http://acm.hdu.edu.cn/showproblem.php?pid=6241
Binary answer to the judgment, the conversion can be limited to the original subject subtree at least up to the point of up and Down points, if up> = down it indicates that the present case is feasible.
When the game think this is too simple to Pass off (fog), in fact, this is proof of correctness, as long as the guarantee for all l <= r and l1 <= x <= r, then the x case of black dots feasible.
#include <map> #include <set> #include <ctime> #include <cmath> #include <queue> #include <stack> #include <vector> #include <string> #include <bitset> #include <cstdio> #include <cstdlib> #include <cstring> #include <sstream> #include <iostream> #include <algorithm> #include <functional> using namespace std; #define For(i, x, y) for(int i=x;i<=y;i++) #define _For(i, x, y) for(int i=x;i>=y;i--) #define Mem(f, x) memset(f,x,sizeof(f)) #define Sca(x) scanf("%d", &x) #define Sca2(x,y) scanf("%d%d",&x,&y) #define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z) #define Scl(x) scanf("%lld",&x) #define Pri(x) printf("%d\n", x) #define Prl(x) printf("%lld\n",x) #define CLR(u) for(int i=0;i<=N;i++)u[i].clear(); #define LL long long #define ULL unsigned long long #define mp make_pair #define PII pair<int,int> #define PIL pair<int,long long> #define PLL pair<long long,long long> #define pb push_back #define fi first #define se second typedef vector<int> VI; int read(){int x = 0,f = 1;char c = getchar();while (c<'0' || c>'9'){if (c == '-') f = -1;c = getchar();} while (c >= '0'&&c <= '9'){x = x * 10 + c - '0';c = getchar();}return x*f;} const double PI = acos(-1.0); const double eps = 1e-9; const int maxn = 2e5 + 10; const int INF = 0x3f3f3f3f; const int mod = 1e9 + 7; int N,M,K; struct Edge{ int to,next; }edge[maxn * 2]; int head[maxn],tot; void init(){ for(int i = 0 ; i <= N; i ++) head[i] = -1; tot = 0; } void add(int u,int v){ edge[tot].to = v; edge[tot].next = head[u]; head[u] = tot++; } LL l[maxn],r[maxn],up[maxn],down[maxn],sz[maxn]; void dfs(int sum,int u,int la){ LL L = 0 ,R = 0; sz[u] = 1; down[u] = l[u]; up[u] = sum - r[u]; for(int i = head[u]; ~i ; i = edge[i].next){ int v = edge[i].to; if(v == la) continue; dfs(sum,v,u); sz[u] += sz[v]; L += down[v]; R += up[v]; } down[u] = max(down[u],L); up[u] = min(up[u],sz[u]); up[u] = min(up[u],R + 1); } bool check(int x){ dfs(x,1,-1); bool flag = true; if(up[1] < x || down[1] > x) return false; for(int i = 1; i <= N ; i ++){ if(up[i] < down[i]) return false; } return true; } int solve(){ int l = 0,r = N; int ans = -1; while(l <= r){ int m = l + r >> 1; if(check(m)){ r = m - 1; ans = m; }else{ l = m + 1; } } return ans; } int main(){ int T; Sca(T); while(T--){ Sca(N); init(); for(int i = 1; i <= N ; i ++) l[i] = r[i] = 0; for(int i = 1; i <= N - 1; i ++){ int u,v; Sca2(u,v); add(u,v); add(v,u); } Sca(M); while(M--){ int p = read();LL num = read(); l[p] = max(l[p],num); } Sca(M); while(M--){ int p = read();LL num = read(); r[p] = max(r[p],num); } Pri(solve()); } return 0; }