Reverse a Road 最短路

Reverse a Road

Time Limit 1000ms

Memory Limit 65536K

description

　　Peter resides in the city of Nanuh, and goes to his working place in this city every weekday.He has been totally annoyed with the road traffic of this city. All the roads in this city are one-way, so he has to drive a longer way than he thinks he need.	One day, the following thought has come up to Peter’s mind: “How about making the sign of one road indicate the opposite direction? I think my act won’t be out as long as I change just one sign. Well, of course I want to make my route to the working place shorter as much as possible. Which road should I alter the direction of?” What a clever guy he is.	You are asked by Peter to write a program that finds the shortest route when the direction of up to one road is allowed to be altered. You don’t have to worry about the penalty for complicity, because you resides in a different country from Peter and cannot be punished by the law of his country. So just help him!
							

input

　　The input consists of a series of datasets, each of which is formatted as follows:
N
S T
M
A1 B1
A2 B2
: : :
AM BM	
，N denotes the number of points. S and T indicate the points where Andrew’s home and working place are located respectively. M denotes the number of roads. Finally, Ai and Bi indicate the starting and ending points of the i-th road respectively. Each point is identified by a unique number from 1 to N. Some roads may start and end at the same point. Also, there may be more than one road connecting the same pair of starting and ending points.	You may assume all the following: 1 <= N <= 1000, 1 <= M <= 10000, and S , T. 
							

output

　　For each dataset, print a line that contains the shortest distance (counted by the number of passed roads) and the road number whose direction should be altered. If there are multiple ways to obtain the shortest distance, choose one with the smallest road number. If no direction change results in a shorter route, print 0 as the road number.	Separate the distance and the road number by a single space. No extra characters are allowed.
							

sample_input

4
1 4
4
1 2
2 3
3 4

4 1
							

sample_output

1 4
							

----------------------------------------------------------

求出起点到每个点的最短路dist[i]，反向求出终点到每个点的最短路dist0[i]，枚举每条边(u,v) dist[v]+dist0[u]+1的最小值即答案。

----------------------------------------------------------

#include <iostream>
#include <cstdio>
#include <queue>
#include <cstring>

using namespace std;

const int maxn=1111;
const int maxm=111111;
const int OO=1e9;

struct EDGE
{
    int to;
    int w;
    int next;
}edges[maxm],edges0[maxm];

struct COCO{
    int u;
    int v;
    int w;
}coco[maxm];
int lk;

int node,src,dest,edge;
int head[maxn],dist[maxn];
int head0[maxn],dist0[maxn],edge0;

void prepare(int _node,int _src=0,int _dest=0)
{
    node=_node,src=_src,dest=_dest;
    for (int i=0; i<=node; i++) head[i]=-1,head0[i]=-1;
    edge=0;
    edge0=0;
    lk=0;
}

void addedge(int u,int v,int c)
{
    edges[edge].w=c,edges[edge].to=v,edges[edge].next=head[u],head[u]=edge++;
}

void addedge0(int u,int v,int c)
{
    edges0[edge0].w=c,edges0[edge0].to=v,edges0[edge0].next=head0[u],head0[u]=edge0++;
}

bool spfa(int node,int src,int head[],EDGE edges[],int dist[])
{
    int i,l,r,u,v,w;
    bool visit[maxn];
    int q[maxn],outque[maxn];
    memset(visit,0,sizeof(visit));
    memset(outque,0,sizeof(outque));
    for (int i=0; i<=node; i++) dist[i]=OO;
    r=0;
    q[r++]=src;
    dist[src]=0;
    visit[src]=true;
    for (l=0; l!=r; ( (++l>=maxn)?(l=0):(1) ))
    {
        u=q[l];
        visit[u]=false;
        outque[u]++;
        if (outque[u]>node) return false;
        for (i=head[u]; i!=-1; i=edges[i].next)
        {
            v=edges[i].to;
            w=edges[i].w;
            if (dist[u]+w<dist[v])
            {
                dist[v]=dist[u]+w;
                if (visit[v]) continue;
                q[r++]=v;
                visit[v]=true;
                if (r>=maxn) r=0;
            }
        }
    }
    return true;
}

void dijkstra(int n,int src,int dist[],int head[],EDGE edges[])
{
    int _min,u,v,w;
    bool visit[maxn]={0};
    for (int i=1;i<=n;i++)
    {
        dist[i]=OO;
    }
    dist[src]=0;
    for (int loop=1; loop<=n; loop++)
    {
        u=0;
        _min=OO;
        for (int i=1; i<=n; i++)
        {
            if (!visit[i]&&dist[i]<_min)
            {
                _min=dist[i];
                u=i;
            }
        }
        if (u==0) return;
        visit[u]=true;
        for (int i=head[u];i!=-1;i=edges[i].next)
        {
            v=edges[i].to;
            w=edges[i].w;
            if (!visit[v]&&dist[u]+w<dist[v])
            {
                dist[v]=dist[u]+w;
            }
        }
    }
}

int main()
{
    int n,s,t,m;
    while (~scanf("%d",&n))
    {
        scanf("%d%d",&s,&t);
        prepare(n,s,t);
        scanf("%d",&m);
        while (m--)
        {
            int u,v;
            scanf("%d%d",&u,&v);
            addedge(u,v,1);
            addedge0(v,u,1);

            coco[lk].u=u;
            coco[lk].v=v;
            coco[lk++].w=1;
        }
        spfa(n,s,head,edges,dist);
        spfa(n,t,head0,edges0,dist0);

        //dijkstra(n,s,dist,head,edges);
        //dijkstra(n,t,dist0,head0,edges0);

        int _min=dist[t];
        int ans=-1;
        for (int i=0;i<lk;i++)
        {
            int u,v,w;
            u=coco[i].u;
            v=coco[i].v;
            w=coco[i].w;
            if (dist[v]+dist0[u]+w<_min)
            {
                _min=dist[v]+dist0[u]+w;
                ans=i;
            }
        }
        printf("%d %d\n",_min,ans+1);
    }

    return 0;
}