CF702E Analysis of Pathes in Functional Graph

倍增练习题。

基环树上倍增一下维护维护最小值和权值和，注意循环的时候$j$这维作为状态要放在外层循环，平时在树上做的时候一个一个结点处理并不会错，因为之前访问的结点已经全部处理过了。

时间复杂度$O(nlogk)$。

Code：

#include <cstdio>
#include <cstring>
using namespace std;
typedef long long ll;

const int N = 1e5 + ;
const int Lg = ;
const int inf = 0x3f3f3f3f;

int n, to[N][Lg];
ll stp, val[N], sum[N][Lg], minn[N][Lg];

template <typename T>
inline void read(T &X) {
    X = ; char ch = ; T op = ;
    for(; ch > ''|| ch < ''; ch = getchar())
        if(ch == '-') op = -;
    for(; ch >= '' && ch <= ''; ch = getchar())
        X = (X << ) + (X << ) + ch - ;
    X *= op;
}

template <typename T>
inline void chkMin(T &x, T y) {
    if(y < x) x = y;
}

template <typename T>
inline T min(T x, T y) {
    return x > y ? y : x;
}

inline void solve(int x) {
    ll resSum = 0LL, resMin = inf, tmp = stp;
    for(int i = ; i >= ; i--)
        if((tmp >> i) & ) {
            resSum += sum[x][i];
            chkMin(resMin, minn[x][i]);
            x = to[x][i];
        }
    printf("%lld %lld\n", resSum, resMin);
}

int main() {
    read(n), read(stp);
    for(int i = ; i <= n; i++) read(to[i][]), to[i][]++;
    for(int i = ; i <= n; i++) read(val[i]);

    memset(minn, 0x3f, sizeof(minn));
    for(int i = ; i <= n; i++)
        sum[i][] = minn[i][] = val[i];
    for(int j = ; j <= ; j++)
        for(int i = ; i <= n; i++)
         {
            to[i][j] = to[to[i][j - ]][j - ];
            minn[i][j] = min(minn[i][j - ], minn[to[i][j - ]][j - ]);
            sum[i][j] = sum[i][j - ] + sum[to[i][j - ]][j - ];
        }

/*    for(int i = 1; i <= n; i++)
        printf("%lld ", sum[i][1]);
    printf("\n");   */

    for(int i = ; i <= n; i++) solve(i);

    return ;
}