poj 3463(最短路和比最短路大1的路的数量)(dijkstra)

求最短路和比最短路大1的路的数量

dij松弛的条件改变下，有四种情况

1.比最短路短2.等于最短路3.长与最短路但短于次短路4.等于次短路

具体见代码：

  1 #include <iostream>

  2 #include <cstdio>

  3 #include <cstring>

  4 #include <algorithm>

  5 #include <queue>

  6 #include <vector>

  7 using namespace std;

  8 const int inf = 0x3f3f3f3f;

  9 const int maxn = 1005;

 10 int n,m,S,F;

 11 struct edge

 12 {

 13     int v,c;

 14     edge(){}

 15     edge(int v, int c):v(v), c(c){}

 16 };

 17 vector <edge> adj[maxn];

 18 

 19 void init()

 20 {

 21     int u,v,c;

 22     scanf("%d%d", &n, &m);

 23     memset(adj, 0, sizeof(adj));

 24     for (int i = 0; i < m; ++i)

 25     {

 26         scanf("%d%d%d", &u, &v, &c);

 27         adj[u].push_back(edge(v,c));

 28     }

 29     scanf("%d%d", &S, &F);

 30 }

 31 

 32 int dij(int src)

 33 {

 34     bool vis[maxn][2];

 35     int dis[maxn][2],cnt[maxn][2];

 36     int v,c;

 37     for (int i = 0; i <= n; ++i)

 38     {

 39         vis[i][0] = vis[i][1] = 0;

 40         dis[i][0] = dis[i][1] = inf;

 41         cnt[i][0] = cnt[i][1] = 0;

 42     }

 43     dis[src][0] = 0;cnt[src][0] = 1;

 44     /*for (int i = 0; i < adj[src].size(); ++i)

 45     {

 46         v = adj[src][i].v;

 47         c = adj[src][i].c;

 48         dis[v][0] = c;

 49         cnt[v][0] = 1;

 50     }

 51     vis[src][0] = 1;*/

 52     for (int i = 1; i < n * 2; ++i)

 53     {

 54         int tmp = inf, k = src, kind=0;

 55         for (int j = 1; j <= n; ++j)

 56         {

 57             if (!vis[j][0] && tmp > dis[j][0])

 58             {

 59                 tmp = dis[j][0];

 60                 k = j;

 61                 kind = 0;

 62             }

 63             else if(!vis[j][1] && tmp > dis[j][1])

 64             {

 65                 tmp = dis[j][1];

 66                 k = j;

 67                 kind = 1;

 68             }

 69         }

 70         if(tmp == inf) break;

 71         vis[k][kind] = 1;

 72         for (int j = 0; j < adj[k].size(); ++j)

 73         {

 74             v = adj[k][j].v;

 75             c = adj[k][j].c;

 76             if (dis[v][0] > tmp + c)

 77             {

 78                 dis[v][1] = dis[v][0];

 79                 dis[v][0] = tmp +c;

 80                 cnt[v][1] = cnt[v][0];

 81                 cnt[v][0] = cnt[k][kind];

 82             }

 83             else if (tmp + c == dis[v][0])

 84                 cnt[v][0] += cnt[k][kind];

 85             else if (dis[v][1] > c + tmp)

 86             {

 87                 dis[v][1] = c + tmp;

 88                 cnt[v][1] = cnt[k][kind];

 89             }

 90             else if (dis[v][1] == tmp + c)

 91                 cnt[v][1] += cnt[k][kind];

 92         }

 93     }

 94     if (dis[F][0] + 1 == dis[F][1])

 95         return cnt[F][0] + cnt[F][1];

 96     else

 97         return cnt[F][0];

 98 }

 99 int main()

100 {

101     int t;

102     scanf("%d", &t);

103     while (t--)

104     {

105         init();

106         printf("%d\n",dij(S));

107     }

108     return 0;

109 }