树上问题总结

Posted 2021-01-22 zenurik

树的基本知识

树的存储

存储树的最常见的方式是使用一个邻接表来保存它的边集。
我们可以把树看做一个由N个点与N-1条边组成的无向连通图，图中的无向边看做成对出现的有向边，以head数组作为表头，ver和edge数组分别存储边的终点和权值，next数组模拟链表指针。

#include <bits/stdc++.h>
#define lowbit(x) x & (-x)
#define INF ~0U>>1
using namespace std;
const int N = 25000 + 10;
int f[N << 1], minf[N << 1], n, L, R;
struct data {
    int a, b, c;
}dat[N];

inline int read() {
    int res = 0, flag = 1;
    char c = getchar();
    for(; !isdigit(c); c = getchar()) if(c == ‘-‘) flag = -1;
    for(; isdigit(c); c = getchar()) res = (res<<3) + (res<<1) + (c^48);
    return res * flag;
}

bool cmp(data x, data y) {
    return x.b < y.b;
}

void updata(int x) {
    for(; x <= n; x += lowbit(x)) {
        minf[x] = f[x];
        for(int i = 1; i < lowbit(x); i <<= 1) minf[x] = min(minf[x], minf[x - i]);
    }
}

int ask(int l, int r) {
    int ans = INF;
    while(l <= r) {
        ans = min(f[r], ans);
        r--;
        for(; r - lowbit(r) >= l; r -= lowbit(r)) ans = min(minf[r], ans);
    }
    return ans;
}

int main() {
    n = read(), L = read(), R = read();
    for(int i = 1; i <= n; i++) dat[i].a = read(), dat[i].b = read(), dat[i].c = read();
    memset(f, -0x3f, sizeof(f));
    f[L - 1] = minf[L - 1] = 0;
    sort(dat + 1, dat + n + 1, cmp);
    for(int i = 1; i <= n; i++) {
        f[dat[i].b] = ask(dat[i].a - 1, dat[i].b - 1) + dat[i].c;
    }
    return 0;
}