CF-503div2-A/B/C

A. New Building for SIS

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

You are looking at the floor plan of the Summer Informatics School's new building. You were tasked with SIS logistics, so you really care about travel time between different locations: it is important to know how long it would take to get from the lecture room to the canteen, or from the gym to the server room.

The building consists of n towers, h floors each, where the towers are labeled from 1 to n, the floors are labeled from 1 to h. There is a passage between any two adjacent towers (two towers i and i + 1 for all i: 1 ≤ i ≤ n - 1) on every floor x, where a ≤ x ≤ b. It takes exactly one minute to walk between any two adjacent floors of a tower, as well as between any two adjacent towers, provided that there is a passage on that floor. It is not permitted to leave the building.

The picture illustrates the first example.

You have given k pairs of locations (ta, fa), (tb, fb): floor fa of tower ta and floor fb of tower tb. For each pair you need to determine the minimum walking time between these locations.

Input

The first line of the input contains following integers:

• n: the number of towers in the building (1 ≤ n ≤ 108),
• h: the number of floors in each tower (1 ≤ h ≤ 108),
• a and b: the lowest and highest floor where it's possible to move between adjacent towers (1 ≤ a ≤ b ≤ h),
• k: total number of queries (1 ≤ k ≤ 104).

Next k lines contain description of the queries. Each description consists of four integers ta, fa, tb, fb (1 ≤ ta, tb ≤ n, 1 ≤ fa, fb ≤ h). This corresponds to a query to find the minimum travel time between fa-th floor of the ta-th tower and fb-th floor of the tb-th tower.

Output

For each query print a single integer: the minimum walking time between the locations in minutes.

Example

input

Copy

3 6 2 3 3
1 2 1 3
1 4 3 4
1 2 2 3

output

Copy

1
4
2
　　询问从t1的f1层到t2的f2层的最短时间，注意特判下在同一座塔的情况。
　　

#include<iostream>
#include<cstring>
#include<cstdio>
#include<map>
#include<set>
#include<queue>
#include<vector>
#include<algorithm>
#include<cmath> 
#include<deque>
#include<list>
using namespace std;
#define LL long long 
#define PI acos(-1.0)
#define inf 0x3f3f3f3f
#define pb push_back
#define mp make_pair
int main()
{
    int n,h,a,b,k,i,j;
    int t1,f1,t2,f2;
    while(scanf("%d%d%d%d%d",&n,&h,&a,&b,&k)!=EOF){
        while(k--){
            scanf("%d%d%d%d",&t1,&f1,&t2,&f2);
            if(t1>t2){
                swap(t1,t2),swap(f1,f2);
            }
            if(t1==t2){
                cout<<abs(f1-f2)<<endl;
            }
            else{
                if(f1>=a&&f1<=b){
                    cout<<t2-t1+abs(f1-f2)<<endl;
                }
                else{
                    cout<<min(abs(a-f1)+t2-t1+abs(a-f2),
                        abs(b-f1)+t2-t1+abs(b-f2))<<endl;
                }
            }
        }
    }
    return 0;
}

B. Badge

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

In Summer Informatics School, if a student doesn't behave well, teachers make a hole in his badge. And today one of the teachers caught a group of $\mathrm{nn students\; doing\; yet\; another\; trick.}$

Let's assume that all these students are numbered from $\mathrm{11 to nn.\; The\; teacher\; came\; to\; student aa and\; put\; a\; hole\; in\; his\; badge.\; The\; student,\; however,\; claimed\; that\; the\; main\; culprit\; is\; some\; other\; student papa.}$

After that, the teacher came to student ${\mathrm{papa and\; made\; a\; hole\; in\; his\; badge\; as\; well.\; The\; student\; in\; reply\; said\; that\; the\; main\; culprit\; was\; student ppappa.}}^{}$

This process went on for a while, but, since the number of students was finite, eventually the teacher came to the student, who already had a hole in his badge.

After that, the teacher put a second hole in the student's badge and decided that he is done with this process, and went to the sauna.

You don't know the first student who was caught by the teacher. However, you know all the numbers ${\mathrm{pipi.\; Your\; task\; is\; to\; find\; out\; for\; every\; student aa,\; who\; would\; be\; the\; student\; with\; two\; holes\; in\; the\; badge\; if\; the\; first\; caught\; student\; was aa.}}^{}$

Input

The first line of the input contains the only integer $\mathrm{nn (1\le n\le 10001\le n\le 1000) —\; the\; number\; of\; the\; naughty\; students.}$

The second line contains $\mathrm{nn integers p1p1,\; ..., pnpn (1\le pi\le n1\le pi\le n),\; where pipi indicates\; the\; student\; who\; was\; reported\; to\; the\; teacher\; by\; student ii.}$

Output

For every student $\mathrm{aa from 11 to nn print\; which\; student\; would\; receive\; two\; holes\; in\; the\; badge,\; if aa was\; the\; first\; student\; caught\; by\; the\; teacher.}$

Examples

input

Copy

3
2 3 2

output

Copy

2 2 3 

input

Copy

3
1 2 3

output

Copy

1 2 3 

Note

The picture corresponds to the first example test case.

When $\mathrm{a=1a=1,\; the\; teacher\; comes\; to\; students 11, 22, 33, 22,\; in\; this\; order,\; and\; the\; student 22 is\; the\; one\; who\; receives\; a\; second\; hole\; in\; his\; badge.}$

When $\mathrm{a=2a=2,\; the\; teacher\; comes\; to\; students 22, 33, 22,\; and\; the\; student 22 gets\; a\; second\; hole\; in\; his\; badge.\; When a=3a=3,\; the\; teacher\; will\; visit\; students 33, 22, 33 with\; student 33 getting\; a\; second\; hole\; in\; his\; badge.}$

For the second example test case it's clear that no matter with whom the teacher starts, that student would be the one who gets the second hole in his badge.

　　直接模拟，找第一个被访问两次的点就好了。

#include<iostream>
#include<cstring>
#include<cstdio>
#include<map>
#include<set>
#include<queue>
#include<vector>
#include<algorithm>
#include<cmath> 
#include<deque>
#include<list>
using namespace std;
#define LL long long 
#define PI acos(-1.0)
#define inf 0x3f3f3f3f
#define pb push_back
#define mp make_pair
bool vis[1100];
int main()
{
    int n,m,i,j,k,p[1111];
    while(scanf("%d",&n)==1){
        for(i=1;i<=n;++i)scanf("%d",p+i);
        for(i=1;i<=n;++i){
            memset(vis,0,sizeof(vis));
            j=i;
            while(!vis[j]){
                vis[j]=1;
                j=p[j];
            }
            printf("%d%c",j,i==n?'
':' ');
        }
    }
    return 0;
}

　　

C. Elections

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

As you know, majority of students and teachers of Summer Informatics School live in Berland for the most part of the year. Since corruption there is quite widespread, the following story is not uncommon.

Elections are coming. You know the number of voters and the number of parties — $\mathrm{nn and mm respectively.\; For\; each\; voter\; you\; know\; the\; party\; he\; is\; going\; to\; vote\; for.\; However,\; he\; can\; easily\; change\; his\; vote\; given\; a\; certain\; amount\; of\; money.\; In\; particular,\; if\; you\; give ii-th\; voter cicibytecoins\; you\; can\; ask\; him\; to\; vote\; for\; any\; other\; party\; you\; choose.}$

The United Party of Berland has decided to perform a statistical study — you need to calculate the minimum number of bytecoins the Party needs to spend to ensure its victory. In order for a party to win the elections, it needs to receive strictly more votes than any other party.

Input

The first line of input contains two integers $\mathrm{nn and mm (1\le n,m\le 30001\le n,m\le 3000) —\; the\; number\; of\; voters\; and\; the\; number\; of\; parties\; respectively.}$

Each of the following $\mathrm{nn lines\; contains\; two\; integers pipi and cici (1\le pi\le m1\le pi\le m, 1\le ci\le 1091\le ci\le 109) —\; the\; index\; of\; this\; voter\text{'}s\; preferred\; party\; and\; the\; number\; of\; bytecoins\; needed\; for\; him\; to\; reconsider\; his\; decision.}$

The United Party of Berland has the index $11.$

Output

Print a single number — the minimum number of bytecoins needed for The United Party of Berland to win the elections.

Examples

input

Copy

1 2
1 100

output

Copy

0

input

Copy

5 5
2 100
3 200
4 300
5 400
5 900

output

Copy

500

input

Copy

5 5
2 100
3 200
4 300
5 800
5 900

output

Copy

600

Note

In the first sample, The United Party wins the elections even without buying extra votes.

In the second sample, The United Party can buy the votes of the first and the fourth voter. This way The Party gets two votes, while parties $\mathrm{33, 44 and 55 get\; one\; vote\; and\; party\; number 22 gets\; no\; votes.}$

In the third sample, The United Party can buy the votes of the first three voters and win, getting three votes against two votes of the fifth party.

　　考虑枚举1的最终得票数，然后计算是否能当选以及最小的费用，选取一个最小的作为答案。
　　

#include<iostream>
#include<cstring>
#include<cstdio>
#include<map>
#include<set>
#include<queue>
#include<vector>
#include<algorithm>
#include<cmath> 
#include<deque>
#include<list>
#include<bits/stdc++.h>
using namespace std;
#define LL long long 
#define PI acos(-1.0)
#define inf 0x3f3f3f3f
#define pb push_back
#define mp make_pair
int p[3333],c[3333],tot[3333],n,m;
LL ans;
priority_queue<int,vector<int>,greater<int> >q[3333];
vector<LL> v[3333];38 int main()
{
    int i,j;
        scanf("%d%d",&n,&m);
        for(i=1;i<=n;++i)scanf("%d%d",p+i,c+i),tot[p[i]]++,q[p[i]].push(c[i]);
        for(i=1;i<=m;++i){
            v[i].push_back(0);
            while(!q[i].empty()){
                v[i].push_back(*(v[i].end()-1)+q[i].top());
                q[i].pop();
            }
        }
        LL ans=1LL*1000000000000000000;
        for(i=tot[1];i<=n;++i){
            LL res=0,buy=0;
            for(j=2;j<=m;++j){
                if(tot[j]>=i){
                    buy+=tot[j]-i+1;
                    res+=v[j][tot[j]-i+1];
                }
            }
            if(buy>i-tot[1]) continue;
            //q[0].clear();
            while(!q[0].empty())q[0].pop();
            //cout<<"i="<<i<<' '<<res<<' '<<buy<<endl;
            for(int ii=2;ii<=m;++ii){
                j=1;
                if(tot[ii]>=i) j=tot[ii]-i+1+1;
                for(;j<v[ii].size();++j){
                    q[0].push(v[ii][j]-v[ii][j-1]);
                }
            }
            while(!q[0].empty()&&buy<i-tot[1]){
                //cout<<"JI"<<q[0].top()<<endl;
                res+=q[0].top();
                q[0].pop();
                buy++;
            }
            if(tot[1]+buy<i)continue;
            //cout<<i<<' '<<res<<endl;
            if(res<ans)ans=res;
        }
        cout<<ans<<endl;
    return 0;
}