Description
Bessie and Elsie have each baked N pies (1≤N≤10^5). Each of the 2N pies has a tastiness value according to Bessie and a (possibly different) tastiness value according to Elsie. Bessie is thinking about giving one of her pies to Elsie. If Elsie receives a pie from Bessie, she will feel obligated to give one of her pies to Bessie. So as to not appear stingy nor flamboyant, Elsie will try to pick a pie that is at least as tasty (in Elsie’s eyes) as the pie she received, but no more than D units tastier (0≤D≤10^9). Such a pie may not exist, in which case Elsie will adopt a pseudonym and exile herself to Japan.
But if Elsie does give Bessie a pie in return, Bessie will similarly try to give Elsie a pie which is at least as tasty but no more than D units tastier (in Bessie’s eyes) as the pie Elsie just gave her. Should this be impossible, Bessie too will exile herself. Otherwise she will give her chosen pie to Elsie. This cycle will continue until one of the cows is exiled, an unhappy outcome, or one of the cows receives a pie which she accords a tastiness value of 0, in which case the gift exchange will end and both cows will be happy.
Bessie和Elsie各自烤了 N(1≤N≤10^5)个馅饼）。Bessie 会这 2N 个馅饼打分，Elsie 也会。二者的打分均为一个<=10^9的非负整数。由于她们口味不同，每个派的两个分数可能不同。 她们想互赠礼物。开始时，Bessie 送给 Elsie 一个馅饼。她们收到礼物（对方做的馅饼）后都会回赠对方一个自己做的馅饼。 她们选择回礼的方法相同。以 Elsie 为例，Elsie 根据自己的打分来选择回礼。回礼的分数至少要大于等于她收到的馅饼的分数，但两个馅饼的分数差不能大于 D(0≤D≤10^9) 。如果有多个馅饼满足要求，Elsie 可以选择其中的任意一个作为回礼。若没有馅饼满足要求，Elsie 会放弃。Bessie 选择回礼的方法同理。 她们之间的礼物交换将持续到一头牛放弃（Bad End），或一头牛收到一个她自己打分为 00 的馅饼，此时礼物交换愉快结束（Happy End）。 请注意，不能把一个馅饼赠送两次，不能把馅饼送给自己。 Bessie 想知道：对于每个她做的馅饼，如果她将这个馅饼作为最开始送给 Elsie 的礼物，她俩至少要互赠多少次馅饼（Bessie 给 Elsie 算一次，Elsie 回赠 Bessie 又算一次），才能 Happy End 。如果不可能 Happy End，请输出 -1。
Input
The first line contains the two integers N and D. The next 2N lines contain two space-separated integers each, respectively denoting the value of a particular pie according to Bessie, and the value of that pie according to Elsie.
The first N lines refer to Bessie’s pies, and the remaining N lines refer to Elsie’s pies.
It is guaranteed that all tastiness values are in the range [0,10^9].
第一行有两个整数 N, D。
在接下来的 2N行中，每行有两个整数，表示对于某个馅饼，Bessie 的打分和 Elsie 的打分。其中，前 N 行表示 Bessie 做的馅饼，后 N行则是 Elsie 做的馅饼。
Output
here should be N lines in the output. Line i should contain a single integer: the minimum number of pies that could be gifted in a happy gift exchange started with Bessie’s pie i. If no gift exchange starting with pie i is happy, then line i should contain the single integer −1 instead.
输出共 N 行，每行一个整数。 第 i 行的整数表示：如果 Bessie 将馅饼 i作为最开始送给 Elsie 的礼物，她俩至少要互赠多少次馅饼才能 Happy End 。如果不可能 Happy End，请在该行输出 -1。
Sample Input
2 1
1 1
5 0
4 2
1 4
Sample Output
3
1
Solution
最短路。两个不是同一个人做的馅饼，如果某个家伙把馅饼i给了对方，对方可以用馅饼j送回去，就把i和j连边。然后跑一遍最短路。答案就是最短距离。新建一个0点，跟所有被评为0的馅饼连边，以0为源点跑最短路。
建图时用二分，或者排序优化。
#include<cstdio>
#include<cstring>
#include<queue>
#include<algorithm>
using namespace std;
const int N=100005;
int n,m,cnt=0,head[N*2],dis[N*2];
struct node {
int to,next;
}edge[1500005];
struct rec {
int x,y,id;
}a[N],b[N];
queue <int> q;
void add(int x,int y) {
edge[cnt].to=y;
edge[cnt].next=head[x];
head[x]=cnt++;
}
bool cmpa(rec x,rec y) {
return x.x<y.x;
}
bool cmpb(rec x,rec y) {
return x.y<y.y;
}
void spfa() {
while(!q.empty()) {
int now=q.front();q.pop();
for(int i=head[now];i!=-1;i=edge[i].next) {
int to=edge[i].to;
if(dis[to]!=-1)continue;
dis[to]=dis[now]+1;
q.push(to);
}
}
}
int main() {
freopen("piepie.in","r",stdin);
freopen("piepie.out","w",stdout);
scanf("%d%d",&n,&m);
memset(head,-1,sizeof(head));
for(int i=1;i<=n*2;i++)
dis[i]=-1;
for(int i=1;i<=n;i++) {
scanf("%d%d",&a[i].x,&a[i].y);
if(a[i].y==0)dis[i]=1,q.push(i);
a[i].id=i;
}
for(int i=1;i<=n;i++) {
scanf("%d%d",&b[i].x,&b[i].y);
if(b[i].x==0)dis[i+n]=1,q.push(i+n);
b[i].id=i+n;
}
sort(a+1,a+1+n,cmpa);
sort(b+1,b+1+n,cmpb);
for(int i=1;i<=n;i++) {
int l=1,r=n,ret=-1;
while(l<=r) {
int mid=(l+r)>>1;
if(b[mid].y>=a[i].y)ret=mid,r=mid-1;
else l=mid+1;
}
if(ret==-1)continue;
for(int j=ret;j<=n;j++) {
if(b[j].y>a[i].y+m)break;
add(b[j].id,a[i].id);
}
}
for(int i=1;i<=n;i++) {
int l=1,r=n,ret=-1;
while(l<=r) {
int mid=(l+r)>>1;
if(a[mid].x>=b[i].x)ret=mid,r=mid-1;
else l=mid+1;
}
if(ret==-1)continue;
for(int j=ret;j<=n;j++) {
if(a[j].x>b[i].x+m)break;
add(a[j].id,b[i].id);
}
}
spfa();
for(int i=1;i<=n;i++)printf("%d\n",dis[i]);
fclose(stdin);
fclose(stdout);
return 0;
}