All Roads Lead to Rome
本题需要记录一共有几条最短路径，并输出最短路中开心值最大的路径或者开心值相等的情况下输出平均开心值最大的路径。
1 #include <iostream>
2 #include <cstdio>
3 #include <cstring>
4 #include <queue>
5 #include <vector>
6 #include <map>
7 #include <algorithm>
8 #define inf 0x3f3f3f3f
9 using namespace std;
10 const int maxn=220;
11 int n,m;
12 string s1,s2,st;
13 int cnt;
14 int fhappy=0;
15 double fhave=0;
16
17 map<string,int>mp1;
18 map<int,string>mp2;
19 struct node{
20 int pos;
21 int cost;
22 node(){}
23 node(int pos,int cost):pos(pos),cost(cost){}
24 friend bool operator < (node a,node b)
25 {
26 return a.cost>b.cost;
27 }
28 }head,tail;
29 vector<node>g[maxn];
30 vector<int> pre[maxn];
31 vector<int> path;
32 vector<int> tpath;
33 int vis[maxn],dis[maxn];
34 int happy[maxn];
35 void dijkstra(int st)
36 {
37 priority_queue<node> q;
38 dis[st]=0;
39 head.pos=st;
40 head.cost=0;
41 q.push(head);
42 while(!q.empty())
43 {
44 head=q.top();
45 q.pop();
46 if(vis[head.pos]) continue;
47 vis[head.pos]=1;
48 int now=head.pos;
49 for(int i=0;i<g[now].size();i++)
50 {
51 tail=g[now][i];
52 int v=tail.pos;
53 int len=tail.cost;
54 if(dis[v]>dis[head.pos]+len)
55 {
56 dis[v]=dis[head.pos]+len;
57 pre[v].clear();
58 pre[v].push_back(head.pos);
59 q.push(tail);
60 }
61 else if(dis[v]==dis[head.pos]+len)
62 {
63 pre[v].push_back(head.pos);
64 q.push(tail);
65 }
66 }
67 }
68 }
69 void dfs(int now)
70 {
71 if(now==0)
72 {
73 cnt++;
74 tpath.push_back(now);
75 int hval=0;
76 for(int i=tpath.size()-2;i>=0;i--)
77 {
78 // cout<<"lala"<<tpath[i]<<endl;
79 hval+=happy[tpath[i]];
80 }
81 double have=1.0*hval/(tpath.size()-1);
82 if(hval>fhappy)
83 {
84 fhappy=hval;
85 fhave=have;
86 path=tpath;
87 }
88 else if(hval==fhappy&&have>fhave)
89 {
90 fhappy=hval;
91 fhave=have;
92 path=tpath;
93 }
94 tpath.pop_back();
95 return;
96 }
97 tpath.push_back(now);
98 for(int i=0;i<pre[now].size();i++)
99 {
100 dfs(pre[now][i]);
101 }
102 tpath.pop_back();
103 }
104 int main()
105 {
106 cin>>n>>m>>st;
107 memset(dis,inf,sizeof(dis));
108 mp1[st]=0;
109 mp2[0]=st;
110 int t;
111 for(int i=1;i<=n-1;i++)
112 {
113 cin>>s1>>happy[i];
114 mp1[s1]=i;
115 mp2[i]=s1;
116 }
117 for(int i=0;i<m;i++)
118 {
119 cin>>s1>>s2>>t;
120 int id1=mp1[s1];
121 int id2=mp1[s2];
122 node tmp;
123 tmp.pos=id2;
124 tmp.cost=t;
125 g[id1].push_back(tmp);
126 tmp.pos=id1;
127 g[id2].push_back(tmp);
128 }
129 int ed=mp1["ROM"];
130 dijkstra(0);
131 dfs(ed);
132 printf("%d %d %d %d
",cnt,dis[ed],fhappy,(int)fhave);
133 for(int i=path.size()-1;i>=0;i--)
134 {
135 cout<<mp2[path[i]];
136 if(i!=0)
137 {
138 printf("->");
139 }
140 }
141 return 0;
142 }