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
