$Noip2014/Luogu2296$ 寻找道路 图论

$Luogu$

$Sol$

首先把不符合条件一的点去掉然后跑$SPFA$就好了叭.

如何判断点是否符合条件一呢?先连反边,记录每个点的入度,然后从终点开始$dfs$,记录每个点被到达的次数,若到达的次数小于它的入读那么就是不满足题意的.

为啥$Noip2014$有$4$道连我都觉得很水的题.

$Code$

#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
#include<algorithm>
#define il inline
#define Rg register
#define go(i,a,b) for(Rg int i=a;i<=b;++i)
#define yes(i,a,b) for(Rg int i=a;i>=b;--i)
#define e(i,u) for(Rg int i=b[u];i;i=a[i].nt)
#define mem(a,b) memset(a,b,sizeof(a))
#define ll long long
#define db double
#define inf 2147483647
using namespace std;
il int read()
{
    Rg int x=0,y=1;char c=getchar();
    while(c<'0'||c>'9'){if(c=='-')y=-1;c=getchar();}
    while(c>='0'&&c<='9'){x=(x<<1)+(x<<3)+c-'0';c=getchar();}
    return x*y;
}
const int N=10010,M=200010;
int n,m,b[N],ct,du[N],s,t,vis[N],dis[N];
bool fl[N];
queue<int>q;
struct nd1{int u,v;}eg[M];
struct nd{int v,nt;}a[M];
il void add(int u,int v){a[++ct]=(nd){v,b[u]};b[u]=ct;}
il void dfs(int u)
{
    e(i,u)
    {
        Rg int v=a[i].v;vis[v]++;
        if(vis[v]>1)continue;
        dfs(v);
    }
}
il void SPFA()
{
    mem(vis,0);mem(dis,63);
    vis[s]=1;dis[s]=0;q.push(s);
    while(!q.empty())
    {
        Rg int u=q.front();q.pop();vis[u]=0;
        e(i,u)
        {
            Rg int v=a[i].v;
            if(!fl[v])continue;
            if(dis[v]>dis[u]+1)
            {
                dis[v]=dis[u]+1;
                if(!vis[v]){vis[v]=1;q.push(v);}
            }
        }
    }
}
il bool cmp(nd1 x,nd1 y){return x.u==y.u?x.v<y.v:x.u<y.u;}
int main()
{
    n=read(),m=read();
    go(i,1,m)
    {
        Rg int u=read(),v=read();
        if(u==v)continue;
        eg[i]=(nd1){u,v};//add(v,u);du[u]++;
    }
    sort(eg+1,eg+m+1,cmp);
    Rg int nm=0;
    go(i,1,m)
    {
        if(eg[i].u==eg[i-1].u && eg[i].v==eg[i-1].v){nm++;continue;}
        add(eg[i].v,eg[i].u);du[eg[i].u]++;
    }
    m-=nm;
    s=read(),t=read();swap(s,t);
    vis[s]=1;dfs(s);
    go(i,1,n)if((vis[i]==du[i]&&vis[i])||i==s)fl[i]=1;
    if(!fl[s]){printf("-1\n");return 0;}
    SPFA();
    if(dis[t]>M)printf("-1\n");
    else printf("%d\n",dis[t]);
    return 0;
}

标签：图论,Noip2014,Luogu2296,int,vis,Rg,il,dis,define
来源： https://www.cnblogs.com/forward777/p/11412563.html