hdu5294||2015多校联合第一场1007 最短路+最大流

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

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(奇门遁甲) 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<li<=100) minute to pass channel i.

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

/**
hdu5294 最短路+最大流
题目大意：给定一个无向图，从起点到终点，只有走最短路，才能在规定时限内到达，问最少去掉几条边使不能到达，最多去掉几条边仍能到达
解题思路：http://blog.sina.com.cn/s/blog_15139f1a10102vnx5.html 和官方题解想的一样
*/
#include<cstdio>
#include<iostream>
#include<queue>
#include<string.h>
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<node; ++i)dis[i]=-1;
dis[q[r++]=src]=0;
for(l=0; l<r; ++l)
for(i=head[u=q[l]]; 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<node; ++i)work[i]=head[i];
while(delta=Dinic_dfs(src,oo))ret+=delta;
}
return ret;
}

///==================================================
const int INF=0x3f3f3f3f;
const int maxm=511111;
const int maxn=2111;

struct EdgeNode
{
int to;
int w;
int next;
};

EdgeNode edges[maxm];
int N,M;
int head1[maxn],edge1;
bool vis[maxn];
queue <int> 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)//单源最短路(s为起点，n为节点总数)
{
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<m;i++)
{
scanf("%d%d%d",&aa[i][0],&aa[i][1],&aa[i][2]);
addedge1(aa[i][0],aa[i][1],aa[i][2]);
addedge1(aa[i][1],aa[i][0],aa[i][2]);
}
spfa(1,n);
memcpy(dis2,dis1,sizeof(dis1));
//printf("dis2->%d\n",dis2
);
spfa(n,n);
// printf("dis1->%d\n",dis1[1]);
int k=0;
for(int i=0;i<m;i++)
{
if(dis2[aa[i][0]]>dis2[aa[i][1]])
swap(aa[i][0],aa[i][1]);
// printf("n-%d:%d-1 %d %d %d\n",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
)
{
bb[k][0]=aa[i][0];
bb[k++][1]=aa[i][1];
// printf("%d %d\n",bb[k-1][0],bb[k-1][1]);
}
}
prepare(n+1,1,n);
for(int i=0;i<k;i++)
{
addedge(bb[i][0],bb[i][1],1);
}
int ans1=Dinic_flow();
init();
for(int i=0;i<k;i++)
{
addedge1(bb[i][0],bb[i][1],1);
addedge1(bb[i][1],bb[i][0],1);
}
spfa(1,n);
//printf("%d\n",dis2
);
int ans2=m-dis1
;
printf("%d %d\n",ans1,ans2);
}
return 0;
}

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(奇门遁甲) 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<li<=100) minute to pass channel i.

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