hdu5294||2015多校聯合第一場1007 最短路+最大流

8 9
1 2 2
2 3 2
2 4 1
3 5 3
4 5 4
5 8 1
1 6 2
6 7 5
7 8 1

2 6

/**
hdu5294 最短路+最大流
題目大意：給定一個無向圖，從起點到終點，只有走最短路，才能在規定時限內到達，問最少去掉幾條邊使不能到達，最多去掉幾條邊仍能到達
解題思路：http://blog.sina.com.cn/s/blog_15139f1a10102vnx5.html 和官方題解想的一樣
*/
#include
#include
#include
#include
using namespace std;
const int oo=1e9;
const int mm=161111;
const int mn=2330;
int node,src,dest,edge;
int ver[mm],flow[mm],_next[mm];
int head[mn],work[mn],dis[mn],q[mn];

void prepare(int _node,int _src,int _dest)
{
    node=_node,src=_src,dest=_dest;
    for(int i=0; i<=node; ++i)head[i]=-1;
    edge=0;
}

void addedge(int u,int v,int c)
{
    ver[edge]=v,flow[edge]=c,_next[edge]=head[u],head[u]=edge++;
    ver[edge]=u,flow[edge]=0,_next[edge]=head[v],head[v]=edge++;
}

bool Dinic_bfs()
{
    int i,u,v,l,r=0;
    for(i=0; i=0; i=_next[i])
            if(flow[i]&&dis[v=ver[i]]<0)
            {
                dis[q[r++]=v]=dis[u]+1;
                if(v==dest)return 1;
            }
    return 0;
}
int Dinic_dfs(int u,int exp)
{
    if(u==dest)return exp;
    for(int &i=work[u],v,tmp; i>=0; i=_next[i])
        if(flow[i]&&dis[v=ver[i]]==dis[u]+1&&(tmp=Dinic_dfs(v,min(exp,flow[i])))>0)
        {
            flow[i]-=tmp;
            flow[i^1]+=tmp;
            return tmp;
        }
    return 0;
}
int Dinic_flow()
{
    int i,ret=0,delta;
    while(Dinic_bfs())
    {
        for(i=0; i que;
int dis1[maxn],dis2[maxn];

void addedge1(int u,int v,int c)
{
    edges[edge1].w=c,edges[edge1].to=v,edges[edge1].next=head1[u],head1[u]=edge1++;
}

void init()
{
    memset(head1,-1,sizeof(head1));
    edge1=0;
}

void spfa(int s,int n)//單源最短路(s為起點，n為節點總數)
{
    int u;
    for (int i=0; i<=n; i++)
        dis1[i]=INF;
    memset(vis,0,sizeof(vis));
    while (!que.empty()) que.pop();
    que.push(s);
    vis[s]=true;
    dis1[s]=0;
    while (!que.empty())
    {
        u=que.front();
        que.pop();
        vis[u]=false;
        for (int i=head1[u]; i!=-1; i=edges[i].next)
        {
            int v=edges[i].to;
            int w=edges[i].w;
            if (dis1[v]>dis1[u]+w)
            {
                dis1[v]=dis1[u]+w;
                if (!vis[v])
                {
                    vis[v]=true;
                    que.push(v);
                }
            }
        }
    }
}
////========================================
int aa[60080][3],bb[60080][3];
int main()
{
    int n,m;
    while(~scanf(%d%d,&n,&m))
    {
        init();
        for(int i=0;i%d
,dis2[n]);
        spfa(n,n);
       // printf(dis1->%d
,dis1[1]);
        int k=0;
        for(int i=0;idis2[aa[i][1]])
                swap(aa[i][0],aa[i][1]);
           // printf(n-%d:%d-1 %d %d %d
,aa[i][1],aa[i][0],dis1[aa[i][1]],aa[i][2],dis2[aa[i][0]]);
            if(dis1[aa[i][1]]+aa[i][2]+dis2[aa[i][0]]==dis2[n])
            {
                bb[k][0]=aa[i][0];
                bb[k++][1]=aa[i][1];
               // printf(%d %d
,bb[k-1][0],bb[k-1][1]);
            }
        }
        prepare(n+1,1,n);
        for(int i=0;i
 


Problem Description

Innocent Wu follows Dumb Zhang into a ancient tomb. Innocent Wu’s at the entrance of the tomb while Dumb Zhang’s at the end of it. The tomb is made up of many chambers, the total number is N. And there are M channels connecting the chambers. Innocent Wu wants
 to catch up Dumb Zhang to find out the answers of some questions, however, it’s Dumb Zhang’s intention to keep Innocent Wu in the dark, to do which he has to stop Innocent Wu from getting him. Only via the original shortest ways from the entrance to the end
 of the tomb costs the minimum time, and that’s the only chance Innocent Wu can catch Dumb Zhang.

Unfortunately, Dumb Zhang masters the art of becoming invisible(奇門遁甲) and tricks devices of this tomb, he can cut off the connections between chambers by using them. Dumb Zhang wanders how many channels at least he has to cut to stop Innocent Wu. And Innocent
 Wu wants to know after how many channels at most Dumb Zhang cut off Innocent Wu still has the chance to catch Dumb Zhang.



 



Input

There are multiple test cases. Please process till EOF.

For each case,the first line must includes two integers, N(<=2000), M(<=60000). N is the total number of the chambers, M is the total number of the channels.

In the following M lines, every line must includes three numbers, and use ai、bi、li as channel i connecting chamber ai and bi(1<=ai,bi<=n), it costs li(0
The entrance of the tomb is at the chamber one, the end of tomb is at the chamber N.



 



Output

Output two numbers to stand for the answers of Dumb Zhang and Innocent Wu’s questions.

 



Sample Input
8 9
1 2 2
2 3 2
2 4 1
3 5 3
4 5 4
5 8 1
1 6 2
6 7 5
7 8 1


Sample Output
2 6


Source 2015 Multi-University Training Contest 1