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 + 5;
const int Lg = 35;
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 = 0; char ch = 0; T op = 1;
    for(; ch > '9'|| ch < '0'; ch = getchar())
        if(ch == '-') op = -1;
    for(; ch >= '0' && ch <= '9'; ch = getchar())
        X = (X << 3) + (X << 1) + ch - 48;
    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 = 34; i >= 0; i--)
        if((tmp >> i) & 1) {
            resSum += sum[x][i];
            chkMin(resMin, minn[x][i]);
            x = to[x][i];
        }
    printf("%lld %lld
", resSum, resMin);
}

int main() {
    read(n), read(stp);
    for(int i = 1; i <= n; i++) read(to[i][0]), to[i][0]++;
    for(int i = 1; i <= n; i++) read(val[i]);

    memset(minn, 0x3f, sizeof(minn));
    for(int i = 1; i <= n; i++) 
        sum[i][0] = minn[i][0] = val[i];
    for(int j = 1; j <= 34; j++)
        for(int i = 1; i <= n; i++)
         {
            to[i][j] = to[to[i][j - 1]][j - 1];
            minn[i][j] = min(minn[i][j - 1], minn[to[i][j - 1]][j - 1]);
            sum[i][j] = sum[i][j - 1] + sum[to[i][j - 1]][j - 1];
        }
    
/*    for(int i = 1; i <= n; i++)
        printf("%lld ", sum[i][1]);
    printf("
");   */

    for(int i = 1; i <= n; i++) solve(i);

    return 0;
}