NOIP模拟（11.07）T3 图

图

题目背景：

11.06 NOIP模拟T3

分析：LCT or 暴力

 

这个题，上来就是50分暴力然后就没有然后了······

讲题，首先我们可以知道最后的最小生成树上的边，要么是A种边建成的最小生成树上的边，要么是B种边建成的最小生成树上的边，否则肯定不对，显然当x为无穷小的时候，最小生成树就是A种边构出的最小生成树，那么随着x的增大，B种边会不断的取代B种边，考虑不同的B种边取代A种边的时间。首先如果两条B边在不同的A边形成的环上，那么这两条边的取代时间互不影响，如果这两条边在现在覆盖到了相同的A边的环，那么肯定是小一些的那一条先进入最小生成树，那么到这里就比较清楚了，我们先对A边和B边各求一个最小生成树，然后从小到大枚举B树的边然后考虑替换掉当前环上k值最大的A边，然后记录替换的x值，最后按替换的x值排序从小到大更改贡献即可。加边删边可以通过LCT来维护，但是暴力跑的比LCT快，因为正解是个随机数据······

 

正解：

Source:

/*created by scarlyw*/#include <cstdio>#include <iostream>#include <cmath>#include <algorithm>#include <string>#include <cstring>#include <cctype>#include <vector>#include <queue>#include <set>#include <ctime>inline char read() {static const int IN_LEN = 1024 * 1024;static char buf[IN_LEN], *s, *t;if (s == t) {t = (s = buf) + fread(buf, 1, IN_LEN, stdin);if (s == t) return -1;}return *s++;}template<class T>inline void R(T &x) {static char c;static bool iosig;for (c = read(), iosig = false; !isdigit(c); c = read()) {if (c == -1) return ;if (c == '-') iosig = true;}for (x = 0; isdigit(c); c = read())x = ((x << 2) + x << 1) + (c ^ '0');if (iosig) x = -x;}const int OUT_LEN = 1024 * 1024;char obuf[OUT_LEN], *oh = obuf;inline void write_char(char c) {if (oh == obuf + OUT_LEN) fwrite(obuf, 1, OUT_LEN, stdout), oh = obuf;*oh++ = c;}template<class T>inline void W(T x) {static int buf[30], cnt;if (x == 0) write_char('0');else {if (x < 0) write_char('-'), x = -x;for (cnt = 0; x; x /= 10) buf[++cnt] = x % 10 + 48;while (cnt) write_char(buf[cnt--]);}}inline void flush() {fwrite(obuf, 1, oh - obuf, stdout);}const int MAXN = 100000 + 10;const int MAXM = 200000 + 10;const int INF = 2000000000;long long mst;long long ans[MAXN];int w[MAXM];struct edges {int u, v, w;inline bool operator < (const edges &a) const {return w < a.w;}} a[MAXM], b[MAXM], ta[MAXN], tb[MAXN];struct node *null;struct node {node *c[2], *fa, *pa;int max, val, rank, size;bool rev;node() {}node(int val) : fa(null), pa(null), max(val), val(val), rank(rand()), size(1), rev(false) {c[0] = c[1] = null;}inline void reverse() {std::swap(c[0], c[1]), rev ^= 1;}inline void push_down() {if (this == null) return ;if (rev) c[0]->reverse(), c[1]->reverse(), rev ^= 1;}inline void maintain() {if (this == null) return ;size = c[0]->size + c[1]->size + 1;max = val;if (w[c[0]->max] > w[max]) max = c[0]->max;if (w[c[1]->max] > w[max]) max = c[1]->max;}inline int relation() {return this == fa->c[1];}inline void rotate() {push_down();node *old_fa = fa;int x = relation();pa = fa->pa, fa->pa = null;if (old_fa->fa != null) old_fa->fa->c[old_fa->relation()] = this;fa = old_fa->fa, old_fa->c[x] = c[x ^ 1];if (c[x ^ 1] != null) c[x ^ 1]->fa = old_fa;c[x ^ 1] = old_fa, old_fa->fa = this, old_fa->maintain(), maintain();}inline void splay(node *target = null) {while (fa != target) {fa->fa->push_down(), fa->push_down();if (fa->fa == target) rotate();else if (fa->relation() == relation()) fa->rotate(), rotate();else rotate(), rotate();}push_down();}inline void expose() {splay();if (c[1] != null) c[1]->pa = this, c[1]->fa = null, c[1] = null, maintain();}inline bool splice() {expose();if (pa == null) return false;pa->expose(), pa->c[1] = this, fa = pa, pa = null, fa->maintain();return true;}inline void access() {expose();while (splice());splay();}inline void evert() {access(), reverse();}} point[MAXN << 1];inline void link(node *u, node *v) {u->evert(), u->pa = v;}inline void cut(node *u, node *v) {u->evert(), v->access(), v->c[0] = null, u->fa = null, u->maintain();}inline int query_max(node *u, node *v) {return u->evert(), v->access(), v->max;}struct query {int x, id;inline bool operator < (const query &a) const {return x < a.x;}} q[MAXN];int n, A, B, m;int father[MAXN];inline void read_in() {R(n), R(A), R(B), R(m);for (int i = 1; i <= A; ++i) R(a[i].u), R(a[i].v), R(a[i].w);for (int i = 1; i <= B; ++i) R(b[i].u), R(b[i].v), R(b[i].w);for (int i = 1; i <= m; ++i) R(q[i].x), q[i].id = i;}inline int get_father(int x) {return (father[x] == x) ? (x) : (father[x] = get_father(father[x]));} inline void kruscal(edges *e, int m, edges *new_e) {std::sort(e + 1, e + m + 1);for (int i = 1; i <= n; ++i) father[i] = i;int cnt = 0;for (int i = 1; i <= m; ++i) { int x = e[i].u, y = e[i].v, fa1 = get_father(x), fa2 = get_father(y);if (fa1 != fa2) {father[fa1] = fa2, new_e[++cnt] = e[i];if (cnt == n - 1) break ;}}}struct data {int a, b, x;data() {}data(int a, int b, int x) : a(a), b(b), x(x) {}inline bool operator < (const data &c) const {return x < c.x;}} modify[MAXN];inline void solve() {null = &point[0], null->fa = null->pa = null->c[0] = null->c[1] = null;null->size = null->val = null->max = null->rev = 0;read_in(), kruscal(a, A, ta), kruscal(b, B, tb), w[0] = -INF;for (int i = 1; i <= n; ++i) point[i] = node(0);for (int i = 1; i < n; ++i) w[i] = ta[i].w;for (int i = 1; i < n; ++i) {point[i + n] = node(i), mst += (long long)ta[i].w;link(&point[ta[i].u], &point[i + n]);link(&point[ta[i].v], &point[i + n]);}for (int i = 1; i < n; ++i) {int x = tb[i].u, y = tb[i].v, cur = query_max(&point[x], &point[y]);modify[i] = data(w[cur], tb[i].w, (tb[i].w - w[cur] + 1) / 2);cut(&point[ta[cur].u], &point[cur + n]);cut(&point[ta[cur].v], &point[cur + n]);link(&point[y], &point[x]);}std::sort(q + 1, q + m + 1), std::sort(modify + 1, modify + n);for (int i = 1, head = 1; i <= m; ++i) {while (head < n && modify[head].x <= q[i].x) {mst -= (long long)modify[head].a, mst += (long long)modify[head].b;head++;}ans[q[i].id] = mst - (long long)(head - 1) * q[i].x + (long long)(n - head) * q[i].x;}for (int i = 1; i <= m; ++i) W(ans[i]), write_char('\n');}int main() {solve();flush();return 0;}

暴力(orz DZYO dalao)：

Source:

//duzhenyu#include<bits/stdc++.h>using namespace std;inline int read(){char ch=getchar();int i=0,f=1;while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}while(isdigit(ch)){i=(i<<1)+(i<<3)+ch-'0';ch=getchar();}return i*f;}inline void W(long long x){static int buf[50];if(!x){putchar('0');return;}if(x<0){putchar('-');x=-x;}while(x){buf[++buf[0]]=x%10;x/=10;}while(buf[0])putchar(buf[buf[0]--]+'0');}const int Maxn=1e5+50; const int INF=0x3f3f3f3f;int n,Acnt,Bcnt,q,anc[Maxn],ht[Maxn],fa[Maxn];long long ans[Maxn],sum;struct E{int x,y,w;friend inline bool operator <(const E &a,const E &b){return a.w<b.w;} }A[Maxn<<1],B[Maxn<<1];struct E2{int x,y,w,low,rpc;inline bool operator =(const E &b){x=b.x,y=b.y,w=b.w;}}Bnow[Maxn];struct Q{int v,id;}qry[Maxn];inline bool cmpv(const Q &a,const Q &b){return a.v<b.v;}inline bool cmpw(const E2 &a,const E2 &b){return a.w<b.w;}inline bool cmplow(const E2 &a,const E2 &b){return a.low<b.low;}inline int getanc(int x){return (anc[x]==x)?(x):(anc[x]=getanc(anc[x]));}typedef pair<int,int> pii;inline void makeroot(int x){static int st[Maxn],w[Maxn],tail;st[tail=1]=x;while(fa[x]){st[++tail]=fa[x];w[tail]=ht[x];x=fa[x];} for(int i=tail;i>1;i--)fa[st[i]]=st[i-1],ht[st[i]]=w[i];fa[st[1]]=0;}inline void Link(int x,int y,int val){makeroot(x);fa[x]=y;ht[x]=val;}inline pii findmx(int x,int y){static int vis[Maxn],vt;++vt;int mx=-INF,mxid=0,xtmp=x,ytmp=y;while(fa[xtmp]){vis[xtmp]=vt;xtmp=fa[xtmp];} while(fa[ytmp]){if(vis[ytmp]==vt)break;if(ht[ytmp]>mx)mx=ht[ytmp],mxid=ytmp;ytmp=fa[ytmp];}++vt;xtmp=y,ytmp=x;while(fa[xtmp]){vis[xtmp]=vt;xtmp=fa[xtmp];} while(fa[ytmp]){if(vis[ytmp]==vt)break;if(ht[ytmp]>mx)mx=ht[ytmp],mxid=ytmp;ytmp=fa[ytmp];}return make_pair(mx,mxid);}int main(){freopen("mst.in","r",stdin);freopen("mst.out","w",stdout);n=read(),Acnt=read(),Bcnt=read(),q=read();for(int i=1;i<=Acnt;i++){A[i].x=read(),A[i].y=read(),A[i].w=read();}for(int i=1;i<=Bcnt;i++){B[i].x=read(),B[i].y=read(),B[i].w=read();}sort(A+1,A+Acnt+1);sort(B+1,B+Bcnt+1);int head=1;for(int i=1;i<=n;i++)anc[i]=i;for(int i=1;i<n;i++){while(getanc(A[head].x)==getanc(A[head].y))++head;anc[getanc(A[head].x)]=getanc(A[head].y);Link(A[head].x,A[head].y,A[head].w);sum+=A[head].w;++head;}head=1;int tot=0;for(int i=1;i<=n;i++)anc[i]=i;for(int i=1;i<n;i++){while(getanc(B[head].x)==getanc(B[head].y))++head;anc[getanc(B[head].x)]=getanc(B[head].y);Bnow[i]=B[head++];}sort(Bnow+1,Bnow+tot+1,cmpw);for(int i=1;i<n;i++){pii t=findmx(Bnow[i].x,Bnow[i].y);if(t.second<=-INF){Bnow[i].low=INF;}else{Bnow[i].low=(Bnow[i].w-t.first+1)/2;Bnow[i].rpc=t.first;fa[t.second]=0;ht[t.second]=0;Link(Bnow[i].x,Bnow[i].y,-INF);}}sort(Bnow+1,Bnow+n,cmplow);for(int i=1;i<=q;i++)qry[i].v=read(),qry[i].id=i;sort(qry+1,qry+q+1,cmpv);int cntB=0;head=0;for(int i=1;i<=q;i++){while(head<n-1&&Bnow[head+1].low<=qry[i].v){++head;cntB++;sum+=Bnow[head].w;sum-=Bnow[head].rpc;}ans[qry[i].id]=sum+1ll*(n-1-2*cntB)*qry[i].v;}for(int i=1;i<=q;i++){W(ans[i]);putchar('\n');}}