题意：与poj3268一样，所有人需要从各点到一点再从一点到各点，求最短路总和。
与poj3268一样，先正向建图跑 dijkstra ，得到该点到其他所有各点的最短路，即这些人回去的最短路，再用反向建的图跑一遍最短路，得到各点到该点的最短路，求和就行了。
1 #include<stdio.h>
2 #include<string.h>
3 #include<algorithm>
4 #include<queue>
5 #include<vector>
6 using namespace std;
7 typedef pair<int,int> pii;
8 typedef long long ll;
9 const int MAXM=1000000;
10 struct cmp{
11 bool operator ()(pii a,pii b){
12 return a.first>b.first;
13 }
14 };
15
16 int head1[MAXM+5],point1[MAXM+5],val1[MAXM+5],next1[MAXM+5],size1;
17 int head2[MAXM+5],point2[MAXM+5],val2[MAXM+5],next2[MAXM+5],size2;
18 int n,m;
19 int dist1[MAXM+5],dist2[MAXM+5];
20
21 void add1(int a,int b,int v){
22 point1[size1]=b;
23 val1[size1]=v;
24 next1[size1]=head1[a];
25 head1[a]=size1++;
26 }
27
28 void add2(int a,int b,int v){
29 point2[size2]=b;
30 val2[size2]=v;
31 next2[size2]=head2[a];
32 head2[a]=size2++;
33 }
34
35 void dij(int s){
36 int i;
37 priority_queue<pii,vector<pii>,cmp>q;
38 memset(dist1,0x3f,sizeof(dist1));
39 memset(dist2,0x3f,sizeof(dist2));
40 dist1[s]=dist2[s]=0;
41 q.push(make_pair(dist1[s],s));
42 while(!q.empty()){
43 pii u=q.top();
44 q.pop();
45 if(u.first>dist1[u.second])continue;
46 for(i=head1[u.second];~i;i=next1[i]){
47 int j=point1[i];
48 if(dist1[j]>dist1[u.second]+val1[i]){
49 dist1[j]=dist1[u.second]+val1[i];
50 q.push(make_pair(dist1[j],j));
51 }
52 }
53 }
54 while(!q.empty())q.pop();
55 q.push(make_pair(dist2[s],s));
56 while(!q.empty()){
57 pii u=q.top();
58 q.pop();
59 if(u.first>dist2[u.second])continue;
60 for(i=head2[u.second];~i;i=next2[i]){
61 int j=point2[i];
62 if(dist2[j]>dist2[u.second]+val2[i]){
63 dist2[j]=dist2[u.second]+val2[i];
64 q.push(make_pair(dist2[j],j));
65 }
66 }
67 }
68 ll ans=0;
69 for(i=1;i<=n;i++){
70 ans+=dist1[i];
71 ans+=dist2[i];
72 }
73 printf("%I64d
",ans);
74 }
75
76 int main(){
77 int t;
78 scanf("%d",&t);
79 for(int q=1;q<=t;q++){
80 scanf("%d%d",&n,&m);
81 int i,a,b,v;
82 memset(head1,-1,sizeof(head1));
83 size1=0;
84 memset(head2,-1,sizeof(head2));
85 size2=0;
86 for(i=1;i<=m;i++){
87 scanf("%d%d%d",&a,&b,&v);
88 add1(a,b,v);
89 add2(b,a,v);
90 }
91 dij(1);
92 }
93
94 return 0;
95 }