Pandaland HDU

枚举一下边
求边的两个点u,v的最短路 但是不能直接从u到v或者v到u
求出最短路再加上边的花费

#include <iostream>#include <algorithm>#include <sstream>#include <string>#include <queue>#include <cstdio>#include <map>#include <set>#include <utility>#include <stack>#include <cstring>#include <cmath>#include <vector>#include <ctime>#include <bitset>using namespace std;#define pb push_back#define sd(n) scanf("%d",&n)#define sdd(n,m) scanf("%d%d",&n,&m)#define sddd(n,m,k) scanf("%d%d%d",&n,&m,&k)#define sld(n) scanf("%lld",&n)#define sldd(n,m) scanf("%lld%lld",&n,&m)#define slddd(n,m,k) scanf("%lld%lld%lld",&n,&m,&k)#define sf(n) scanf("%lf",&n)#define sff(n,m) scanf("%lf%lf",&n,&m)#define sfff(n,m,k) scanf("%lf%lf%lf",&n,&m,&k)#define ss(str) scanf("%s",str)#define ans() printf("%d",ans)#define ansn() printf("%d\n",ans)#define anss() printf("%d ",ans)#define lans() printf("%lld",ans)#define lanss() printf("%lld ",ans)#define lansn() printf("%lld\n",ans)#define fansn() printf("%.10f\n",ans)#define r0(i,n) for(int i=0;i<(n);++i)#define r1(i,e) for(int i=1;i<=e;++i)#define rn(i,e) for(int i=e;i>=1;--i)#define rsz(i,v) for(int i=0;i<(int)v.size();++i)#define szz(x) ((int)x.size())#define mst(abc,bca) memset(abc,bca,sizeof abc)#define lowbit(a) (a&(-a))#define all(a) a.begin(),a.end()#define pii pair<int,int>#define pli pair<ll,int>#define pll pair<ll,ll>#define mp(aa,bb) make_pair(aa,bb)#define lrt rt<<1#define rrt rt<<1|1#define X first#define Y second#define PI (acos(-1.0))#define sqr(a) ((a)*(a))typedef long long ll;typedef unsigned long long ull;const ll mod = 1000000000+7;const double eps=1e-9;const int inf=0x3f3f3f3f;const ll infl = 10000000000000000;const int maxn=  8000+10;const int maxm = 10000+10;//Pretests passedint in(int &ret){    char c;    int sgn ;    if(c=getchar(),c==EOF)return -1;    while(c!='-'&&(c<'0'||c>'9'))c=getchar();    sgn = (c=='-')?-1:1;    ret = (c=='-')?0:(c-'0');    while(c=getchar(),c>='0'&&c<='9')ret = ret*10+(c-'0');    ret *=sgn;    return 1;}int ans;int dis[maxn];map<pii,int>ma;int cnt ;bool vis[maxn];int u[maxn],v[maxn],w[maxn];vector<int>g[maxn];void dij(int st,int ed){    mst(dis,0x3f);    mst(vis,0);    priority_queue<pii>q;    dis[st] = 0;    q.push(mp(0,st));    while(!q.empty())    {        pii pp = q.top();q.pop();        int ut = pp.Y;        if(vis[ut])continue;        vis[ut] = 1;        int check = -pp.X;        if(check>ans)break;        rsz(i,g[ut])        {            int j = g[ut][i];            int nt = u[j]==ut ? v[j] : u[j];            if(nt==st&&ut==ed)continue;            if(ut==st&&nt==ed)continue;            if(!vis[nt]&&dis[nt]>dis[ut]+w[j])            {                dis[nt] = dis[ut]+w[j];                q.push(mp(-dis[nt],nt));            }        }    }}int main(){#ifdef LOCAL    freopen("input.txt","r",stdin);//    freopen("output.txt","w",stdout);#endif // LOCAL    int t;    sd(t);    r1(cas,t)    {        printf("Case #%d: ",cas);        int n;        sd(n);        ans = inf;        ma.clear();        cnt = 0 ;        r0(i,maxn)g[i].clear();        for(int i=1;i<=n;++i)        {            int x1,y1,x2,y2;            sdd(x1,y1);            sdd(x2,y2);            pii x = mp(x1,y1) , y = mp(x2,y2);            if(!ma.count(x))ma[x] = ++cnt;            if(!ma.count(y))ma[y] = ++cnt;            int a = ma[x] , b = ma[y];            u[i] = a,v[i] = b;            g[a].pb(i);            g[b].pb(i);            sd(w[i]);        }        r1(i,n)        {            dij(u[i],v[i]);            ans = min(ans,dis[v[i]]+w[i]);        }        if(ans==inf)ans = 0;        ansn();    }    return 0;}