HDU 5682 zxa and leaf

zxa and leaf

链接

题意：

　　给树上所有点赋一个权值，一条边的权值是两个端点的权值差，使最大的边的权值最小。其中k个叶子节点已经赋值。

分析：

　　二分一个答案mid，然后dp一遍，求每个点的取值范围。

代码：

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<iostream>
#include<cmath>
#include<cctype>
#include<set>
#include<queue>
#include<vector>
#include<map>
using namespace std;
typedef long long LL;

inline int read() {
    int x=0,f=1;char ch=getchar();for(;!isdigit(ch);ch=getchar())if(ch=='-')f=-1;
    for(;isdigit(ch);ch=getchar())x=x*10+ch-'0';return x*f;
}

const int N = 100005;
struct Edge{ int to, nxt; } e[N];
int head[N], fa[N], L[N], R[N], w[N], En, n;

inline void add_edge(int u,int v) {
    ++En; e[En].to = v, e[En].nxt = head[u]; head[u] = En;
    ++En; e[En].to = u, e[En].nxt = head[v]; head[v] = En;
}
void dfs(int u,int x) {
    for (int i = head[u]; i; i = e[i].nxt) {
        int v = e[i].to;
        if (v == fa[u]) continue;
        fa[v] = u;
        dfs(v, x);
        L[u] = max(L[u], L[v] - x);
        R[u] = min(R[u], R[v] + x);
    }
}
bool check(int x) {
    for (int i = 1; i <= n; ++i) 
        if (w[i]) L[i] = R[i] = w[i];
        else L[i] = 0, R[i] = 1e9;
    dfs(1, x);
    for (int i = 1; i <= n; ++i) 
        if (L[i] > R[i]) return 0;
    return 1;
}
void solve() {
    n = read();int k = read();
    for (int i = 1; i < n; ++i) {
        int u = read(), v = read();
        add_edge(u, v);
    }
    for (int i = 1; i <= k; ++i) w[read()] = read();
    int l = 0, r = 1e9, ans;
    while (l <= r) {
        int mid = (l + r) >> 1;
        if (check(mid)) ans = mid, r = mid - 1;
        else l = mid + 1;
    }
    printf("%d
", ans);
    En = 0;
    for (int i = 1; i <= n; ++i) head[i] = w[i] = fa[i] = 0;
}
int main() {
    for (int T = read(); T --; solve()) ;
    return 0;
}