[Cattle off] twelve bridge problem solving report

The meaning of problems

There is a \ (n-\) points \ (m \) with the right edges of the directed graph \ ((n-\ Le 50000, \ m \ Le 200000) \) ,
which \ (K \) edges is constant to go through \ ((k \ Le 12) \) ,

From node \ (1 \) starting requirements: after all must go through the side, back to the node \ (1 \) is the shortest path.

Thinking

First, although the relatively large number of points, but most of it is useless, we only need to focus on a few sides that must endpoint (or bridge) is.

Therefore, the beginning of the first respectively \ (1 \) nodes and endpoints all the bridges as a starting point, running several times \ (Dijkstra \) , and the shortest distance between the number of points to build a new map.

Then, consider the shape pressure DP, provided \ (f [u] [i ] \) of the current point \ (U \) , after a set of edges (bridge set) of \ (I \) shortest path when.

But since this is an undirected graph, there is no topological order, can not be directly dp,
so every time we enumerate a set of edges, the first update all selected endpoints to the bridge \ (f \) values, and the these points added to the priority queue, and then running side dijkstra, the rest of all the points have been updated.

Code

#include<bits/stdc++.h>
#define ll long long
#define mkp make_pair
using namespace std;
const int N=5e4+7;
const int M=2e5+7;
const int K=12+7;
const int L=4096+7;
int n,m,k,all,poi,num[N],cnt,bdg[2*K][2*K],bi[K],rec[L];
ll f[2*K+1][L],dis[N],mp[2*K+1][2*K+1],ans,inf;
int lst[N],nxt[2*M],to[2*M],tot;
ll len[2*M];
bool b[N];
struct bridge{
    int x,y;
    ll w;
}e[M];
void add(int x,int y,ll w){
    nxt[++tot]=lst[x];
    to[tot]=y;
    len[tot]=w;
    lst[x]=tot;
}
void read(){
    cin>>n>>m>>k; num[1]=++cnt; int x,y;
    for(int i=1;i<=m;i++){
        scanf("%d%d%lld",&x,&y,&e[i].w);
        if(!num[x]) num[x]=++cnt;
        if(!num[y]) num[y]=++cnt;
        x=num[x];
        y=num[y];
        e[i].x=x;
        e[i].y=y;
        add(x,y,e[i].w);
        add(y,x,e[i].w);
        if(i<=k){ bdg[x][y]=bdg[y][x]=i; poi=cnt; }
    }
    for(int i=1;i<=k;i++){ bi[i]=1<<i-1; rec[bi[i]]=i; }
    all=(1<<k)-1;
    memset(f,127,sizeof(f));
    memset(mp,-1,sizeof(mp));
    inf=f[0][0];
}
priority_queue<pair<ll,int> >h;
void dijk(int st){
    memset(dis,127,sizeof(dis));
    memset(b,0,sizeof(b));
    dis[st]=0;
    h.push(mkp(0,st));
    while(!h.empty()){
        int u=h.top().second; h.pop();
        if(b[u]) continue;
        b[u]=1;
        for(int i=lst[u];i;i=nxt[i])
            if(dis[to[i]]>dis[u]+len[i]){
                dis[to[i]]=dis[u]+len[i];
                h.push(mkp(-dis[to[i]],to[i]));
            }
    }
    for(int i=1;i<=2*k+1;i++)
        if(dis[i]!=inf) mp[st][i]=dis[i];
}
void print(int x){
    for(int i=1;i<=k;i++){
        printf("%d",x&1);
        x>>=1;
    }
}
void run(int i){
    while(!h.empty()){
        int u=h.top().second; h.pop();
        if(b[u]) continue;
        b[u]=1;
        for(int v=1;v<=n;v++)
            if(mp[u][v]!=-1&&f[v][i]>f[u][i]+mp[u][v]){
                f[v][i]=f[u][i]+mp[u][v];
                h.push(mkp(-f[v][i],v));
            }
    }
}
int main(){
//  freopen("bridge.in","r",stdin);
//  freopen("bridge.out","w",stdout);
    read();
    for(int i=1;i<=2*k+1;i++) dijk(i);
    f[1][0]=0; h.push(mkp(0,1));
    n=poi;
    ans=inf;
    for(int i=0;i<=all;i++){
        for(int j=i;j;j-=j&-j){
            int t=rec[j&-j];
            int x=e[t].x,y=e[t].y;
            f[x][i]=min(f[x][i],f[y][i^bi[t]]+e[t].w);
            f[y][i]=min(f[y][i],f[x][i^bi[t]]+e[t].w);
            h.push(mkp(-f[x][i],x));
            h.push(mkp(-f[y][i],y));
        }
        memset(b,0,sizeof(b));
        run(i);
    }
    for(int i=1;i<=n;i++)
        if(f[i][all]!=inf) ans=min(ans,f[i][all]+mp[i][1]);
    printf("%lld\n",ans);
    return 0;
}