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IDProblemSubmitterResultTimeMemoryLanguageFile sizeSubmit timeJudge time
#385165#2161. The Cost of Speed Limitsbulijiojiodibuliduo#WA 2ms11580kbC++172.0kb2024-04-10 16:11:402024-04-10 16:11:41

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  • [2024-04-10 16:11:41]
  • 评测
  • 测评结果:WA
  • 用时:2ms
  • 内存:11580kb
  • [2024-04-10 16:11:40]
  • 提交

answer

#include <bits/stdc++.h>
using namespace std;
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=n-1;i>=a;i--)
#define pb push_back
#define eb emplace_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
typedef vector<int> VI;
typedef basic_string<int> BI;
typedef long long ll;
typedef pair<int,int> PII;
typedef double db;
mt19937 mrand(random_device{}()); 
const ll mod=1000000007;
int rnd(int x) { return mrand() % x;}
ll powmod(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
ll gcd(ll a,ll b) { return b?gcd(b,a%b):a;}
// head

const int N=101000;
int n,m,c,sz[N];
vector<PII> e[N],ch[N];
ll dp[N],pd[25][N];
VI wt;
void dfs1(int u,int f) {
	sz[u]=1;
	for (auto [v,w]:e[u]) if (v!=f) {
		dfs1(v,u);
		ch[u].pb(mp(v,w));
		sz[u]+=sz[v];
	}
	sort(all(ch[u]),[&](PII a,PII b) {
		return sz[a.fi]>sz[b.fi];
	});
}
const ll inf=1ll<<50;
void dfs2(int u,int dep) {
	//printf("?????????????? %d %d\n",u,dep);
	rep(j,0,SZ(ch[u])) {
		int v=ch[u][j].fi;
		int w=ch[u][j].se;
		if (j==0) {
			dfs2(v,dep);
			dp[u]+=dp[v];
			rep(k,0,m) pd[dep][k]=min(pd[dep][k],dp[v])+(wt[w]<=wt[k]?wt[k]-wt[w]:inf);
		} else {
			memset(pd[dep+1],0,m*sizeof(int));
			dfs2(v,dep+1);
			dp[u]+=dp[v];
			rep(k,0,m) pd[dep][k]+=min(pd[dep+1][k],dp[v])+(wt[w]<=wt[k]?wt[k]-wt[w]:inf);
		}
	}
	dp[u]+=(SZ(e[u])>1?SZ(e[u]):0)*c;
	//printf("%d : %lld\n",u,dp[u]);
	//rep(k,0,m) printf("%lld ",pd[dep][k]); puts("");
}
int main() {
	scanf("%d%d",&n,&c);
	rep(i,1,n) {
		int u,v,s;
		scanf("%d%d%d",&u,&v,&s);
		e[u].pb(mp(v,s));
		e[v].pb(mp(u,s));
		wt.pb(s);
	}
	sort(all(wt)); wt.erase(unique(all(wt)),wt.end());
	m=SZ(wt);
	rep(u,1,n+1) {
		for (auto &[v,w]:e[u]) w=lower_bound(all(wt),w)-wt.begin();
	}
	dfs1(1,0);
	dfs2(1,0);
	ll ans=dp[1];
	rep(k,0,m) ans=min(ans,pd[0][k]);
	printf("%lld\n",ans);
}

Details

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Test #1:

score: 0
Wrong Answer
time: 2ms
memory: 11580kb

input:

13 20
1 8 101
1 9 30
1 2 100
1 3 100
2 4 75
2 5 70
2 6 82
2 7 77
3 10 73
3 11 69
3 12 83
3 13 79

output:

273

result:

wrong answer 1st lines differ - expected: '272', found: '273'