Codeforce 101B. Buses(线段树or树状数组+离散化)

最新推荐文章于 2020-01-16 17:18:48 发布

还是太年轻

最新推荐文章于 2020-01-16 17:18:48 发布

阅读量463

点赞数

CC 4.0 BY-SA版权

分类专栏：线段树树状数组

本文链接：https://blog.youkuaiyun.com/albertluf/article/details/79919748

线段树同时被 2 个专栏收录

27 篇文章

订阅专栏

树状数组

7 篇文章

订阅专栏

Little boy Gerald studies at school which is quite far from his house. That's why he has to go there by bus every day. The way from home to school is represented by a segment of a straight line; the segment contains exactly n + 1 bus stops. All of them are numbered with integers from 0 to n in the order in which they follow from Gerald's home. The bus stop by Gerald's home has number 0 and the bus stop by the school has number n.

There are m buses running between the house and the school: the i-th bus goes from stop si to ti (si < ti), visiting all the intermediate stops in the order in which they follow on the segment. Besides, Gerald's no idiot and he wouldn't get off the bus until it is still possible to ride on it closer to the school (obviously, getting off would be completely pointless). In other words, Gerald can get on the i-th bus on any stop numbered from si to ti - 1 inclusive, but he can get off the i-th bus only on the bus stop ti.

Gerald can't walk between the bus stops and he also can't move in the direction from the school to the house.

Gerald wants to know how many ways he has to get from home to school. Tell him this number. Two ways are considered different if Gerald crosses some segment between the stops on different buses. As the number of ways can be too much, find the remainder of a division of this number by 1000000007 (109 + 7).

Input

The first line contains two space-separated integers: n and m (1 ≤ n ≤ 109, 0 ≤ m ≤ 105). Then follow m lines each containing two integers si, ti. They are the numbers of starting stops and end stops of the buses (0 ≤ si < ti ≤ n).

Output

Print the only number — the number of ways to get to the school modulo 1000000007 (109 + 7).

题解：
线段树求解。开始的时候按照左端点排序wa了。
因为有可能左端点小的右端点可能很大，因此
计算之后可能又会有点落到这个区间上，因此
按右端点排序。
代码：

#include<cstdio>
#include<string.h>
#include<algorithm>
using namespace std;
#define ll long long
const int maxn=1e6+5;
const ll mod=1e9+7;
ll n,m,p[maxn*2];
struct node1
{
	ll l,r;
}c[maxn];
bool cmp(node1 a,node1 b)
{
	if(a.r!=b.r)return a.r<b.r;
	return a.l<b.l;
}
struct node
{
	ll left,right,mid;
	ll x;
}tree[maxn*4];
void build(ll l,ll r,int rt)
{
	tree[rt].left=l;
	tree[rt].right=r;
	tree[rt].mid=(l+r)>>1;
	if(l==r)return;
	build(l,tree[rt].mid,rt<<1);
	build(tree[rt].mid+1,r,rt<<1|1);
}
ll query(ll l,ll r,int rt)
{
	if(l<=tree[rt].left&&r>=tree[rt].right)
		return tree[rt].x%mod;
	ll ans=0;
	if(l<=tree[rt].mid)
		ans+=query(l,r,rt<<1);
	ans%=mod;
	if(r>tree[rt].mid)
		ans+=query(l,r,rt<<1|1);
	return ans%mod;
}
void add(ll L,ll C,int rt)
{
	if(tree[rt].left==tree[rt].right)
	{
		tree[rt].x=(tree[rt].x+C)%mod;
		return;
	}
	if(L<=tree[rt].mid)
		add(L,C,rt<<1);
	else
		add(L,C,rt<<1|1);
	tree[rt].x=(tree[rt<<1].x+tree[rt<<1|1].x)%mod;
}
int main()
{
	scanf("%lld%lld",&n,&m);
	int tol=1;
	for(int i=0;i<m;i++)
	{
		scanf("%lld%lld",&c[i].l,&c[i].r);
		p[tol++]=c[i].l;
		p[tol++]=c[i].r;
	}
	sort(p+1,p+tol+1);
	sort(c,c+m,cmp);
	build(1,tol,1);
	ll sum=0;
	for(int i=0;i<m;i++)
	{
		ll ans=0;
		ll l=lower_bound(p+1,p+tol+1,c[i].l)-p;
		ll r=lower_bound(p+1,p+tol+1,c[i].r)-p;
		if(c[i].l==0)ans++;
		if(r>=l)ans+=query(l,r-1,1);
		add(r,ans,1);
		if(c[i].r==n)sum=query(r,r,1);
	}
	printf("%lld\n",sum);
	return 0;
}

因为只有点的修改，所以可以用更简单的树状数组。

#include<bits/stdc++.h>
using namespace std;
#define ll long long
const int maxn=1e6+5;
const ll mod=1e9+7;
ll p[maxn],bit[maxn],tol;
struct node1
{
    ll l,r;
}c[maxn];
bool cmp(node1 a,node1 b)
{
    if(a.r!=b.r)return a.r<b.r;
    return a.l<b.l;
}
ll sum(ll i)
{
    ll s=0;
    while(i>0)
    {
        s=(s+bit[i])%mod;
        i-=i&-i;
    }
    return s%mod;
}
void add(ll i,ll x)
{
    while(i<=tol)
    {
        bit[i]=(bit[i]+x)%mod;
        i+=i&-i;
    }
}
int main()
{
    ll n,m;scanf("%lld%lld",&n,&m);
    tol=1;
    for(int i=0;i<m;i++)
    {
        scanf("%lld%lld",&c[i].l,&c[i].r);
        p[tol++]=c[i].l;
        p[tol++]=c[i].r;
    }
    sort(p+1,p+tol+1);
    sort(c,c+m,cmp);
    ll s=0;
    for(int i=0;i<m;i++)
    {
        int l=lower_bound(p+1,p+tol+1,c[i].l)-p;
        int r=lower_bound(p+1,p+tol+1,c[i].r)-p;
        ll ans=0;
        if(c[i].l==0)ans++;
        ans+=sum(r-1)-sum(l-1);
        ans=(ans+mod)%mod;
        add(r,ans);
        if(c[i].r==n)s=sum(r)-sum(r-1);
    }
    printf("%lld\n",(s+mod)%mod);
    return 0;
}