HDU3191 【输出次短路条数】

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=3191

How Many Paths Are There

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2128    Accepted Submission(s): 749

Problem Description

  oooccc1 is a Software Engineer who has to ride to the work place every Monday through Friday. For a long period, he went to office with the shortest path because he loves to sleep late…Time goes by, he find that he should have some changes as you could see, always riding with the same path is boring.
  One day, oooccc1 got an idea! Why could I take another path? Tired at all the tasks he got, he got no time to carry it out. As a best friend of his, you’re going to help him!
  Since oooccc1 is now getting up earlier, he is glad to take those paths, which are a little longer than the shortest one. To be precisely, you are going to find all the second shortest paths.
  You would be given a directed graph G, together with the start point S which stands for oooccc’1 his house and target point E presents his office. And there is no cycle in the graph. Your task is to tell him how long are these paths and how many there are.

 

Input

There are some cases. Proceed till the end of file.
The first line of each case is three integers N, M, S, E (3 <= N <= 50, 0 <= S , E <N)
N stands for the nodes in that graph, M stands for the number of edges, S stands for the start point, and E stands for the end point.
Then M lines follows to describe the edges: x y w. x stands for the start point, and y stands for another point, w stands for the length between x and y.
All the nodes are marked from 0 to N-1.

 

Output

For each case,please output the length and count for those second shortest paths in one line. Separate them with a single space.

题目大意：给定一张有向图，求出起点到终点的次短路条数。

思路：很简单的一道题次短路条数模板题， 但是 WA了 看讨论说是不能用优先队列dij，但是我不想改了 上个错误模板代码。

代码如下：

  1 #include<stdio.h>
  2 #include<string.h>
  3 #include<queue>
  4 #define mem(a, b) memset(a, b, sizeof(a))
  5 const int MAXN = 55;
  6 const int MAXM = 5000;
  7 const int inf = 0x3f3f3f3f;
  8 using namespace std;
  9 
 10 int n, m, st, ed;
 11 int head[MAXN], cnt;
 12 int dis[2][MAXN], num[2][MAXN], vis[2][MAXN];
 13 
 14 struct Edge
 15 {
 16     int to, next, w;
 17 }edge[MAXM];
 18 
 19 struct Node
 20 {
 21     int id, dis, p;
 22     bool operator < (const Node &a)const 
 23     {
 24         return dis > a.dis;
 25     }
 26 }no;
 27 
 28 void add(int a, int b, int c)
 29 {
 30     cnt ++;
 31     edge[cnt].to = b;
 32     edge[cnt].w = c;
 33     edge[cnt].next = head[a];
 34     head[a] = cnt;
 35 }
 36 
 37 void dij()
 38 {
 39     mem(vis, 0);
 40     priority_queue<Node> Q;
 41     for(int i = 0; i < n; i ++)
 42     {
 43         dis[0][i] = dis[1][i] = inf;
 44         num[0][i] = num[1][i] = 0;
 45     }
 46     dis[0][st] = 0;
 47     num[0][st] = 1;
 48     no.p = 0, no.id = st, no.dis = 0;
 49     Q.push(no);
 50     while(!Q.empty())
 51     {
 52         Node a = Q.top();
 53         Q.pop();
 54         if(vis[a.p][a.id])
 55             continue;
 56         vis[a.p][a.id] = 1;
 57         for(int i = head[a.id]; i != -1; i = edge[i].next)
 58         {
 59             int to = edge[i].to;
 60             if(dis[0][to] > dis[a.p][a.id] + edge[i].w)
 61             {
 62                 dis[1][to] = dis[0][to];
 63                 dis[0][to] = dis[a.p][a.id] + edge[i].w;
 64                 num[1][to] = num[0][to];
 65                 num[0][to] = num[a.p][a.id];
 66                 no.p = 0, no.dis = dis[0][to], no.id = to;
 67                 Q.push(no);
 68                 no.p = 1, no.dis = dis[1][to], no.id = to;
 69                 Q.push(no);
 70             }
 71             else if(dis[0][to] == dis[a.p][a.id] + edge[i].w)
 72                 num[0][to] += num[a.p][a.id];
 73             else if(dis[1][to] > dis[a.p][a.id] + edge[i].w)
 74             {
 75                 dis[1][to] = dis[a.p][a.id] + edge[i].w;
 76                 num[1][to] = num[a.p][a.id];
 77                 no.p = 1, no.dis = dis[1][to], no.id = to;
 78                 Q.push(no);
 79             }
 80             else if(dis[1][to] == dis[a.p][a.id] + edge[i].w)
 81                 num[1][to] += num[a.p][a.id];
 82         }
 83     }
 84 }
 85 
 86 int main()
 87 {
 88     while(scanf("%d%d%d%d", &n, &m, &st, &ed) != EOF)
 89     {
 90         mem(head, -1), cnt = 0;
 91         for(int i = 1; i <= m; i ++)
 92         {
 93             int a, b, c;//有向图 
 94             scanf("%d%d%d", &a, &b, &c);
 95             add(a, b, c);
 96         }
 97         dij();
 98         printf("%d %d\n", dis[1][ed], num[1][ed]);
 99     }
100     return 0;
101 }

HDU3191