dfs | Security Badges

Description You are in charge of the security for a large building, with n rooms and m doors between the rooms. The rooms and doors are conveniently numbered from 1 to n, and from 1 to m, respectively.

Door i opens from room ai to room bi , but not the other way around. Additionally, each door has a security code that can be represented as a range of numbers [ci , di ].

There are k employees working in the building, each carrying a security badge with a unique, integer-valued badge ID between 1 and k. An employee is cleared to go through door i only when the badge ID x satisfies ci ≤ x ≤ di .

Your boss wants a quick check of the security of the building. Given s and t, how many employees can go from room s to room t?

Input The first line of input contains three space-separated integers n, m, and k (2 ≤ n ≤ 1,000; 1 ≤ m ≤ 5,000; 1 ≤ k ≤ 109 ). The second line of input contains two space-separated integers s and t (1 ≤ s, t ≤ n; s 6= t). Each of the next m lines contains four space-separated integers ai , bi , ci , and di (1 ≤ ai , bi ≤ n; 1 ≤ ci ≤ di ≤ k; ai 6= bi), describing door i. For any given pair of rooms a, b there will be at most one door from a to b (but there may be both a door from a to b and a door from b to a).

Output Print, on a single line, the number of employees who can reach room t starting from room s.

题意 共有n间房间，m扇门，k位员工（编号为1~k）。每扇门连接两个房间，可以允许通过该门的员工编号范围称作code，用闭区间[ci,di]表示。求从房间s出发，最终共有多少人员可以到达房间t。

思路 dfs求出所有起点为s终点为t的路径，对于每一条路径，求出所有code的交集；对于所有路径，求出所有答案的并集。两种效率低下的朴素思路分别为：1.依次对每位员工判断是否能到达终点 2.对每条路径判断有哪些员工可以到达终点。针对第一种思路进行优化，可以将所有员工分为若干区间，同区间内的员工具有共同的性质，即能够一起通过某条路径到达终点，或一起被卡在某条路径的半途。

  1 #include <iostream>
  2 #include <stdio.h>
  3 #include <stdlib.h>
  4 #include <memory.h>
  5 #include <algorithm>
  6
  7 using namespace std;
  8
  9 //邻接表模板
 10 typedef struct adjnode
 11 {
 12     int end;
 13     int c;
 14     int d;
 15     struct adjnode *next;
 16 } Node;
 17 typedef struct adjlist
 18 {
 19     Node *head;
 20 } List;
 21 typedef struct _Graph
 22 {
 23     int vertices;
 24     int edges;
 25     List *array;
 26 }Graph;
 27
 28 Graph* Create(int v)
 29 {
 30     Graph *graph=new Graph;
 31
 32     graph->vertices=v;
 33     graph->edges=0;
 34
 35     graph->array=new List[v+1]; //0~v
 36
 37     for(int i=0;i<=v;i++){
 38         graph->array[i].head=NULL;
 39     }
 40
 41     return graph;
 42 }
 43
 44 void AddEdge(Graph *graph,int begin,int end,int ci,int di)
 45 {
 46     Node *newnode=new Node;
 47
 48     newnode->end=end;
 49     newnode->next=graph->array[begin].head;
 50     newnode->c=ci;
 51     newnode->d=di;
 52     graph->array[begin].head=newnode;
 53
 54     graph->edges++;
 55 }
 56
 57
 58 int t,ans=0;
 59 int edges[10005];
 60
 61 bool dfs(Graph *g,int cur,int l,int r,bool vis[1005]){
 62     vis[cur]=true;
 63
 64     if(cur==t){
 65         return true;
 66     }
 67
 68     Node *p=g->array[cur].head;
 69     while(p){
 70         if(!vis[p->end] && p->c<=l && p->d>=r){
 71             if(dfs(g,p->end,l,r,vis))
 72                 return true;
 73         }
 74         p=p->next;
 75     }
 76
 77     return false;
 78 }
 79
 80 int main()
 81 {
 82     int n,m,k,s;
 83     cin>>n>>m>>k>>s>>t;
 84
 85     Graph *hotel=Create(n);
 86
 87     int a,b,u,v;
 88     for(int i=0;i<m;i++){
 89         cin>>a>>b>>u>>v;
 90
 91         AddEdge(hotel,a,b,u,v);
 92         edges[i*2]=u;
 93         edges[i*2+1]=v+1;  //[a,b]∪[b+1,c]等同于[a,c]
 94     }
 95
 96     sort(edges,edges+2*m);
 97     int cnt=unique(edges,edges+2*m)-edges;
 98
 99     bool vis[1005];
100
101     for(int i=1;i<cnt;i++){
102         memset(vis,false,sizeof(vis));
103         if(dfs(hotel,s,edges[i-1],edges[i]-1,vis))  //规定受检区间左闭右开
104             ans+=(edges[i]-edges[i-1]);
105     }
106
107     cout<<ans<<endl;
108
109     return 0;
110 }

 

 

 

来源 2017-2018 acm-icpc northwest regional contest

参考 https://blog.csdn.net/yz467796454/article/details/78753171 ； 官方题解