HDU 6326 Problem H. Monster Hunter （貪心+並查集）*
Problem H. Monster HunterTime Limit: 6000/3000 MS (Java/Others) Memory Limit: 524288/524288 K (Java/Others)Total Submission(s): 773 Accepted Submission(s): 194 Problem Description Little Q is fighting against scary monsters in the game ``Monster Hunter''. The battlefield consists of n intersections, labeled by 1,2,...,n , connected by n−1 bidirectional roads. Little Q is now at the 1 -th intersection, with X units of health point(HP). Input The first line of the input contains an integer T(1≤T≤2000) , denoting the number of test cases.In each test case, there is one integer n(2≤n≤100000) in the first line, denoting the number of intersections. For the next n−1 lines, each line contains two integers ai,bi(0≤ai,bi≤109) , describing monsters at the 2,3,...,n -th intersection. For the next n−1 lines, each line contains two integers u and v , denoting a bidirectional road between the u -th intersection and the v -th intersection. It is guaranteed that ∑n≤106 . Output For each test case, print a single line containing an integer, denoting the minimum initial HP. Sample Input 1 4 2 6 5 4 6 2 1 2 2 3 3 4 Sample Output 3 Source Recommend chendu | We have carefully selected several similar problems for you: |
#include<iostream>
#include<algorithm>
#include<string>
#include<map>//int dx[4]={0,0,-1,1};int dy[4]={-1,1,0,0};
#include<set>//int gcd(int a,int b){return b?gcd(b,a%b):a;}
#include<vector>
#include<cmath>
#include<queue>
#include<string.h>
#include<stdlib.h>
#include<cstdio>
#define mod 1e9+7
#define ll long long
#define MAX 1000000000
#define ms memset
using namespace std;
const int maxn=1e5+5;
/*
題目大意:一顆樹，每個怪獸有a和b屬性，
先減去a再加上b，
如果不是樹形那麼這題就好辦了，
但關鍵是要先打下父節點才能打子節點。
用並查集來代替這個結構，
再用優先佇列把點壓入並擬定優先順序。
這樣按順序更新父節點即可。
*/
int n;
struct node
{
ll a,b, id;
node(){}
bool operator<(const node &x) const
{
if (b-a>=0&&x.b-x.a<0) return false;
if (b-a<0&&x.b-x.a>=0) return true;
return b-a>0?a>x.a:b<x.b;
}
bool operator!=(const node& y) const
{
return a!=y.a||b!=y.b;
}
};
node s[maxn];///點集合
int fa[maxn],f[maxn];///用dfs來補充fa。
int Find(int x) { return x==f[x]?x:f[x]=Find(f[x]); }
priority_queue<node> pq;
vector<int> sons[maxn];
void dfs(int u,int v)
{
for(int i=0;i<sons[u].size();i++)
{
int p=sons[u][i];
if(p==v) continue;
fa[p]=u;dfs(p,u);
}
}
int main()
{
int t;scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
for(int i=1;i<=n;i++) sons[i].clear();
s[1].a=s[1].b=0,s[1].id=1;
fa[1]=f[1]=1;
for(int i=2;i<=n;i++)
{
scanf("%lld%lld",&s[i].a,&s[i].b);
s[i].id=f[i]=i;
pq.push(s[i]);
}
for(int i=0;i<n-1;i++)
{
int x,y;
scanf("%d%d",&x,&y);
sons[x].push_back(y);
sons[y].push_back(x);
}
dfs(1,0);
while(!pq.empty())
{
int x=pq.top().id;
if(pq.top()!=s[x]||x==1) { pq.pop();continue; }///已經被修改過的點就不用單獨修改了
int y=Find(fa[x]); f[x]=y; pq.pop();
s[y].a+=max(0ll,s[x].a-s[y].b);
s[y].b=s[x].b+max(0ll,s[y].b-s[x].a);
pq.push(s[y]);
}
printf("%lld\n",s[1].a);
}
return 0;
}