Codeforces 1296F Berland Beauty
传送门：https://codeforces.com/contest/1296/problem/F
题意：
给你一个n个点的树，树有边权
现在有m个限制条件，表示点u到点v路径上的最小值是w
现在让我们构造这个树的边权，如果不能构造输出-1
题解：
因为n只有5000，所以对于每一个限制条件，我们都在路径上保存一下这条边的最大边权，记录一下这个限制条件中路径上经过的边的编号
最后再check一遍m次限制路径上的边权的最小值是否等于w即可
代码：
/**
* ┏┓ ┏┓
* ┏┛┗━━━━━━━┛┗━━━┓
* ┃ ┃
* ┃ ━ ┃
* ┃ ＞ ＜ ┃
* ┃ ┃
* ┃... ⌒ ... ┃
* ┃ ┃
* ┗━┓ ┏━┛
* ┃ ┃ Code is far away from bug with the animal protecting
* ┃ ┃ 神兽保佑,代码无bug
* ┃ ┃
* ┃ ┃
* ┃ ┃
* ┃ ┃
* ┃ ┗━━━┓
* ┃ ┣┓
* ┃ ┏┛
* ┗┓┓┏━┳┓┏┛
* ┃┫┫ ┃┫┫
* ┗┻┛ ┗┻┛
*/
// warm heart, wagging tail,and a smile just for you!
//
// _ooOoo_
// o8888888o
// 88" . "88
// (| -_- |)
// O\ = /O
// ____/`---'\____
// .' \| |// `.
// / \||| : |||// \
// / _||||| -:- |||||- \
// | | \ - /// | |
// | \_| ''\---/'' | |
// \ .-\__ `-` ___/-. /
// ___`. .' /--.--\ `. . __
// ."" '< `.___\_<|>_/___.' >'"".
// | | : `- \`.;`\ _ /`;.`/ - ` : | |
// \ \ `-. \_ __\ /__ _/ .-` / /
// ======`-.____`-.___\_____/___.-`____.-'======
// `=---='
// ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
// 佛祖保佑 永无BUG
#include <set>
#include <map>
#include <stack>
#include <cmath>
#include <queue>
#include <cstdio>
#include <string>
#include <bitset>
#include <vector>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long LL;
typedef pair<int, int> pii;
typedef unsigned long long uLL;
#define ls rt<<1
#define rs rt<<1|1
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define bug printf("*********\n")
#define FIN freopen("input.txt","r",stdin);
#define FON freopen("output.txt","w+",stdout);
#define IO ios::sync_with_stdio(false),cin.tie(0)
#define debug1(x) cout<<"["<<#x<<" "<<(x)<<"]\n"
#define debug2(x,y) cout<<"["<<#x<<" "<<(x)<<" "<<#y<<" "<<(y)<<"]\n"
#define debug3(x,y,z) cout<<"["<<#x<<" "<<(x)<<" "<<#y<<" "<<(y)<<" "<<#z<<" "<<z<<"]\n"
const int maxn = 3e5 + 5;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7;
const double Pi = acos(-1);
LL gcd(LL a, LL b) {
return b ? gcd(b, a % b) : a;
}
LL lcm(LL a, LL b) {
return a / gcd(a, b) * b;
}
double dpow(double a, LL b) {
double ans = 1.0;
while(b) {
if(b % 2)ans = ans * a;
a = a * a;
b /= 2;
} return ans;
}
LL quick_pow(LL x, LL y) {
LL ans = 1;
while(y) {
if(y & 1) {
ans = ans * x % mod;
} x = x * x % mod;
y >>= 1;
} return ans;
}
struct EDGE {
int v, nxt, w;
} edge[maxn << 1];
int head[maxn], tot;
void init() {
memset(head, -1, sizeof(head));
tot = 0;
}
void add_edge(int u, int v, int w) {
edge[tot].v = v;
edge[tot].w = w;
edge[tot].nxt = head[u];
head[u] = tot++;
}
int to[maxn];
int lim[maxn];
int ans[maxn];
vector<int> vec[maxn];
bool dfs(int u, int fa, int id) {
if(u == to[id]) return true;
for(int i = head[u]; i != -1; i = edge[i].nxt) {
int v = edge[i].v;
if(v == fa) continue;
if(dfs(v, u, id)) {
int val = edge[i].w;//第几条边，保存一下最大值
ans[val] = max(ans[val], lim[id]);
vec[id].push_back(val);
return true;
}
}
return false;
}
int main() {
#ifndef ONLINE_JUDGE
FIN
#endif
int n;
init();
scanf("%d", &n);
for(int i = 1, u, v; i < n; i++) {
scanf("%d%d", &u, &v);
add_edge(u, v, i);
add_edge(v, u, i);
}
int m;
scanf("%d", &m);
for(int i = 1; i <= m; i++) {
int u, v, w;
scanf("%d%d%d", &u, &v, &w);
to[i] = v;
lim[i] = w;
dfs(u, 0, i);
}
int flag = 1;
for(int i = 1; i <= m; i++) {
int Minn = INF;
for(auto it : vec[i]) {
Minn = min(Minn, ans[it]);
}
if(Minn != lim[i]) {
flag = 0;
break;
}
}
if(flag) {
for(int i = 1; i < n; i++) {
printf("%d%c", max(1, ans[i]), i == n - 1 ? '\n' : ' ');
}
} else {
printf("-1\n");
}
return 0;
}