bzoj 2387: [Ceoi2011]Traffic

bzoj 2387: [Ceoi2011]Traffic

题目描述

 

The center of Gdynia is located on an island in the middle of the Kacza river. Every morning thousands of cars drive through this island from the residential districts on the western bank of the river (using bridge connections to junctions on the western side of the island) to the industrial areas on the eastern bank (using bridge connections from junctions on the eastern side of the island).   

The island resembles a rectangle, whose sides are parallel to the cardinal directions. Hence, we view it as an A × B rectangle in a Cartesian coordinate system, whose opposite corners are in points (0, 0) and (A,B). On the island, there are n junctions numbered from 1 to n. The junction number i has coordinates (xi, yi). If a junction has coordinates of the form (0, y), it lies on the western side of the island. Similarly, junctions with the coordinates (A, y) lie on the eastern side. Junctions are connected by streets. Each street is a line segment connecting two junctions. Streets can be either unidirectional or bidirectional. No two streets may have a common point (except for, possibly, a common end in a junction). There are are no bridges or tunnels. You should not assume anything else about the shape of the road network. In particular, there can be streets going along the river bank or junctions with no incoming or outgoing streets. Because of the growing traffic density, the city mayor has hired you to check whether the current road network on the island is sufficient.

He asked you to write a program which determines how many junctions on the eastern side of the island are reachable from each junction on the western side.   

格丁尼亚的中心位于Kacza河中的一座岛屿。每天清晨，成千上万辆汽车通过岛屿从西岸的住宅区 （由桥连接岛的西部）到东岸的工业区（由桥连接岛的东部）。

该岛类似于矩形，它的边平行于主方向。故可将它看作是笛卡尔坐标系中的一个A*B的矩形，它的对角分别为（0, 0）和（A, B）。 岛上有n个交通节点，编号为1…n（junction, 此处可理解为广义的路口），第i个节点坐标为(xi, yi)。 如果一个节点的坐标为(0, y)，它就位于岛的西岸。类似的，坐标为(A, y)的节点位于岛的东岸。 各个节点由街道连接，每条街道用线段连接两个节点。街道有单向行驶或双向行驶之分。除端点外任意两条街道都没有公共点。也没有桥梁或者隧道。 你不能对道路网络形状做任何其他假设。沿河岸的街道或节点可能没有入口或者出口街道。

由于交通堵塞日趋严重，市长聘请你测试岛上当前的道路网是否足够。要求你写一个程序确定从岛的西岸的每个节点能够到达东岸的多少个节点。

 

 

输入

 

The first line of the standard input contains four integers n, m, A and B (1 <= n <= 300 000, 0 <= m <= 900 000, 1 <= A,B <= 10^9). They denote the number of junctions in the center of Gdynia, the number of streets and dimensions of the island, respectively. In each of the following n lines there are two integers xi, yi (0 <= xi <= A, 0 <= yi <= B) describing the coordinates of the junction number i. No two junctions can have the same coordinates. The next m lines describe the streets. Each street is represented in a single line by three integers ci, di, ki (1 <= ci, di <= n, ci ≠di, ki ∈ {1, 2}). Their meaning is that junctions ci and di are connected with a street. If ki = 1, then this is a unidirectional street from ci to di. Otherwise, the street can be driven in both directions. Each unordered pair {ci, di} can appear in the input at most once. You can assume that there is at least one junction on the western side of the island from which it is possible to reach some junction on the eastern side of the island. Additionally, in test cases worth at least 30 points, n,m <= 6 000.

第1行包含4个整数n, m, A, B(1≤n≤300000, 0≤m≤900000,1≤A,B≤10^9)，分别表示格丁尼亚市中心的节点数，街道数和岛屿的尺寸。

接下来的n行，每行包含两个整数xi，yi (0≤xi≤A,0≤yi≤B)，表示第i个节点的坐标。任意两个节点的坐标都不相同。

再往下的m行表示街道，每条街道用3个整数ci, di, ki（1≤ci, di≤n, ci≠di, ki∈{1,2}），表示节点ci、di有街道连接。如果ki=1,表示从ci到di的街道是单向的，否则，这条街道可以双向行驶。

每个无序对{ci, di}最多出现1次。

你可以假设西岸节点中至少有1个能够到达东岸的一些节点。

至少有30分的测试数据，n, m≤6000。

 

 

输出

 

Your program should write to the standard output one line for each junction on the western side of the island. This line should contain the number of junctions on the eastern side that are reachable from that junction. The output should be ordered according to decreasing y-coordinates of the junctions.

为每个西岸节点输出1行，包括从这个节点出发能够到达东岸的节点数目，

输出根据这些节点的y坐标降序排序。

---

solution

orz Jessie

首先这是一个平面图；

如果把不能到达的点排除掉，那么左边能到的右边点一定是连续的一段，右边到左边也是一样

于是我们把左边能到的右边点按y从小到大编号，

然后按编号从小到大一个个bfs，遇到有打过编号的点就返回（现在一定不如之前）

再从大到小做一次，即可得出每一个点能到的区间[l,r]

由于每个点最多被编号1次，所以稳得很

#include<cstdio>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#define maxn 300005
using namespace std;
int n,m,A,B,head[maxn],HEAD[maxn],l[maxn],r[maxn]; 
int t1,t2,ts,tt,t,tot,flag[maxn];
queue<int>q;
struct node{
    int v,nex;
}E[1800005],e[1800005];
struct no{
    int id,v,x,y;
}S[maxn],T[maxn],s[maxn];
bool cmp(no a,no b){return a.v<b.v;}
bool Cmp(no a,no b){return a.v>b.v;}
void lj(int t1,int t2){
    E[++tot].v=t2;E[tot].nex=HEAD[t1];HEAD[t1]=tot;
    e[tot].v=t1;e[tot].nex=head[t2];head[t2]=tot;
}
void add(int t1,int t2){
    E[++tot].v=t2;E[tot].nex=HEAD[t1];HEAD[t1]=tot;
    e[tot].v=t2;e[tot].nex=head[t1];head[t1]=tot;
    E[++tot].v=t1;E[tot].nex=HEAD[t2];HEAD[t2]=tot;
    e[tot].v=t1;e[tot].nex=head[t2];head[t2]=tot;
}
int main(){
    cin>>n>>m>>A>>B;
    for(int i=1;i<=n;i++){
        scanf("%d%d",&s[i].x,&s[i].y);
        if(s[i].x==0)S[++ts]=(no){i,s[i].y};
        if(s[i].x==A)T[++tt]=(no){i,s[i].y}; 
    }
    for(int i=1;i<=m;i++){
        scanf("%d%d%d",&t1,&t2,&t);
        if(t==1)lj(t1,t2);
        else add(t1,t2);
    }
    sort(T+1,T+tt+1,cmp);
    for(int i=1;i<=ts;i++){
        q.push(S[i].id);flag[S[i].id]=1;
    }
    while(!q.empty()){
        int x=q.front();q.pop();
        for(int i=HEAD[x];i;i=E[i].nex){
            if(!flag[E[i].v]){flag[E[i].v]=1;q.push(E[i].v);}
        }
    }
    int cnt=0;
    for(int i=1;i<=tt;i++){
        if(flag[T[i].id])T[i].x=++cnt;
    }
    for(int k=1;k<=tt;k++){
        if(T[k].x>0){
            q.push(T[k].id);
            while(!q.empty()){
                int x=q.front();q.pop();
                for(int i=head[x];i;i=e[i].nex){
                    if(l[e[i].v]==0){l[e[i].v]=T[k].x;q.push(e[i].v);}
                }
            }
        }
    }
    for(int k=tt;k>=1;k--){
        if(T[k].x>0){
            q.push(T[k].id);
            while(!q.empty()){
                int x=q.front();q.pop();
                for(int i=head[x];i;i=e[i].nex){
                    if(r[e[i].v]==0){r[e[i].v]=T[k].x;q.push(e[i].v);}
                }
            }
        }
    }
    sort(S+1,S+ts+1,Cmp);
    for(int i=1;i<=ts;i++){
        if(l[S[i].id]==0)puts("0");
        else printf("%d\n",r[S[i].id]-l[S[i].id]+1);
    }
    return 0;
}