Codeforces 702E Analysis of Pathes in Functional Graph（倍增）

给你一个图，每个点都有一条出边
然后边有权值，让你输出走k步，从i出发走的总距离和其中最小的边是多少
k很大，图可以预处理，倍增法

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代码：

#include <map>#include <set>#include <stack>#include <queue>#include <cmath>#include <string>#include <vector>#include <cstdio>#include <cctype>#include <cstring>#include <sstream>#include <cstdlib>#include <iostream>#include <algorithm>#pragma comment(linker,"/STACK:102400000,102400000")using namespace std;#define   MAX           100005#define   MAXN          1000005#define   maxnode       205#define   sigma_size    26#define   lson          l,m,rt<<1#define   rson          m+1,r,rt<<1|1#define   lrt           rt<<1#define   rrt           rt<<1|1#define   middle        int m=(r+l)>>1#define   LL            long long#define   ull           unsigned long long#define   mem(x,v)      memset(x,v,sizeof(x))#define   lowbit(x)     (x&-x)#define   pii           pair<int,int>#define   bits(a)       __builtin_popcount(a)#define   mk            make_pair#define   limit         10000//const int    prime = 999983;const int    INF   = 0x3f3f3f3f;const LL     INFF  = 0x3f3f;const double pi    = acos(-1.0);const double inf   = 1e18;const double eps   = 1e-4;const LL    mod    = 1e9+7;const ull    mx    = 133333331;/*****************************************************/inline void RI(int &x) {      char c;      while((c=getchar())<'0' || c>'9');      x=c-'0';      while((c=getchar())>='0' && c<='9') x=(x<<3)+(x<<1)+c-'0'; }/*****************************************************/int pre[MAX][40];int w[MAX][40];LL s[MAX][40];LL ans1;int ans2;void solve(int u,LL k){    if(k==0){        return ;    }    int t=0;    while((1LL<<t)<=k) t++;    t--;    ans1+=s[u][t];    ans2=min(ans2,w[u][t]);    solve(pre[u][t],k-(1LL<<t));}int main(){    //freopen("in.txt","r",stdin);    int n;    LL k;    while(cin>>n>>k){        for(int i=0;i<n;i++) scanf("%d",&pre[i][0]);        for(int i=0;i<n;i++) scanf("%d",&w[i][0]),s[i][0]=w[i][0];        for(int i=1;(1LL<<i)<=k;i++){            for(int j=0;j<n;j++){                pre[j][i]=pre[pre[j][i-1]][i-1];                w[j][i]=min(w[j][i-1],w[pre[j][i-1]][i-1]);                s[j][i]=s[j][i-1]+s[pre[j][i-1]][i-1];            }        }        for(int i=0;i<n;i++){            ans1=0;            ans2=INF;            solve(i,k);            printf("%I64d %d\n",ans1,ans2);        }    }    return 0;}