CF1005F Berland and the Shortest Paths 最短路树计数

Posted 2020-11-30 liubainian

问题描述

LG-CF1005F

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题解

由题面显然可得，所求即最短路树。

所以跑出最短路树，计数，输出方案即可。

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(mathrm{Code})

#include<bits/stdc++.h>
using namespace std;

template <typename Tp>
void read(Tp &x){
    x=0;char ch=1;int fh;
    while(ch!='-'&&(ch>'9'||ch<'0')) ch=getchar();
    if(ch=='-') ch=getchar(),fh=-1;
    else fh=1;
    while(ch>='0'&&ch<='9') x=(x<<1)+(x<<3)+ch-'0',ch=getchar();
    x*=fh;
}

const int maxn=200007;
const int maxm=400007;
const int INF=0x3f3f3f3f;

int n,m,k;
int Head[maxn],to[maxm],Next[maxm],tot,w[maxm];

void add(int x,int y){
    to[++tot]=y,Next[tot]=Head[x],Head[x]=tot,w[tot]=1;
}

int dis[maxn];
bool vis[maxn];

priority_queue< pair<int,int> > q;
#define pii(x,y) make_pair(x,y)

void dijkstra(){
    memset(dis,0x3f,sizeof(dis));
    q.push(pii(0,1));dis[1]=0;
    while(q.size()){
        int x=q.top().second;q.pop();
        if(vis[x]) continue;vis[x]=1;
        for(int i=Head[x];i;i=Next[i]){
            int y=to[i];
            if(dis[y]>dis[x]+w[i]){
                dis[y]=dis[x]+w[i];
                q.push(pii(-dis[y],y));
            }
        }
    }
}

int ans=1;

vector<int>g[maxn];
int val[maxn];
void build(){
    for(int x=1;x<=n;x++){
        for(int i=Head[x];i;i=Next[i]){
            int y=to[i];
            if(dis[y]==dis[x]+w[i]){
                val[y]++;
                g[y].push_back((i+1)>>1);
            }
        }
    }
    for(int i=1;i<=n;i++){
        if(val[i]) ans=ans*val[i];
        if(ans>=k){
            ans=k;return;
        }
    }
}

bool v[maxm];
int md;

void dfs(int x){
    if(x==n+1){
        for(int i=1;i<=tot;i+=2) printf("%d",v[(i+1)>>1]);
        puts("");++md;
        if(md==ans) exit(0);return;
    }
    for(int i=0;i<g[x].size();i++){
        v[g[x][i]]=1;dfs(x+1);v[g[x][i]]=0;
    }
}

int main(){
    read(n);read(m);read(k);
    for(int i=1,x,y;i<=m;i++){
        read(x);read(y);
        add(x,y);add(y,x);
    }
    dijkstra();build();
    printf("%d
",ans);
    dfs(2);
    return 0;
}