codeforces 361 div2 E. Mike and Geometry Problem

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假如每个点被n1条线段覆盖求k个线段覆盖它的方案数显然C（n1,k）
点太多那就离散化以线段的形式来求最多20万条线段求覆盖线段数更合理

#include<cstdio>#include<iostream>#include<algorithm>#include<vector>#define mod 1000000007ll#define LL long long#define maxn 200010using namespace std;struct node{   LL l,r;}no[maxn];LL d[2*maxn],f[maxn],cnt = 0,sum[2*maxn] = {0},n,k;void init(){    f[0] = 1;    for(int i=1;i<maxn;i++)    {        f[i] = f[i-1]*(LL)i;        f[i] %= mod;    }}LL modPow(LL n,LL m){    LL ans = 1;    while(m)    {        if(m&1ll)ans*=n;        ans%=mod;        n*=n;        n%=mod;        m>>=1;    }    return ans;}LL C(LL n,LL m){    return  f[n]*modPow(f[m]*f[n-m]%mod,mod-2)%mod;}int main(){    init();    scanf("%I64d %I64d",&n,&k);    for(int i=0;i<n;i++) scanf("%I64d %I64d",&no[i].l,&no[i].r);    for(int i=0;i<n;i++)    {        d[cnt++] = no[i].l;        d[cnt++] = no[i].r+1;    }    sort(d,d+cnt);    cnt = unique(d,d+cnt)-d;    for(int i=0;i<n;i++)    {        LL pre = lower_bound(d,d+cnt,no[i].l)-d;        sum[pre]++;        pre = lower_bound(d,d+cnt,no[i].r+1)-d;        sum[pre]--;    }    LL sum1 = sum[0],ans = 0;    for(int i=1;i<cnt;i++)    {        LL temp = 0;        if(sum1>=k) temp = (d[i]-d[i-1])%mod*C(sum1,k)%mod;        ans+=temp;        ans%=mod;        sum1+=sum[i];    }    printf("%I64d\n",ans);    return 0;}

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