[UVALive]7338 树链剖分+线段树

诶秒出正解的题目写起来就是爽

有一个结论：在生成树上添加一条新边E{a,b,v}使得新图的最小生成树改变，充要条件是E和生成树形成的环中，v不是权值最大的边。

对最小生成树进行树链剖分，按树链剖分序建线段树

对于第n~m条边E{a,b,v}，将a，b路径上的最长边求出来e{a0,b0,v0}，v-v0即为该边能减小最多的值，增加最多的值为INF

对于前1~n-1条边E{a,b,v}，能增加最多的值即为使E在第i条边与生成树连接的环中，第i条边权值的最小值-v，减少最多的值为INF

本机出了数据平均要跑5s+，UVALive评测及2.8秒就飞过去了

诶，别人的评测机真牛逼

Notice：1、注意最后答案取值最大超过1e17，答案要开long long

      2、输出的case冒号后面有个空格，否则PE

（诶为了1A，写了造数据和暴力对拍）

正解：

#include 
#include 
#include 
#include 

#define N 40050
#define M 400050

#define INF (1<<30) 
#define mid ((l+r)>>1)
#define ls l,mid,(t<<1)
#define rs mid+1,r,(t<<1^1) 

#define L (t<<1)
#define R (t<<1^1)

using namespace std;
typedef long long LL;

int tr[4*N],ag[4*N],S,P;
int head[N],cnt,tot;
int son[N],D[N],siz[N],w[N],top[N],A[N],fa[N];
int vson[N];
int a[M],b[M],v[M],n,m;

inline int read() {
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}

vector E[N],Q[N];
struct Edge{ int a,b,v,next; }e[M];
void add_Edge(int a,int b,int v) {
	e[++cnt].a = a;
	e[cnt].b = b;
	e[cnt].v = v;
	e[cnt].next = head[a];
	head[a] = cnt;
}


void dfs1(int u,int f) {
	D[u] = D[f] + 1; siz[u] = 1; fa[u] = f;
	for (int i=0;i siz[ son[u] ]) son[u] = v , vson[u] = Q[u][i];
	}
}

void dfs2(int u,int h,int q) {
	top[u] = h; w[u] = ++tot; A[tot] = q;
	if (son[u]) dfs2(son[u],h,vson[u]);
	for (int i=0;i= ll && r <= rr) { ag[t] = min(ag[t],V); return ;  }
	push_down(t);
	if (ll <= mid)	update(ls); 
	if (rr > mid)   update(rs);
}

void query(int l,int r,int t) {
	if (l >= ll && r <= rr) { 
		S = max(tr[t],S); P = min(ag[t],P); 
		return ; 
	}
	push_down(t);
	if (ll <= mid) query(ls); 
	if (rr >  mid) query(rs);
	tr[t] = max(tr[L],tr[R]);
}

void jump(int &x,int &y) {
	if (top[x] == top[y]) {
		if (D[x] > D[y]) swap(x,y);
		ll = w[ son[x] ]; rr = w[y]; y = x;
	} else {
		if (D[ top[x] ] < D[ top[y] ]) swap(x,y);
		ll = w[ top[x] ]; rr = w[x];
		x = fa[ top[x] ];
	}
}

void solve(int cas) {
	long long ans = 0LL;
	n = read(); m = read();
	for (int i=1;i<=m;i++) {
		a[i] = read(); b[i] = read(); v[i] = read();
		add_Edge(a[i],b[i],v[i]); 
		add_Edge(b[i],a[i],v[i]);
		if (i < n) {
			E[ a[i] ].push_back( b[i] );
			Q[ a[i] ].push_back( v[i] );
			E[ b[i] ].push_back( a[i] );
			Q[ b[i] ].push_back( v[i] );
		}
	}
	dfs1(1,1); dfs2(1,0,0);

	build(1,n,1);
	for (int i=n;i<=m;i++) {
		int x,y;int p = INF;
		x = a[i] ; y = b[i]; V = v[i];
		while (x != y) {
			S = 0;
			jump(x,y); query(1,n,1);
			p = min(p,V - S);
		}
		x = a[i] ; y = b[i]; V = v[i];
		while (x != y) 
			jump(x,y) , update(1,n,1);
		ans += 1LL*-i + 1LL*i*i*p;
	}
	
	for (int i=1;i<=n-1;i++) {
		int x = a[i] , y = b[i]; jump(x,y); P = INF;
		query(1,n,1);
		if (P == INF) ans += -1LL*i - 1LL*i*i; 
				 else ans += 1LL*i*(P-v[i]) - 1LL*i*i;
	}
	printf("Case %d: ",cas);
	printf("%lld\n",ans);
	
	for (int i=1;i<=n;i++) E[i].clear() , Q[i].clear();
	
}

void init() {
	cnt = tot = 0;
	memset(e,0,sizeof(e));
	memset(head,0,sizeof(head));
	memset(son,0,sizeof(son));
	memset(D,0,sizeof(D)); 
	//memset(ag,127/3,sizeof(ag));
	/*
	memset(tr,0,sizeof(tr));
	memset(vson,0,sizeof(vson));
	memset(A,0,sizeof(A));
	memset(a,0,sizeof(a));
	memset(b,0,sizeof(b));
	memset(v,0,sizeof(v));
	memset(siz,0,sizeof(siz));
	memset(w,0,sizeof(w));
	memset(top,0,sizeof(top));
	memset(fa,0,sizeof(fa));
	*/
}

int main() {
	int T = read();
	for (int i=1;i<=T;i++) {
		init();
		solve(i);
	}
	return 0;
}

造数据

#include 
#include 
#include 
#include 
#include 
#include 

#define N 1000050

using namespace std;

const int maxn = 10000;
const int maxm = 100000;
const int V = 1001;
const int cas = 25;

struct Edge{ int a,b,v,p; }e[N];
int fa[N],n,m,cnt;
bool cmp(Edge p1,Edge p2) {	return p1.v < p2.v; }
int get(int x) { return fa[x] == x ? x : fa[x] = get(fa[x]); }
int main()
{
	freopen("problem6.in","w",stdout);
	srand(time(0));	
	int T = rand() % cas + 1; 
	T = 25;
	printf("%d\n",T);
	while (T--) {
		memset(e,0,sizeof(e)); cnt = 0;
		n = rand() * rand() % maxn + 2; m = rand() * rand() % maxm + n;
		printf("%d %d\n",n,m);
		for (int i=1;i<=n-1;i++) {
			e[++cnt].a = i+1;
			e[cnt].b = rand() * rand() % i + 1;
			e[cnt].v = rand() * rand() % V + 1;
		}
		for (int i=n;i<=m;i++) {
			e[++cnt].a = rand() * rand() % (n-1) + 2;
			e[cnt].b = rand() * rand() % (e[cnt].a-1) + 1;
			e[cnt].v = rand() % V ;
		}
		sort(e+1,e+cnt+1,cmp);
		for (int i=1;i<=n;i++) fa[i] = i;
		for (int i=1;i<=cnt;i++) 
			if (get( e[i].a ) != get( e[i].b )) {
				fa[ get(e[i].a) ] = get( e[i].b );
				e[i].p = 1;
			}
		for (int i=1;i<=cnt;i++) if ( e[i].p) 
			printf("%d %d %d\n",e[i].a,e[i].b,e[i].v);
		for (int i=1;i<=cnt;i++) if (!e[i].p)
			printf("%d %d %d\n",e[i].a,e[i].b,e[i].v);
	}
	
	fclose(stdout);
	return 0;
}

时间复杂度极其垃圾的暴力

#include 
#include 
#include 
#include 

#define N 1000050
#define INF 1001

using namespace std;
typedef long long LL;

struct Edge{ int a,b,v,p;}e[N];
int d[N],fa[N],n,m;
bool cmp(int p1,int p2) {
	return e[p1].v == e[p2].v ? e[p1].p > e[p2].p : e[p1].v < e[p2].v;
}

int get(int x) { return fa[x] == x ? x : fa[x] = get( fa[x] ); }

LL go(int x) {
	int tmp = e[x].v , p1 = INF , p2 = -1;
	for (int j=-1;j<=INF;j++) {
		e[x].v = j;
		for (int i=1;i<=m;i++) d[i] = i; 
		for (int i=1;i<=n;i++) fa[i] = i;
		sort(d+1,d+m+1,cmp);
		bool flag = true;
		for (int i=1;i<=m;i++)
			if ( get( e[ d[i] ].a ) != get( e[ d[i] ].b ) ) {
				if (!e[ d[i] ].p) flag = false;
				fa[ get(e[ d[i] ].a ) ] = get( e[ d[i] ].b );
			}
		if (flag) p1 = min(p1,j) , p2 = max(p2,j);
		e[x].v = tmp;
	}
	LL r = 0;
	if (p1 == -1) r += -1LL*x*x; else r += 1LL*x*x*(tmp-p1);
	if (p2 == INF) r += -1LL*x; else r += 1LL*x*(p2-tmp);
	return r;
	//return 1LL*x*(p2-tmp) + 1LL*x*x*(tmp-p1);
}

void solve() {
	memset(e,0,sizeof(e));
	LL ans = 0LL;
	scanf("%d%d",&n,&m);
	for (int i=1;i<=m;i++) {
		scanf("%d%d%d",&e[i].a,&e[i].b,&e[i].v);
		if (i <= n-1) e[i].p = 1;
	}
	for (int i=1;i<=m;i++) ans += go(i);
	cout << ans << endl;
}

int main()
{
	freopen("problem6.in","r",stdin);
	freopen("problem6_cmp.out","w",stdout);
	int T = 0; scanf("%d",&T);
	for (int i=1;i<=T;i++) {
		printf("Case %d: ",i); solve();
	}
	fclose(stdin); fclose(stdout);
	return 0;
}