NOIP 模拟 $84\; \rm 寻找道路$

题解 \(by\;zj\varphi\)

先把所有到 1 节点距离为 0 的节点加到队列里，然后进行 bfs。

bfs 每次遍历一个点的出边时，先遍历权值为 0 的边，再遍历权值为 1 的边，这样可以保证每次一个点被转移的时候一定是最小距离。

Code

#include<bits/stdc++.h>
#define ri signed
#define pd(i) ++i
#define bq(i) --i
#define func(x) std::function<x>
namespace IO{
    char buf[1<<21],*p1=buf,*p2=buf;
    #define gc() p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?(-1):*p1++
    #define debug1(x) std::cerr << #x"=" << x << ' '
    #define debug2(x) std::cerr << #x"=" << x << std::endl
    #define Debug(x) assert(x)
    struct nanfeng_stream{
        template<typename T>inline nanfeng_stream &operator>>(T &x) {
            bool f=false;x=0;char ch=gc();
            while(!isdigit(ch)) f|=ch=='-',ch=gc();
            while(isdigit(ch)) x=(x<<1)+(x<<3)+(ch^48),ch=gc();
            return x=f?-x:x,*this;
        }
    }cin;
}
using IO::cin;
namespace nanfeng{
    #define FI FILE *IN
    #define FO FILE *OUT
    template<typename T>inline T cmax(T x,T y) {return x>y?x:y;}
    template<typename T>inline T cmin(T x,T y) {return x>y?y:x;}
    static const int N=1e6+7,MOD=1e9+7;
    struct edge{int v,nxt,w;}e[N<<1];
    int first[N],t=1,dep[N],dis[N],que[N],st[N],hd=1,tl,n,m;
    bool vis[N];
    auto add=[](int u,int v,int w) {e[t]={v,first[u],w},first[u]=t++;};
    func(void(int)) dfs=[](int x) {
        dep[x]=dis[x]=0;
        vis[x]=true;
        que[++tl]=x;
        for (ri i(first[x]),v;i;i=e[i].nxt) {
            if (vis[v=e[i].v]||e[i].w) continue;
            dfs(v);
        }
    };
    inline int main() {
        FI=freopen("path.in","r",stdin);
        FO=freopen("path.out","w",stdout);
        cin >> n >> m;
        memset(dis+1,-1,sizeof(int)*n);
        for (ri i(1),u,v,w;i<=m;pd(i)) cin >> u >> v >> w,add(u,v,w);
        dfs(1);
        for (ri i(1);i<=n;pd(i)) {
            int cnt=0,lsp=dep[que[hd]],lss=dis[que[hd]];
            while(hd<=tl&&dep[que[hd]]==lsp&&dis[que[hd]]==lss) st[++cnt]=que[hd++]; 
            for (ri j(1);j<=cnt;pd(j)) {
                int x=st[j];
                for (ri k(first[x]),v;k;k=e[k].nxt) {
                    if (vis[v=e[k].v]||e[k].w) continue;
                    dep[v]=dep[x]+1;
                    dis[v]=(dis[x]<<1)%MOD;
                    vis[que[++tl]=v]=true;
                }
            }
            for (ri j(1);j<=cnt;pd(j)) {
                int x=st[j];
                for (ri k(first[x]),v;k;k=e[k].nxt) {
                    if (vis[v=e[k].v]||!e[k].w) continue;
                    dep[v]=dep[x]+1;
                    dis[v]=((dis[x]<<1)+1)%MOD;
                    vis[que[++tl]=v]=true;
                }
            }
        }
        for (ri i(2);i<=n;pd(i)) printf("%d ",dis[i]);
        return 0;
    }
}
int main() {return nanfeng::main();}