HDU 5294 Tricks Device (2015 Multi-school first game 7th) Max Stream + shortest circuit

Title Link: http://acm.hdu.edu.cn/showproblem.php?pid=5294

Test Instructions: give you n a tomb, m path, a person in the Tomb No. 1th (beginning), another person in n

Tomb (end), the beginning of the person only through the shortest path to catch up with the end of the person, and the end of the person can cut off any path.

The first question--the end of the person to make the starting point that the person can not catch up the number of paths to be cut, the minimum number of paths to output

The second question--the beginning that man can catch up with the end of the person's case, the end that person can cut the maximum number of paths, output the maximum number of paths

Ideas: to make the starting point that the person can not catch up, as long as the shortest path does not exist on the good, so long as the shortest path to the flow of each road is 1, and the maximum flow of the ball out 1-n is, can cut the least number of paths;

If the beginning of the person can catch up, so long as his shortest path exists, and in these shortest path to find the least number of paths, then the maximum number of paths is the least M-path number of the path number.

Code:

#include  <cstdio> #include  <cstdlib> #include  <cmath> #include  <cstring > #include  <iostream> #include  <queue> #include  <algorithm> #include  < vector>using namespace std; #define  ll __int64#define inf 0x3f3f3f3fconst  int maxn=20005;struct qnode{int v;int c;qnode (int _v=0,int _c=0): V (_v), C (_ c) {}bool operator < (Const qnode &r) const{return c>r.c;}}; Bool vis[2005];int dist[2005];int mp[2005][2005];void dijkstra (Int n,int start)/ /Find the shortest path size {memset (vis,false,sizeof (VIS)); for (int i=1;i<=n;i++) Dist[i]=inf;priority_queue<qnode>que , while (!que.empty ()) Que.pop ();d Ist[start]=0;que.push (Qnode (start,0,0)); Qnode tmp;while (!que.empty ()) {tmp= Que.top (); Que.pop (); Int u=tmp.v;if (Vis[u]) continue;vis[u]=true;for (int i=1;i<=n;i++)           {            if (!vis[i]&&mp[u][i]<inf &&dist[i]>dist[u]+mp[u][i])              {                dist[i]=dist[ u]+mp[u][i];                 Que.push (Qnode (i,dist[i));            }  &NBSP;&NBSP;&NBSP;&NBSP;&NBSP;&NBSP;&NBSP;}}}INT&NBSP;MP2[2005][2005];VOID&NBSP;DIJKSTRA1 (int n,int  Start)//Find the shortest possible route {memset (vis,false,sizeof (VIS)); for (int i=1;i<=n;i++) Dist[i]=inf;priority_queue<qnode >que;while (!que.empty ()) Que.pop ();d Ist[start]=0;que.push (Qnode (start,0,0)); Qnode tmp;while (!que.empty ( ) {tmp=que.top (); Que.pop (); Int u=tmp.v;int gg=tmp.flag+1;if (Vis[u]) continue;vis[u]=true;for (int  i=1;i<=n;i++)         {             if (!vis[i]&&mp2[u][i]<inf&&dist[i]>dist[u]+mp2[u][i])              {                 dist[i]=dist[u]+mp2[u][i];                 que.push (Qnode (i,dist[i));             }        }}}struct  edge{    int from,to,cap,flow;    edge () {}     edge (INT&NBSP;F,INT&NBSP;T,INT&NBSP;C,INT&NBSP;FL): From (f), to (t), Cap (c), Flow (fl) {}};struct  dinic{    int n,m,s,t;    vector<edge> edges;     vector<int>&nBsp g[maxn];    int cur[maxn];    int d[maxn];     bool vis[maxn];    void init (int n,int s,int t)      {        this->n=n, this->s=s, this->t= T;        edges.clear ();         for (int i=0;i<n;i++)  g[i].clear ();     }    void  addedge (INT&NBSP;FROM,INT&NBSP;TO,INT&NBSP;CAP)     {         edges.push_back ( edge (from,to,cap,0)  );         edges.push_back ( edge (to,from,0,0)  )         m =  edges.size ();         g[from].push_back (m-2);       &nBsp; g[to].push_back (m-1);     }    bool bfs ()      {        queue<int> Q;         q.push (s);         memset (vis,0,sizeof (VIS));         d[s]=0;        vis[s]= True;        while (! Q.empty ())         {             int x=q.front ();  q.pop ();             for (Int i=0;i<g[x].size (); ++i)              {                 edge& e=edges[g[x][i]];                 if (!vis[e.to] && e.cap >e.flow)                 {                      d[e.to]=1+d[x];                     vis[e.to]=true;                     q.push (e.to);                 }             }        }         return vis[t];    }    int dfs (int  x,int a)     {        if (x==t | |  a==0)  return a;        int flow=0,f;         for (Int& i=cur[x];i<g[x].size (); ++i)          {            edge&  e=edges[g[x][i]];            if (d[e.to]==d[x ]+1 &&  (F=dfs (E.to,min (a,e.cap-e.flow)  )  ) >0)              {                 e.flow +=f;                 edges[G[x][i]^1].flow -=f;                 flow +=f;                a-=f;                 if (a==0)   break;            }         }        return flow;    }     int max_flow ()     {         int ans=0;        while (BFS ())          {            memset (cur,0, sizeof (cur));             ans += dfs (s), INF);         }        return  ans;  &Nbsp; }}dc;int ggg[2005][2005];int main () {    int n,m,i,j,a,b,c;     while (~scanf ("%d%d", &n,&m))     {         dc.init (n,1,n);         memset (Ggg,0,sizeof (GGG));         for (i=1;i<=n;i++)              for (j=1;j<=n;j++)                  mp[i][j]=mp2[i][j]=inf;        for (i=0;i<m;i++)         {        &NBSP;&NBSP;&NBSP;&NBSP;&NBSP;SCANF ("%d%d%d", &a,&b,&c);             if (mp[a][b]>c)              {                mp[a][b]=c;                 mp[b][a]=c;                 ggg[a][b]=1;                 ggg[b][a]=1;             }             else if (mp[a][b]==c)              {                 ggg[a][b]++;                 ggg[b][a]++;            }         }         dijkstra (n,1);         for (i=1;i<=n;i++)         {             for (j=1;j<=n;j++)              {                 if (Mp[i][j]+dist[i]==dist[j])                  {                     for (int k=0;k<ggg[i][j];k++)                          dc . Addedge (i,j,1);                     mp2[i][j]=1;                 }            }         }        dijkstra1 (n,1);         printf ("%d %d\n", Dc.max_flow (), m-dist[n]);    }     return 0;}

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HDU 5294 Tricks Device (2015 Multi-school first game 7th) Max Stream + shortest circuit