In the age of television, not many people attend theater performances. Antique Comedians of Malidinesia are aware of this fact. They want to propagate theater and, most of all, Antique Comedies. They have printed invitation cards with all the necessary information and with the programme. A lot of students were hired to distribute these invitations among the people. Each student volunteer has assigned exactly one bus stop and he or she stays there the whole day and gives invitation to people travelling by bus. A special course was taken where students learned how to influence people and what is the difference between influencing and robbery.
The transport system is very special: all lines are unidirectional and connect exactly two stops. Buses leave the originating stop with passangers each half an hour. After reaching the destination stop they return empty to the originating stop, where they wait until the next full half an hour, e.g. X:00 or X:30, where 'X' denotes the hour. The fee for transport between two stops is given by special tables and is payable on the spot. The lines are planned in such a way, that each round trip (i.e. a journey starting and finishing at the same stop) passes through a Central Checkpoint Stop (CCS) where each passenger has to pass a thorough check including body scan.
All the ACM student members leave the CCS each morning. Each volunteer is to move to one predetermined stop to invite passengers. There are as many volunteers as stops. At the end of the day, all students travel back to CCS. You are to write a computer program that helps ACM to minimize the amount of money to pay every day for the transport of their employees.
The transport system is very special: all lines are unidirectional and connect exactly two stops. Buses leave the originating stop with passangers each half an hour. After reaching the destination stop they return empty to the originating stop, where they wait until the next full half an hour, e.g. X:00 or X:30, where 'X' denotes the hour. The fee for transport between two stops is given by special tables and is payable on the spot. The lines are planned in such a way, that each round trip (i.e. a journey starting and finishing at the same stop) passes through a Central Checkpoint Stop (CCS) where each passenger has to pass a thorough check including body scan.
All the ACM student members leave the CCS each morning. Each volunteer is to move to one predetermined stop to invite passengers. There are as many volunteers as stops. At the end of the day, all students travel back to CCS. You are to write a computer program that helps ACM to minimize the amount of money to pay every day for the transport of their employees.
Input
The input consists of N cases. The first line of the input contains only positive integer N. Then follow the cases. Each case begins with a line containing exactly two integers P and Q, 1 <= P,Q <= 1000000. P is the number of stops including CCS and Q the number of bus lines. Then there are Q lines, each describing one bus line. Each of the lines contains exactly three numbers - the originating stop, the destination stop and the price. The CCS is designated by number 1. Prices are positive integers the sum of which is smaller than 1000000000. You can also assume it is always possible to get from any stop to any other stop.
Output
For each case, print one line containing the minimum amount of money to be paid each day by ACM for the travel costs of its volunteers.
Sample Input
2 2 2 1 2 13 2 1 33 4 6 1 2 10 2 1 60 1 3 20 3 4 10 2 4 5 4 1 50
Sample Output
46 210
题意：有n个站点， 有m条公交车路线，公交车路线只从一个起点站直接到达终点站，单向，每条路线都有费用；
有n个人从1出发，他们每个人要到一个且只能有一个公交站点，每个人的站点都不是重复的，可以路过但不算到达， 然后返回点1。，求所有人来回的最小费用之和；
思路：一个简单的最短路，但是本题要求往返之和，因此SPFA要走两边，另外由于题的数据量比较大，因此开了一个vector来存放边。
代码：
1 #include <cstdio> 2 #include <fstream> 3 #include <algorithm> 4 #include <cmath> 5 #include <deque> 6 #include <vector> 7 #include <queue> 8 #include <string> 9 #include <cstring> 10 #include <map> 11 #include <stack> 12 #include <set> 13 #include <sstream> 14 #include <iostream> 15 #define mod 998244353 16 #define eps 1e-6 17 #define ll long long 18 #define INF 0x3f3f3f3f 19 using namespace std; 20 21 //以键对做基础类型开ve数组，相当与一个三元数组 22 vector<pair<int,int>> ve[1000002]; 23 //dis存放由起点到其他道路之间的最短路，数据太大所以开long long 24 ll dis[1000002]; 25 //vis判断某点是否已达最短 26 bool vis[1000002]; 27 //a,b,c存放从a到b的距离c的数据 28 int a[1000002],b[1000002],c[1000002]; 29 int n,m; 30 //最后的和太大，依此开long long 31 ll ans; 32 void spfa() 33 { 34 //qu用来存放需要遍历的点 35 queue<int> qu; 36 //初始dis为最大数 37 for(int i=1;i<=n;i++) 38 { 39 dis[i]=INF; 40 } 41 //初始vis为0，表示所有点都未被记录 42 memset(vis,0,sizeof(vis)); 43 //1到1的距离为0 44 dis[1]=0; 45 //1不需要记录 46 vis[1]=1; 47 //第一个点为1，入队 48 qu.push(1); 49 //当队伍为空时退出 50 while(!qu.empty()) 51 { 52 //出队 53 int s=qu.front(); 54 qu.pop(); 55 //k的记录被推反，重新记录 56 vis[s]=0; 57 //遍历与s相连的其他点 58 for(int i=0;i<ve[s].size();i++) 59 { 60 int u=ve[s][i].first; 61 int v=ve[s][i].second; 62 //如果1到s之间距离加上s到u将距离比1到u点距离小则更新dis 63 if(dis[u]>dis[s]+v) 64 { 65 //更新dis[u] 66 dis[u]=dis[s]+v; 67 //如果u点已被标记则不入队 68 if(!vis[u]) 69 { 70 //由于u点距离被更新，因此需要重新遍历与u连接的点，所以将u点入队，且标记u点 71 vis[u]=1; 72 qu.push(u); 73 } 74 } 75 } 76 } 77 //累加所有最短路的和 78 for(int i=1;i<=n;i++) 79 { 80 ans+=dis[i]; 81 } 82 } 83 int main() 84 { 85 //t个测试 86 int t; 87 scanf("%d",&t); 88 while(t--) 89 { 90 //n个点，m条边 91 scanf("%d %d",&n,&m); 92 ans=0; 93 for(int i=1;i<=m;i++) 94 { 95 scanf("%d %d %d",&a[i],&b[i],&c[i]); 96 //将由a开始的正向数据存入ve中 97 ve[a[i]].push_back(make_pair(b[i],c[i])); 98 } 99 //最短路遍历求和 100 spfa(); 101 //清除ve的数据 102 for(int i=1;i<=n;i++) 103 { 104 ve[i].clear(); 105 } 106 for(int i=1;i<=m;i++) 107 { 108 //将由其他点开始的反向数据存入ve中 109 ve[b[i]].push_back(make_pair(a[i],c[i])); 110 } 111 //最短路遍历求和 112 spfa(); 113 //输出结果 114 printf("%lld\n",ans); 115 //清除ve的数据 116 for(int i=1;i<=n;i++) 117 { 118 ve[i].clear(); 119 } 120 } 121 }