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题目链接:
Time Limit: 12000/6000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 2024 Accepted Submission(s): 628
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <bits/stdc++.h>
#include <stack>
#include <map>
using namespace std;
#define For(i,j,n) for(int i=j;i<=n;i++)
#define mst(ss,b) memset(ss,b,sizeof(ss));
#define lson o<<1
#define rson o<<1|1
typedef long long LL;
template<class T> void read(T&num) {
char CH; bool F=false;
for(CH=getchar();CH<‘0‘||CH>‘9‘;F= CH==‘-‘,CH=getchar());
for(num=0;CH>=‘0‘&&CH<=‘9‘;num=num*10+CH-‘0‘,CH=getchar());
F && (num=-num);
}
int stk[70], tp;
template<class T> inline void print(T p) {
if(!p) { puts("0"); return; }
while(p) stk[++ tp] = p%10, p/=10;
while(tp) putchar(stk[tp--] + ‘0‘);
putchar(‘\n‘);
}
const LL mod=1e9+7;
const double PI=acos(-1.0);
const LL inf=1e18;
const int N=1e6+10;
const int maxn=2e6+5;
const double eps=1e-12;
int n,m,cnt,s,e,vis[maxn],c[maxn],head[maxn];
LL dis[2][maxn];
struct Edge
{
int to,next,val;
}edge[maxn];
queue<int>qu;
inline void add_edge(int from,int to,int va)
{
edge[cnt].to=to;
edge[cnt].next=head[from];
edge[cnt].val=va;
head[from]=cnt++;
}
void solve(int s,int e,int f)
{
while(!qu.empty())qu.pop();
mst(vis,0);
for(int i=0;i<maxn;i++)dis[f][i]=inf;
dis[f][s]=0;
qu.push(s);
vis[s]=1;
while(!qu.empty())
{
int fr=qu.front();qu.pop();
for(int i=head[fr];i!=-1;i=edge[i].next)
{
int x=edge[i].to;
if(dis[f][x]>dis[f][fr]+edge[i].val)
{
dis[f][x]=dis[f][fr]+edge[i].val;
if(!vis[x])qu.push(x),vis[x]=1;
}
}
vis[fr]=0;
}
}
int main()
{
int t,Case=0;
read(t);
while(t--)
{
printf("Case #%d: ",++Case);
mst(head,-1);
cnt=0;
read(n);read(m);
int x,h,t;
for(int i=1;i<=m;i++)
{
read(t);read(h);
for(int j=1;j<=h;j++)
{
read(x);
add_edge(x,n+i,t);
add_edge(n+i,x,t);
}
}
solve(1,n,0);
solve(n,1,1);
LL ans=inf;
for(int i=1;i<=n;i++)ans=min(ans,max(dis[0][i],dis[1][i]));
if(ans==inf)printf("Evil John\n");
else
{
printf("%lld\n",ans/2);
int num=0;
for(int i=1;i<=n;i++)if(max(dis[0][i],dis[1][i])==ans)c[++num]=i;
for(int i=1;i<num;i++)printf("%d ",c[i]);
printf("%d\n",c[num]);
}
}
return 0;
}
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原文地址:http://www.cnblogs.com/zhangchengc919/p/5928373.html