2015ACM/ICPC亚洲区沈阳站-重现赛（感谢东北大学） Meeting（SPAF+拆点）

本文探讨了在大规模数据集上使用SPFA算法进行路径寻优的问题，通过优化建边方式减少时间消耗，并解决了农场中两个角色如何快速会面的具体案例。

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Meeting

Time Limit: 12000/6000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 3618 Accepted Submission(s): 1165

Problem Description

Bessie and her friend Elsie decide to have a meeting. However, after Farmer John decorated his
fences they were separated into different blocks. John's farm are divided into

n blocks labelled from

1 to

n .
Bessie lives in the first block while Elsie lives in the

n -th one. They have a map of the farm
which shows that it takes they

ti minutes to travel from a block in

Ei to another block
in

Ei where

Ei (1≤i≤m) is a set of blocks. They want to know how soon they can meet each other
and which block should be chosen to have the meeting.

Input

The first line contains an integer

T (1≤T≤6) , the number of test cases. Then

T test cases
follow.

The first line of input contains

n and

m .

2≤n≤105 . The following

m lines describe the sets

Ei (1≤i≤m) . Each line will contain two integers

ti(1≤ti≤109) and

Si (Si>0) firstly. Then

Si integer follows which are the labels of blocks in

Ei . It is guaranteed that

∑mi=1Si≤106 .

Output

For each test case, if they cannot have the meeting, then output "Evil John" (without quotes) in one line.

Otherwise, output two lines. The first line contains an integer, the time it takes for they to meet.
The second line contains the numbers of blocks where they meet. If there are multiple
optional blocks, output all of them in ascending order.

Sample Input

Sample Output

  
   Case #1: 3
3 4
Case #2: Evil John

   
    
     Hint
    
In the first case, it will take Bessie 1 minute travelling to the 3rd block, and it will take Elsie 3 minutes travelling to the 3rd block. It will take Bessie 3 minutes travelling to the 4th block, and it will take Elsie 3 minutes travelling to the 4th block. In the second case, it is impossible for them to meet.

Source

2015ACM/ICPC亚洲区沈阳站-重现赛（感谢东北大学）

123

【思路】

两遍SPFA 一个从1 一个从n 开始

问题在于数据边太大，一般暴力肯定会超时，因此在建边上优化，对于每个集合建立一个出口点，每次扫这个出口点， n个集合就是n个点

这样大大优化建边

【注意】

可能会爆 int 用long long

【代码】

#include <iostream>
#include <stdio.h>
#include <algorithm>
#include <cmath>
#include <math.h>
#include <cstring>
#include <string>
#include <queue>
#include <stack>
#include <stdlib.h>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <vector>
#define mem(a,b) memset(a,b,sizeof(a))
#define findx(x) lower_bound(b+1,b+1+bn,x)-b



#define FIN      freopen("input.txt","r",stdin)
#define FOUT     freopen("output.txt", What a Ridiculous Election"w",stdout)
#define S1(n)    scanf("%d",&n)
#define SL1(n)   scanf("%I64d",&n)
#define S2(n,m)  scanf("%d%d",&n,&m)
#define SL2(n,m)  scanf("%I64d%I64d",&n,&m)
#define Pr(n)     printf("%d\n",n)
#define lson rt << 1, l, mid
#define rson rt << 1|1, mid + 1, r

using namespace std;
typedef long long ll;

const ll  INF=1e18+10;
const int maxn=2e6+5;
const int MOD=1000000007;
const int mod=1e9+7;
int dir[5][2]={0,1,0,-1,1,0,-1,0};
int n,m;
ll dis[maxn],dis1[maxn];
int head[maxn];
int ans[maxn],tol;
ll cont;
int vis[maxn];
struct Edge{
    int v,w,next;
}edge[maxn];
void init()
{
    cont=0;
    memset(head,-1,sizeof(head));
}
void add(int u,int v,int w)
{
    edge[++cont].v=v;
    edge[cont].w=w;
    edge[cont].next=head[u];
    head[u]=cont;
}

void SPFA(int st)
{
    queue<int> Q;
    for(int i=0;i<=n+m+3;i++)
        dis[i]=INF;
    mem(vis,0);
    dis[st]=0;
    Q.push(st);
    vis[st]=1;
    while(!Q.empty())
    {
        ll u=Q.front();
        Q.pop();
        vis[u]=0;
        for(int t=head[u];t!=-1;t=edge[t].next)
        {
            ll v=edge[t].v;
            ll w=edge[t].w;
            if(dis[v]>dis[u]+w)
            {
                dis[v]=dis[u]+w;
                if(!vis[v])
                {
                    vis[v]=1;
                    Q.push(v);
                }
            }
        }
    }
}
int main()
{
    int T,cot=0;
    cin>>T;
    int te,num,x;
    while(T--)
    {
        scanf("%d %d",&n,&m);
        init();
        for(int i=1;i<=m;i++)
        {
            scanf("%d %d",&te,&num);
            for(int j=1;j<=num;j++)
            {
                scanf("%d",&x);
                add(x,i+n,te);
                add(i+n,x,te);
            }
        }
        SPFA(1);
        printf("Case #%d: ",++cot);
        if(dis[n]==INF)
        {
            printf("Evil John\n");
            continue;
        }
        for(int i=0;i<=maxn;i++)
            dis1[i]=dis[i];
        SPFA(n);
        tol=0;
        ll res=INF;

        for(int i=1;i<=n;i++)
        {
            res=min(res,max(dis[i],dis1[i]));
        }
        printf("%lld\n",res/2);
        tol=1;
        for(int i=1;i<=n;i++)
        {
            if(max(dis[i],dis1[i])==res)
                ans[tol++]=i;
        }
        for(int i=1;i<tol;i++)
        {
            if(i==1)
                printf("%d",ans[i]);
            else
                printf(" %d",ans[i]);
        }
        printf("\n");
    }
    return 0;
}