| Time Limit: 8000MS | Memory Limit: 262144K | |
| Total Submissions: 17538 | Accepted: 5721 |
Description
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.
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
22 21 2 132 1 334 61 2 102 1 601 3 203 4 102 4 54 1 50
Sample Output
46210 题意：意思很简单，给一个无向图，顶点1—n,求顶点1到其他所有顶点的来回费用最小，即求顶点1到其他顶点的最小费用以及其他所有顶点到顶点1的最小费用的和。 思路：这个题给的数据是100W，如果用邻接矩阵存可能会超内存，如果用Dij可能会超时，所以就用邻接表+SPFA，建图的时候要正逆向建图，正向建图求顶点1到其他顶点的最短路，逆向建图求其他顶点到顶点1的最短路，所以两次SPFA求以顶点1为源点的最短路径的和就是最小花费。
1 #include2 #include 3 #include 4 using namespace std; 5 6 const int maxn = 1000010; 7 const int INF = 0x3f3f3f3f; 8 typedef long long LL; 9 10 struct edge 11 { 12 int to,w; 13 struct edge *next; 14 }; 15 16 struct edge *map1[maxn],*map2[maxn]; 17 int n,m; 18 LL ans; 19 LL dis[maxn]; 20 int inque[maxn]; 21 22 void SPFA(int flag) 23 { 24 queue que; 25 while(!que.empty()) 26 que.pop(); 27 for(int i = 1; i <= n; i++) 28 { 29 dis[i] = INF; 30 inque[i] = 0; 31 } 32 33 dis[1] = 0; 34 inque[1] = 1; 35 que.push(1); 36 while(!que.empty()) 37 { 38 int u = que.front(); 39 que.pop(); 40 inque[u] = 0; 41 42 struct edge *tmp; 43 if(flag == 1) 44 tmp = map1[u]; 45 else tmp = map2[u]; 46 while(tmp) 47 { 48 int v = tmp->to; 49 int w = tmp->w; 50 if(dis[v] > dis[u]+w) 51 { 52 dis[v] = dis[u] + w; 53 if(!inque[v]) 54 { 55 inque[v] = 1; 56 que.push(v); 57 } 58 } 59 tmp = tmp->next; 60 } 61 } 62 for(int i = 1; i <= n; i++) 63 ans += dis[i]; 64 } 65 66 int main() 67 { 68 int t; 69 scanf("%d",&t); 70 while(t--) 71 { 72 int u,v,w; 73 struct edge *tmp1,*tmp2; 74 memset(map1,0,sizeof(map1)); 75 memset(map2,0,sizeof(map2)); 76 scanf("%d %d",&n,&m); 77 for(int i = 0; i < m; i++) 78 { 79 scanf("%d %d %d",&u,&v,&w); 80 tmp1 = new edge;//为tmp1开辟空间；正向建图 81 tmp1->to = v; 82 tmp1->w = w; 83 tmp1->next = NULL; 84 if(map1[u] == NULL) 85 map1[u] = tmp1; 86 else 87 { 88 tmp1->next = map1[u]; 89 map1[u] = tmp1; 90 } 91 92 tmp2 = new edge;//为tmp2开辟空间；逆向建图 93 tmp2->to = u; 94 tmp2->w = w; 95 tmp2->next = NULL; 96 97 if(map2[v] == NULL) 98 map2[v] = tmp2; 99 else100 {101 tmp2->next = map2[v];102 map2[v] = tmp2;103 }104 }105 /*for(int i = 1; i <= n; i++)//输出邻接表。106 {107 printf("%d: ",i);108 tmp1 = map1[i];109 while(tmp1)110 {111 printf("%d ",tmp1->to);112 tmp1 = tmp1->next;113 }114 printf("\n");115 }*/116 ans = 0;117 SPFA(1);//正向寻找1顶点到其他所有顶点的最短距离118 SPFA(0);//逆向寻找1顶点到其他所有顶点的最短距离119 printf("%lld\n",ans);120 }121 return 0;122 }