| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 5842 | Accepted: 2031 |
Description
Tour operator Your Personal Holiday organises guided bus trips across the Benelux. Every day the bus moves from one city S to another city F. On this way, the tourists in the bus can see the sights alongside the route travelled. Moreover, the bus makes a number of stops (zero or more) at some beautiful cities, where the tourists get out to see the local sights.
Different groups of tourists may have different preferences for the sights they want to see, and thus for the route to be taken from S to F. Therefore, Your Personal Holiday wants to offer its clients a choice from many different routes. As hotels have been booked in advance, the starting city S and the final city F, though, are fixed. Two routes from S to F are considered different if there is at least one road from a city A to a city B which is part of one route, but not of the other route.
There is a restriction on the routes that the tourists may choose from. To leave enough time for the sightseeing at the stops (and to avoid using too much fuel), the bus has to take a short route from S to F. It has to be either a route with minimal distance, or a route which is one distance unit longer than the minimal distance. Indeed, by allowing routes that are one distance unit longer, the tourists may have more choice than by restricting them to exactly the minimal routes. This enhances the impression of a personal holiday.
For example, for the above road map, there are two minimal routes from S = 1 to F = 5: 1 → 2 → 5 and 1 → 3 → 5, both of length 6. There is one route that is one distance unit longer: 1 → 3 → 4 → 5, of length 7.
Now, given a (partial) road map of the Benelux and two cities S and F, tour operator Your Personal Holiday likes to know how many different routes it can offer to its clients, under the above restriction on the route length.
Input
The first line of the input file contains a single number: the number of test cases to follow. Each test case has the following format:
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One line with two integers N and M, separated by a single space, with 2 ≤ N ≤ 1,000 and 1 ≤ M ≤ 10, 000: the number of cities and the number of roads in the road map.
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M lines, each with three integers A, B and L, separated by single spaces, with 1 ≤ A, B ≤ N, A ≠ B and 1 ≤ L ≤ 1,000, describing a road from city A to city B with length L.
The roads are unidirectional. Hence, if there is a road from A to B, then there is not necessarily also a road from B to A. There may be different roads from a city A to a city B.
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One line with two integers S and F, separated by a single space, with 1 ≤ S, F ≤ N and S ≠ F: the starting city and the final city of the route.
There will be at least one route from S to F.
Output
For every test case in the input file, the output should contain a single number, on a single line: the number of routes of minimal length or one distance unit longer. Test cases are such, that this number is at most 109 = 1,000,000,000.
Sample Input
2 5 8 1 2 3 1 3 2 1 4 5 2 3 1 2 5 3 3 4 2 3 5 4 4 5 3 1 5 5 6 2 3 1 3 2 1 3 1 10 4 5 2 5 2 7 5 2 7 4 1
Sample Output
3 2
Hint
The first test case above corresponds to the picture in the problem description.
Source
1 /* 2 求s到t的最短路与次短路（这里要求只比最短路多1）的条数之和 3 4 联想到最小，次小的一种更新关系： 5 if(x<最小)更新最小，次小 6 else if(==最小)更新方法数 7 else if(x<次小)更新次小 8 else if(x==次小)更新方法数 9 10 同时记录s到u最短，次短路及方法数 11 用一个堆每次取最小的，更新完后再入堆 12 还是那个原理，第一次遇到的就是最优的，然后vi标记为真 13 方法数注意是加法原理，不是乘法 14 \ 15 -- u -- v 所以是加法原理 16 */ 17 18 19 #include<iostream> 20 #include<queue> 21 using namespace std; 22 23 #define inf 1000000000 24 25 int n,m; 26 int edgenum; 27 int st,et; 28 29 int head[1010]; 30 int dis[1010][2]; 31 int cnt[1010][2]; 32 int vst[1010][2]; 33 34 typedef struct edge 35 { 36 int from,to; 37 int dis; 38 int next; 39 }Edge; 40 Edge edge[1010*10]; 41 42 typedef struct Node 43 { 44 int v; 45 int flag; //表示节点类型，0表示最短，1表示次短 46 Node(int a,int b) 47 { 48 v=a; 49 flag=b; 50 } 51 bool operator < (const Node &a)const 52 { 53 return dis[v][flag]>dis[a.v][a.flag]; 54 } 55 }Node; 56 57 void addedge(int a,int b,int c) 58 { 59 edge[edgenum].from=a;edge[edgenum].to=b;edge[edgenum].dis=c;edge[edgenum].next=head[a];head[a]=edgenum++; 60 } 61 62 int Dijkstra() 63 { 64 memset(vst,0,sizeof(vst)); 65 memset(cnt,0,sizeof(cnt)); 66 for(int i=1;i<=n;i++) 67 dis[i][0]=dis[i][1]=inf; 68 priority_queue<Node>Q; 69 Q.push(Node(st,0)); 70 dis[st][0]=0; 71 cnt[st][0]=1; 72 while(!Q.empty()) 73 { 74 Node top=Q.top(); 75 Q.pop(); 76 int v=top.v; 77 int flag=top.flag; 78 if(vst[v][flag]) 79 continue; 80 vst[v][flag]=1; //从v点（flag短路）出发的路线已被全部找完 81 for(int i=head[v];i!=-1;i=edge[i].next) //遍历 82 { 83 int to=edge[i].to; 84 int dist=dis[v][flag]+edge[i].dis; 85 if(dist<dis[to][0]) //如果找到路径比原最短路还短 86 { 87 if(dis[to][0]!=inf) //原来已经找到了最短路的情况下 88 { 89 dis[to][1]=dis[to][0]; //原最短路置为次短路 90 cnt[to][1]=cnt[to][0]; //原次短路条数置为次短路条数 91 Q.push(Node(to,1)); //入队：为什么要入队？ 92 } //因为当前点的距离已被更新，当前点之后的点也必须随之更新 93 dis[to][0]=dist; //更新最短路 94 cnt[to][0]=cnt[v][flag]; //更新最短路条数 95 Q.push(Node(to,0)); //入队 96 } 97 else if(dist==dis[to][0]) //如果找到路径和原最短路相同 98 { 99 cnt[to][0]+=cnt[v][flag]; //增加最短路径条数，但不入队 100 } 101 else if(dist<dis[to][1]) //如果找到路径比原次短路短 102 { 103 dis[to][1]=dist; //更新次短路 104 cnt[to][1]=cnt[v][flag]; //增加次短路条数 105 Q.push(Node(to,1)); 106 } 107 else if(dist==dis[to][1]) //如果找到路径和次短路相同 108 { 109 cnt[to][1]+=cnt[v][flag]; //增加方法数 110 } 111 } 112 } 113 if(dis[et][0]+1==dis[et][1]) 114 return cnt[et][0]+cnt[et][1]; 115 return cnt[et][0]; 116 } 117 118 int main() 119 { 120 int t; 121 scanf("%d",&t); 122 while(t--) 123 { 124 edgenum=0; 125 memset(head,-1,sizeof(head)); 126 scanf("%d%d",&n,&m); 127 while(m--) 128 { 129 int a,b,c; 130 scanf("%d%d%d",&a,&b,&c); 131 addedge(a,b,c); 132 } 133 scanf("%d%d",&st,&et); 134 printf("%d\n",Dijkstra()); 135 } 136 return 0; 137 }