少线段是落在当前要画的线段内部的。

思路：因为每次画的线段的长度是递增的，所以当前画的线段不可能被其他线段包含，那么统计小于左端点的点的个数x和小于等于右端点的点的个数y，ans=y-x。分别用树状数组维护，没写过树状数组了，都忘了，又复习了一下。

代码：

#include <iostream>
#include <functional>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#pragma comment (linker,"/STACK:102400000,102400000")
#define pi acos(-1.0)
#define eps 1e-6
#define lson rt<<1,l,mid
#define rson rt<<1|1,mid+1,r
#define FRE(i,a,b)  for(i = a; i <= b; i++)
#define FREE(i,a,b) for(i = a; i >= b; i--)
#define FRL(i,a,b)  for(i = a; i < b; i++)
#define FRLL(i,a,b) for(i = a; i > b; i--)
#define mem(t, v)   memset ((t) , v, sizeof(t))
#define sf(n)       scanf("%d", &n)
#define sff(a,b)    scanf("%d %d", &a, &b)
#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
#define pf          printf
#define DBG         pf("Hi\n")
typedef long long ll;
using namespace std;

#define INF 0x3f3f3f3f
#define mod 1000000009
const int maxn = 1005;
const int MAXN = 400005;
const int MAXM = 200010;
const int N = 1005;

typedef pair<int,int>P;
int bit[2][MAXN];
int n;
int pos[MAXN],num;
int L[MAXN],R[MAXN],op[MAXN];
P pir[MAXN];

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

void add(int i,int x,int c) //c=0时表示维护的左端点，c=1时表示维护的右端点
{
    while (i<MAXN)
    {
        bit[c][i]+=x;
        i+=lowbit(i);
    }
}

int sum(int i,int c)
{
    int s=0;
    while (i>0)
    {
        s+=bit[c][i];
        i-=lowbit(i);
    }
    return s;
}

int main()
{
#ifndef ONLINE_JUDGE
    freopen("C:/Users/lyf/Desktop/IN.txt","r",stdin);
#endif
    int i,j,cas=0;
    while (~scanf("%d",&n))
    {
        int tot=0;num=0;
        memset(bit,0,sizeof(bit));
        for (i=1;i<=n;i++)
        {
            scanf("%d%d",&op[i],&L[i]);
            if (op[i]==0)
            {
                R[i]=L[i]+(++tot);
                pos[num++]=L[i];
                pos[num++]=R[i];
            }
        }
        tot=0;
        sort(pos,pos+num);
        num=unique(pos,pos+num)-pos;
        for (i=1;i<=n;i++)  //离散化
        {
            if (op[i]==0)
            {
                L[i]=lower_bound(pos,pos+num,L[i])-pos+1;
                R[i]=lower_bound(pos,pos+num,R[i])-pos+1;
                pir[++tot]=make_pair(L[i],R[i]);
            }
        }
        printf("Case #%d:\n",++cas);
        for (i=1;i<=n;i++)
        {
            if (op[i]==0)
            {
                printf("%d\n",sum(R[i],1)-sum(L[i]-1,0));
                add(L[i],1,0);
                add(R[i],1,1);
            }
            else
            {
                add(pir[L[i]].first,-1,0);
                add(pir[L[i]].second,-1,1);
            }
        }
    }
    return 0;
}

Segment Game (hdu 5372 树状数组+离散化)

原文：http://blog.csdn.net/u014422052/article/details/47440871