Codeforces Round #436 (Div. 2) F. Cities Excursions 路径上的第k个点（倍增）

原文处

传送门

题意：先形成一个有向图，问你从s开始，字典序最小的路径上第k个点是哪个。

做法：学习别人的倍增做法，关键处标出来了。

//china no.1#pragma comment(linker, "/STACK:1024000000,1024000000")#include <vector>#include <iostream>#include <string>#include <map>#include <stack>#include <cstring>#include <queue>#include <list>#include <stdio.h>#include <set>#include <algorithm>#include <cstdlib>#include <cmath>#include <iomanip>#include <cctype>#include <sstream>#include <functional>#include <stdlib.h>#include <time.h>#include <bitset>using namespace std;#define pi acos(-1)#define s_1(x) scanf("%d",&x)#define s_2(x,y) scanf("%d%d",&x,&y)#define s_3(x,y,z) scanf("%d%d%d",&x,&y,&z)#define s_4(x,y,z,X) scanf("%d%d%d%d",&x,&y,&z,&X)#define S_1(x) scan_d(x)#define S_2(x,y) scan_d(x),scan_d(y)#define S_3(x,y,z) scan_d(x),scan_d(y),scan_d(z)#define PI acos(-1)#define endl '\n'#define srand() srand(time(0));#define me(x,y) memset(x,y,sizeof(x));#define foreach(it,a) for(__typeof((a).begin()) it=(a).begin();it!=(a).end();it++)#define close() ios::sync_with_stdio(0); cin.tie(0);#define FOR(x,n,i) for(int i=x;i<=n;i++)#define FOr(x,n,i) for(int i=x;i<n;i++)#define fOR(n,x,i) for(int i=n;i>=x;i--)#define fOr(n,x,i) for(int i=n;i>x;i--)#define W while#define sgn(x) ((x) < 0 ? -1 : (x) > 0)#define bug printf("***********\n");#define db double#define ll long long#define mp make_pair#define pb push_backtypedef long long LL;const int INF=0x3f3f3f3f;const LL LINF=0x3f3f3f3f3f3f3f3fLL;const int dx[]={-1,0,1,0,1,-1,-1,1};const int dy[]={0,1,0,-1,-1,1,-1,1};const int maxn=4e3+10;const int maxx=4e5+10;const double EPS=1e-8;const double eps=1e-8;const int mod=1e9+7;template<class T>inline T min(T a,T b,T c) { return min(min(a,b),c);}template<class T>inline T max(T a,T b,T c) { return max(max(a,b),c);}template<class T>inline T min(T a,T b,T c,T d) { return min(min(a,b),min(c,d));}template<class T>inline T max(T a,T b,T c,T d) { return max(max(a,b),max(c,d));}template <class T>inline bool scan_d(T &ret){char c;int sgn;if (c = getchar(), c == EOF){return 0;}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;}inline bool scan_lf(double &num){char in;double Dec=0.1;bool IsN=false,IsD=false;in=getchar();if(in==EOF) return false;while(in!='-'&&in!='.'&&(in<'0'||in>'9'))in=getchar();if(in=='-'){IsN=true;num=0;}else if(in=='.'){IsD=true;num=0;}else num=in-'0';if(!IsD){while(in=getchar(),in>='0'&&in<='9'){num*=10;num+=in-'0';}}if(in!='.'){if(IsN) num=-num;return true;}else{while(in=getchar(),in>='0'&&in<='9'){num+=Dec*(in-'0');Dec*=0.1;}}if(IsN) num=-num;return true;}void Out(LL a){if(a < 0) { putchar('-'); a = -a; }if(a >= 10) Out(a / 10);putchar(a % 10 + '0');}void print(LL a){ Out(a),puts("");}//freopen( "in.txt" , "r" , stdin );//freopen( "data.txt" , "w" , stdout );//cerr << "run time is " << clock() << endl;vector<int>G[maxn],g[maxn];bool vis[maxn];vector<pair<pair<int,int>,int> >Q[maxx];int n,m,q;int st[maxn][13];int ans[maxx];void dfs(int x){    if(vis[x]) return ;    vis[x]=1;    for(int i=0;i<g[x].size();i++)        dfs(g[x][i]);}int main(){    //freopen( "in.txt" , "r" , stdin );    //freopen( "out.txt" , "w" , stdout );    W(S_3(n,m,q))    {        me(ans,0);        FOR(0,maxx,i)            Q[i].clear();        FOR(0,maxn,i)            G[i].clear(),g[i].clear();        FOR(1,m,i)        {            int u,v;            S_2(u,v);            G[u].pb(v);            g[v].pb(u);        }        int s,t,k;        FOR(1,q,i)        {            S_3(s,t,k);            Q[t].pb(mp(mp(s,k),i));        }        FOR(1,n,i)            sort(G[i].begin(),G[i].end());//字典序最小        FOR(1,n,i) if(Q[i].size())        {            me(vis,0);            dfs(i);//联通 标记            me(st,0);            FOR(0,12,j) st[n+1][j]=n+1;            FOR(1,n,j)            //点可能到t点的路径有很多，所以我现在找一条字典序最小的路径            //并对倍增数组初始化            {                if(i==j) st[j][0]=n+1;                else if(vis[j])                {                    for(int k=0;k<G[j].size();k++)                    {                        if(!vis[G[j][k]]) continue;                        st[j][0]=G[j][k];                        break;                    }                }            }            FOR(1,12,j)                FOR(1,n,u) if(vis[u])                {                    st[u][j]=st[st[u][j-1]][j-1];                }                //标记  倍增            for(auto it:Q[i])            {                int s=it.first.first;                int k=it.first.second-1;                int id=it.second;                if(vis[s])                {                    if(st[s][12]==n+1)                    {                        FOR(0,12,j)                            if(k&(1<<j))                                s=st[s][j];                        if(s!=n+1)                            ans[id]=s;                        else ans[id]=-1;//跑过了，k太大                    }                    else ans[id]=-1; //成环或发现更小字典序                }                else ans[id]=-1;//未连通            }        }        FOR(1,q,i)            print(ans[i]);    }}