Codeforces Round #436 (Div. 2) F. Cities Excursions

先对每个点dfs一次标记以点i为起点可以到达的点
然后用类似floyd的方法对每个i->j找出i能到的最小的点o并且o有路径能到达j
将o作为i->j的路径的第一个点
然后在这个图上进行倍增
假设有环那么倍增会一直绕着环跑 这种情况判断一下边界就可以了

#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 mp make_pair#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=  3000+10;const int maxm = 100+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;}vector<int>g[maxn];bool can[maxn][maxn];short int fa[maxn][maxn][13];void dfs(int u,int f){    if(can[f][u])return;    can[f][u]=1;    int sz = g[u].size();    r0(i,sz)    {        int v = g[u][i];        dfs(v,f);    }}int main(){#ifdef LOCAL    freopen("input.txt","r",stdin);//    freopen("output.txt","w",stdout);#endif // LOCAL    int n,m,q;    sddd(n,m,q);    r1(i,m)    {        int u,v;        sdd(u,v);        g[u].pb(v);    }    r1(i,n)dfs(i,i);    for(int i=1;i<=n;++i)    {        sort(all(g[i]));        for(int j = 1;j<=n;++j)        {            if(j!=i)            {                int sz = g[i].size();                r0(k,sz)                {                    int o = g[i][k];                    if(can[o][j])                    {                        fa[i][j][0] = o;                        break;                    }                }            }        }    }    for(int k = 1;k<13;++k)    {        for(int i=1;i<=n;++i)        {            for(int j=1;j<=n;++j)            {                fa[i][j][k] = fa[ fa[i][j][k-1] ][j][k-1];            }        }    }    r1(i,q)    {        int a,b,c;        sddd(a,b,c);        --c;        if(fa[a][b][12]||!can[a][b])        {            puts("-1");            continue;        }        for(int i=12;i>=0;--i)        {            if(fa[a][b][i]&&(1<<i)<=c)            {                c -= (1<<i);                a = fa[a][b][i];            }        }        if(c)puts("-1");        else printf("%d\n",a);    }    return 0;}