HDOJ 5294 Tricks Device 最短路(记录路径)+最小割

最短路记录路径,同一时候求出最短的路径上最少要有多少条边,

然后用在最短路上的边又一次构图后求最小割.

Tricks Device

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 1584    Accepted Submission(s): 388

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

 

Author

FZUACM

 

Source

2015 Multi-University Training Contest 1

 

/* ***********************************************
Author        :CKboss
Created Time  :2015年07月24日 星期五 10时07分09秒
File Name     :HDOJ5294.cpp
************************************************ */

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
#include <cmath>
#include <cstdlib>
#include <vector>
#include <queue>
#include <set>
#include <map>

using namespace std;

typedef pair<int,int> pII;

const int INF=0x3f3f3f3f;
const int maxn=2200;

int n,m;

/*************EDGE********************/

struct Edge
{
	int to,next,cost,cap,flow;
}edge[maxn*60],edge2[maxn*60];

int Adj[maxn],Size;
int Adj2[maxn],Size2;

void Add_Edge(int u,int v,int c)
{
	edge[Size].to=v;
	edge[Size].next=Adj[u];
	edge[Size].cost=c;
	Adj[u]=Size++;
}

/********spfa************/

int dist[maxn];
bool inQ[maxn];

vector<int> Pre[maxn];

int spfa(Edge* edge,int* Adj)
{
    memset(dist,63,sizeof(dist));
    memset(inQ,false,sizeof(inQ));
    dist[1]=0;
    queue<int> q;
    inQ[1]=true;q.push(1);

    while(!q.empty())
    {
        int u=q.front();q.pop();

        for(int i=Adj[u];~i;i=edge[i].next)
        {
            int v=edge[i].to;
            if(dist[v]>dist[u]+edge[i].cost)
            {
				Pre[v].clear();
				Pre[v].push_back(u);

                dist[v]=dist[u]+edge[i].cost;

                if(!inQ[v])
                {
                    inQ[v]=true;
                    q.push(v);
                }
            }
			else if(dist[v]==dist[u]+edge[i].cost)
			{
				Pre[v].push_back(u);
			}
        }

        inQ[u]=false;

    }
    return dist[n];
}

/********************rebuild************************/

void Add_Edge2(int u,int v,int w,int rw=0)
{
	edge2[Size2].cost=1;
	edge2[Size2].to=v; edge2[Size2].cap=w; edge2[Size2].next=Adj2[u];
	edge2[Size2].flow=0; Adj2[u]=Size2++;

	edge2[Size2].cost=1;
	edge2[Size2].to=u; edge2[Size2].cap=w; edge2[Size2].next=Adj2[v];
	edge2[Size2].flow=0; Adj2[v]=Size2++;
}

bool used[maxn];
int edges;

void rebuild()
{
	memset(used,false,sizeof(used));
	queue<int> q;
	q.push(n); used[n]=true;
	edges=0;

	while(!q.empty())
	{
		int v=q.front(); q.pop();
		for(int i=0,sz=Pre[v].size();i<sz;i++)
		{
			int u=Pre[v][i];
			/// u--->v
			//cout<<u<<" ---> "<<v<<endl;
			edges++;
			Add_Edge2(u,v,1);

			if(used[u]==false)
			{
				used[u]=true; q.push(u);
			}
		}
	}
}

/************************max_flow*******************************/

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

int sap(int start,int end,int N,Edge* edge=edge2)
{
	memset(gap,0,sizeof(gap));
	memset(dep,0,sizeof(dep));
	memcpy(cur,Adj2,sizeof(Adj2));

	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;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;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;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=Adj2[u];~i;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 init()
{
	memset(Adj,-1,sizeof(Adj)); Size=0;
	memset(Adj2,-1,sizeof(Adj2)); Size2=0;
	for(int i=1;i<=n;i++) Pre[i].clear();
}

int main()
{
	//freopen("in.txt","r",stdin);
	//freopen("out.txt","w",stdout);

	while(scanf("%d%d",&n,&m)!=EOF)
	{
		init();
		for(int i=0,u,v,c;i<m;i++)
		{
			scanf("%d%d%d",&u,&v,&c);
			Add_Edge(u,v,c); Add_Edge(v,u,c);
		}
		spfa(edge,Adj);
		rebuild();
		int max_flow=sap(1,n,n);
		int min_short_path=spfa(edge2,Adj2);
		printf("%d %d\n",max_flow,m-min_short_path);
	}

    return 0;
}