两遍Dijkstra算法 HDU-3499
Flight
Recently, Shua Shua had a big quarrel with his GF. He is so upset that he decides to take a trip to some other city to avoid meeting her. He will travel only by air and he can go to any city if there exists a flight and it can help him reduce the total cost to the destination. There's a problem here: Shua Shua has a special credit card which can reduce half the price of a ticket ( i.e. 100 becomes 50, 99 becomes 49. The original and reduced price are both integers. ). But he can only use it once. He has no idea which flight he should choose to use the card to make the total cost least. Can you help him?
InputThere are no more than 10 test cases. Subsequent test cases are separated by a blank line.
The first line of each test case contains two integers N and M ( 2 <= N <= 100,000
0 <= M <= 500,000 ), representing the number of cities and flights. Each of the following M lines contains "X Y D" representing a flight from city X to city Y with ticket price D ( 1 <= D <= 100,000 ). Notice that not all of the cities will appear in the list! The last line contains "S E" representing the start and end city. X, Y, S, E are all strings consisting of at most 10 alphanumeric characters.
OutputOne line for each test case the least money Shua Shua have to pay. If it's impossible for him to finish the trip, just output -1.Sample Input
4 4
Harbin Beijing 500
Harbin Shanghai 1000
Beijing Chengdu 600
Shanghai Chengdu 400
Harbin Chengdu
4 0
Harbin Chengdu
Sample Output
800
-1
Hint
In the first sample, Shua Shua should use the card on the flight from
Beijing to Chengdu, making the route Harbin->Beijing->Chengdu have the
least total cost 800. In the second sample, there's no way for him to get to
Chengdu from Harbin, so -1 is needed.
题意：给出两个城市间的飞机票价格，可以使其中之一价格减半，给出目的地和出发地，求将某一个航班机票价格减半后的最小花费。
注意：花费会超int，所以用long long，在设置INF时不能设成0x3f3f3f3f，可以用const long long inf=999999999999,但是这样无法用memset将dis数组初始化，可以用fill，写法：
fill(dis,dis+maxn,inf)。注意飞机是单行的。
思路：枚举每条路径上的每一个航班花费，将其减半，使用双向的Dijkstra()，求出从出发地到某条道路的dis,以及从目的地到该道路的dis,将该道路的权值减半，for循环枚举全部。
1 #include<iostream>
2 #include<cstdio>
3 #include<cstring>
4 #include<queue>
5 #include<algorithm>
6 #include<map>
7 using namespace std;
8 const int maxn=1000010;
9 const long long inf=99999999999;
10 int head[maxn],vis[maxn];
11 long long dis1[maxn],dis2[maxn];
12 int n,m,cnt;
13 struct node{
14 int pos;
15 long long cost;
16 node(){}
17 node(int pos,long long cost):pos(pos),cost(cost){}
18 friend bool operator < (node a,node b)
19 {
20 return a.cost>b.cost;
21 }
22 };
23 struct edge{
24 int to;
25 int next;
26 long long w;
27 }e[maxn];
28 void init1()
29 {
30 memset(head,-1,sizeof(head));
31 memset(vis,0,sizeof(vis));
32 fill(dis1,dis1+maxn,inf);
33 cnt=0;
34 }
35 void init2()
36 {
37 cnt=0;
38 memset(head,-1,sizeof(head));
39 memset(vis,0,sizeof(vis));
40 fill(dis2,dis2+maxn,inf);
41 }
42 void add(int x,int y,long long w)
43 {
44 e[cnt].to=y;
45 e[cnt].w=w;
46 e[cnt].next=head[x];
47 head[x]=cnt++;
48 }
49 void dijkstra(int sx)
50 {
51 priority_queue<node> q;
52 q.push(node(sx,0));
53 dis1[sx]=0;
54 while(!q.empty())
55 {
56 node u=q.top();
57 q.pop();
58 if(vis[u.pos])
59 continue;
60 vis[u.pos]=1;
61 for(int i=head[u.pos];i!=-1;i=e[i].next)
62 {
63 int v=e[i].to;
64 if(dis1[v]>dis1[u.pos]+e[i].w)
65 {
66 dis1[v]=dis1[u.pos]+e[i].w;
67 q.push(node(v,dis1[v]));
68 }
69 }
70 }
71 }
72 void dijkstra2(int sx)
73 {
74 priority_queue<node> q;
75 q.push(node(sx,0));
76 dis2[sx]=0;
77 while(!q.empty())
78 {
79 node u=q.top();q.pop();
80 if(vis[u.pos])
81 continue;
82
83 vis[u.pos]=1;
84 for(int i=head[u.pos];i!=-1;i=e[i].next)
85 {
86 int v=e[i].to;
87 if(dis2[v]>dis2[u.pos]+e[i].w)
88 {
89 dis2[v]=dis2[u.pos]+e[i].w;
90 q.push(node(v,dis2[v]));
91 }
92 }
93 }
94 }
95 int u[maxn],v[maxn];
96 long long wei[maxn];
97 string a,b;
98 int main()
99 {
100 int sx,ex;
101 int n,m;
102 while(~scanf("%d%d",&n,&m))
103 {
104 init1();
105 int id=0;
106 map<string,int >mp;
107 for(int i=1;i<=m;i++)
108 {
109 cin>>a>>b>>wei[i];
110 if(!mp.count(a))
111 mp[a]=id++;
112 if(!mp.count(b))
113 mp[b]=id++;
114 u[i]=mp[a];
115 v[i]=mp[b];
116 add(u[i],v[i],wei[i]);
117 }
118 cin>>a>>b;
119 if(!mp.count(a))
120 mp[a]=id++;
121 if(!mp.count(b))
122 mp[b]=id++;
123 sx=mp[a];
124 ex=mp[b];
125 dijkstra(sx);
126 if(dis1[ex]==inf)
127 {
128 printf("-1\n");
129 }
130 else
131 {
132 long long ans=inf;
133 init2();
134 for(int i=1;i<=m;i++)
135 {
136 add(v[i],u[i],wei[i]);
137 }
138 dijkstra2(ex);
139 for(int i=1;i<=m;i++)
140 {
141 ans=min(ans,dis1[u[i]]+dis2[v[i]]+wei[i]/2);
142 }
143 cout<<ans<<endl;
144 }
145 }
146 return 0;
147 }
