题意

A从1要追在N的 B 只能走最短的路

问B最少切断多少条路可以让A不能过来

问B最多切断多少条路A还是能过来

对于1 求出1到N的所有最短路的路径，对其求最小割

对于2 求出长度最小的最短路即可

如何求所有最短路呢  利用这个

void dfs(int s,int ans)

{

    if(s==N)

    {

        ANS2=min(ans,ANS2);

        return ;

    }

    for(EDGE *p=Graph[s].first;p;p=p->next)

    {

        if(dist[s]+p->w==dist[p->to])

        {

            if(VISIT[s]==0) { VISIT[s]=1;nn++;}

            if(VISIT[p->to]==0) { VISIT[p->to]=1;nn++;}

            addedge(s,p->to,1,0);

            dfs(p->to,ans+1);

        }

    }

}

即这个判断来判断所有路径

 if(dist[s]+p->w==dist[p->to])

代码如下

#include <cstdio>

#include <cstdlib>

#include <cmath>

#include <cstring>

#include <ctime>

#include <algorithm>

#include <iostream>

#include <sstream>

#include <string>

#define oo 0x13131313

using namespace std;

const int maxn=5000;

int visit[maxn];

int dist[maxn];

int VISIT[maxn];

int N,m;

struct EDGE{

    int w;

    int to;

    EDGE *next;

}E[300000],*EE;

struct node{

    EDGE *first;

}Graph[maxn];

//网络流

const int MAXN=4000+5;

const int MAXM=300000;

const int INF=0x3f3f3f3f;

int nn=0;

int ANS2=0x3f3f3f3f;

struct Edge

{

    int to,next,cap,flow;

    void get(int a,int b,int c,int d)

    {

        to=a;next=b;cap=c;flow=d;

    }

}edge[MAXM];

int tol;

int head[MAXN];

int gap[MAXN],dep[MAXN],pre[MAXN],cur[MAXN];

void init()

{

    tol=0;

    memset(head,-1,sizeof(head));

}

//单向图三个参数，无向图四个参数

void addedge(int u,int v,int w,int rw=0)

{

    edge[tol].get(v,head[u],w,0);head[u]=tol++;

    edge[tol].get(u,head[v],rw,0);head[v]=tol++;

}

int sap(int start,int end,int N)

{

    memset(gap,0,sizeof(gap));

    memset(dep,0,sizeof(dep));

    memcpy(cur,head,sizeof(head));

    int u=start;

    pre[u]=-1;

    gap[0]=N;

    int ans=0;

    while(dep[start]<N)

    {

        if(u==end)

        {

            int Min=INF;

            for(int i=pre[u];i!=-1;i=pre[edge[i^1].to])

                if(Min>edge[i].cap-edge[i].flow)

                   Min=edge[i].cap-edge[i].flow;

            for(int i=pre[u];i!=-1;i=pre[edge[i^1].to])

            {

                edge[i].flow+=Min;

                edge[i^1].flow-=Min;

            }

            u = start;

            ans+=Min;

            continue;

        }

        bool flag=false;

        int v;

        for(int i=cur[u];i !=-1;i=edge[i].next)

        {

            v=edge[i].to;

            if(edge[i].cap-edge[i].flow&&dep[v]+1==dep[u])

            {

                flag=true;

                cur[u]=pre[v]=i;

                break;

            }

        }

        if(flag)

        {

            u=v;

            continue;

        }

        int Min=N;

        for(int i=head[u];i!=-1;i=edge[i].next)

            if(edge[i].cap-edge[i].flow&&dep[edge[i].to]<Min)

        {

            Min=dep[edge[i].to];

            cur[u]=i;

        }

        gap[dep[u]]--;

        if(!gap[dep[u]]) return ans;

        dep[u]=Min+1;

        gap[dep[u]]++;

        if(u!=start) u=edge[pre[u]^1].to;

    }

    return ans;

}

void Link(int u,int v,int w)

{

    EE->next=Graph[u].first;EE->to=v;EE->w=w;Graph[u].first=EE++;

}

void input()

{

    ANS2=m;

    memset(E,0,sizeof(E));

    memset(Graph,0,sizeof(Graph));

    EE=E+1;

    int a,b,c;

    for(int i=1;i<=m;i++)

    {

        scanf("%d%d%d",&a,&b,&c);

        Link(a,b,c);

        Link(b,a,c);

    }

}

void finddist(int &s)

{

    int ans=2000000000;

    for(int i=1;i<=N;i++)

    if(visit[i]==0)

    {

        if(dist[i]<ans)

        {

            ans=dist[i];

            s=i;

        }

    }

    visit[s]=1;

}

void Dijkstra(int s)

{

    dist[s]=0;

    for(int k=1;k<=N-1;k++)

    {

        finddist(s);

        for(EDGE *p=Graph[s].first;p;p=p->next)

            if(dist[s]+p->w<dist[p->to]) dist[p->to]=dist[s]+p->w;

    }

}

void CSH()

{

    memset(visit,0,sizeof(visit));

    for(int i=0;i<maxn;i++)

    dist[i]=1000000000;

}

void dfs(int s,int ans)

{

    if(s==N)

    {

        ANS2=min(ans,ANS2);

        return ;

    }

    for(EDGE *p=Graph[s].first;p;p=p->next)

    {

        if(dist[s]+p->w==dist[p->to])

        {

            if(VISIT[s]==0) { VISIT[s]=1;nn++;}

            if(VISIT[p->to]==0) { VISIT[p->to]=1;nn++;}

            addedge(s,p->to,1,0);

            dfs(p->to,ans+1);

        }

    }

}

void solve()

{

     nn=0;

     memset(VISIT,0,sizeof(VISIT));

     CSH();

     Dijkstra(1);

      init();

     dfs(1,0);

     printf("%d %d\n",sap(1,N,nn),m-ANS2);

}

int main()

{

  // freopen("a.in","r",stdin);

  // freopen("a.out","w",stdout);

    while(cin>>N>>m)

    {

        input();

        solve();

    }

}