D.

題意：給2組數(shù)據(jù)a和b數(shù)組，每次有2種操作：(+,l,r,x)把a(bǔ)數(shù)組第l個(gè)到第r個(gè)元素全置為x，(?,l,r)查詢[l,r]之間哪些位置滿足a[i]>=b[i]（i>=l && i<=r）并把這些位置的數(shù)量統(tǒng)計(jì)

一直想很久，沒想到什么有效的方案，直到看到題解才明白過來，原來線段樹套平衡樹還有這種情況：里面其實(shí)不是平衡樹，只是有序表。

然后這題就轉(zhuǎn)換為區(qū)間查找數(shù)對應(yīng)排名

由于此題不用對2個(gè)數(shù)組都修改，其中1個(gè)b樹可作為固定的線段樹套有序表以節(jié)省時(shí)間，另外1個(gè)表a樹則單純使用線段樹的方法先修改，再更新對應(yīng)b樹結(jié)點(diǎn)的排名

這里查找排名如果全部logn查找會因?yàn)槌?shù)太大直接卡，注意每個(gè)結(jié)點(diǎn)都含有序表并且上下有包含關(guān)系

那咱們可以在b樹自底向上更新父結(jié)點(diǎn)排名對應(yīng)左右子樹里的排名，用歸并排序的方法，占用空間才o(nlogn)，時(shí)間也是o(nlogn)

順帶把會改變的a樹1個(gè)個(gè)結(jié)點(diǎn)查詢b樹查出排名，修改時(shí)先查出根結(jié)點(diǎn)對應(yīng)位置，再根據(jù)位置子樹表一邊向下更新一邊轉(zhuǎn)移到子樹對應(yīng)位置

#include<stdio.h> #include<string.h> #include<stdlib.h> #include<queue> #include<stack> #include<math.h> #include<vector> #include<map> #include<set> #include<stdlib.h> #include<cmath> #include<string> #include<algorithm> #include<iostream> using namespace std; typedef __int64 ll; int max(int a,int b){return a>b?a:b;} int min(int a,int b){return a<b?a:b;} int cnt; const int N=100010,M=262150,E=1768950; int n,m,i,a[N],b[N],x,l,r; int st[M],en[M],v[M],tag[M],pl[E],pr[E],pool[E],cur; ll ans,sum;void build(int x,int l,int r) {tag[x]=-1;if(l==r){st[x]=cur+1;pool[++cur]=b[l];en[x]=cur;v[x]=(a[l]>=b[l]);return;}int mid=((l+r)>>1);build(x<<1,l,mid);build((x<<1)|1,mid+1,r);v[x]=v[x<<1]+v[(x<<1)|1];int al=st[x<<1],ar=en[x<<1],bl=st[(x<<1)|1],br=en[(x<<1)|1];st[x]=cur+1;while(al<=ar&&bl<=br)pool[++cur]=pool[al]<pool[bl]?pool[al++]:pool[bl++];while(al<=ar)pool[++cur]=pool[al++];while(bl<=br)pool[++cur]=pool[bl++];en[x]=cur;al=st[x<<1],bl=st[x<<1|1];for(int i=st[x];i<=cur;i++){while(al<=ar&&pool[al]<=pool[i])al++;while(bl<=br&&pool[bl]<=pool[i])bl++;pl[i]=al-1,pr[i]=bl-1;if(pl[i]<st[x<<1])pl[i]=0;if(pr[i]<st[(x<<1)|1])pr[i]=0;} }inline void rankpt(int x,int p) {v[x]=(p?p-st[x]+1:0);tag[x]=p; } inline void pushdown(int x) {if(tag[x]<0)return;int p=tag[x];rankpt(x<<1,pl[p]);rankpt((x<<1)|1,pr[p]);tag[x]=-1; }void update(int x,int a,int b,int p) {if(l<=a && b<=r){rankpt(x,p);return;}pushdown(x);int mid=(a+b)>>1;if(l<=mid)update(x<<1,a,mid,pl[p]);if(r>mid)update((x<<1)|1,mid+1,b,pr[p]);v[x]=v[x<<1]+v[(x<<1)|1]; }void query(int x,int a,int b) {if(l<=a && b<=r){ans+=v[x];return;}pushdown(x);int mid=((a+b)>>1);if(l<=mid)query(x<<1,a,mid);if(r>mid)query((x<<1)|1,mid+1,b);v[x]=v[x<<1]+v[(x<<1)|1]; }inline int lower(int x){//lower_bound(pool+st[1],pool+ed[1]+1,x);int l=st[1],r=en[1],mid,t=0;while(l<=r)if(pool[mid=(l+r)>>1]<=x)l=(t=mid)+1;else r=mid-1;return t; }int seeda, seedb, C = ~(1<<31), MM = (1<<16)-1; int rnd(int last) {seeda = (36969 + (last >> 3)) * (seeda & MM) + (seeda >> 16);seedb = (18000 + (last >> 3)) * (seedb & MM) + (seedb >> 16);return (C & ((seeda << 16) + seedb)) % 1000000000; }int main() {int t,ku;scanf("%d",&t);while(t--){scanf("%d%d%d%d",&n,&m,&seeda,&seedb);for(i=1;i<=n;i++)scanf("%d",a+i);for(i=1;i<=n;i++)scanf("%d",b+i);ans=sum=cur=0;build(1,1,n);for(i=1;i<=m;i++){l=rnd(ans)%n+1,r=rnd(ans)%n+1,ku=rnd(ans)+1;int kkk=lower(ku);if(l>r)l^=r^=l^=r;if((l+r+ku)&1)update(1,1,n,lower(ku));else{ans=0;query(1,1,n);sum=(sum+(ll)i*ans)%1000000007;}}printf("%I64d\n",sum);}return 0; } View Code

F.

題意，有1個(gè)圖（可能不是全部聯(lián)通的）Gi表示去掉第i個(gè)點(diǎn)的圖權(quán)值（1個(gè)連通圖的權(quán)值=所有點(diǎn)的積 多個(gè)連通圖的權(quán)值=連通圖權(quán)值之和）

求1*G1+2*G2+3*G3+...+n*Gn

求點(diǎn)雙聯(lián)通分量，求出割點(diǎn)，然后求出此點(diǎn)以外點(diǎn)雙聯(lián)通連通圖的權(quán)值

#include <stdio.h> #include <stdlib.h> #include <string.h> #include <math.h> #include <ctype.h> #include <iostream> #include <algorithm> #include <vector> #include <stack> #include <queue> #include <deque> #include <set> #include <map>using namespace std;#define MAXN 100005 #define MOD 1000000007int inv(int x) {int res=1, mod=MOD-2;do{if(mod & 1) res = (long long)res * x % MOD;x = (long long)x * x % MOD;}while(mod >>= 1);return res; }vector<int> G[MAXN]; int w[MAXN]; int z[MAXN]; int cz[MAXN]; int pre[MAXN]; bool iscut[MAXN]; int ncc_no[MAXN]; vector<int> ncc[MAXN]; int ncc_size[MAXN]; int ncc_cnt; int clock;int tarjan(int u, int fa) {ncc[ncc_no[u] = ncc_cnt].push_back(u);ncc_size[ncc_cnt] = ((long long)ncc_size[ncc_cnt] * w[u]) % MOD;int lowu = pre[u] = ++clock;int child, i, v, lowv, nowsize, tmp;for(child=0, i=G[u].size(); i--;){if(!pre[v = G[u][i]]){child++;nowsize = ncc_size[ncc_cnt];lowv = tarjan(v, u);lowu = min(lowv, lowu);if(lowv >= pre[u]){iscut[u] = true;tmp = (long long)ncc_size[ncc_cnt] * inv(nowsize) % MOD;z[u] = (z[u] + tmp) % MOD;cz[u] = (long long)cz[u] * tmp % MOD;}}else if(pre[v] < pre[u] && v != fa)lowu = min(lowu, pre[v]);}if(fa < 0 && child == 1)iscut[u] = false;return lowu; }int main() {int t, n, m, x, y, i, sum;scanf("%d", &t);while(t-- && scanf("%d%d", &n, &m) > 0){for(i=1; i<=n; i++){scanf("%d", w+i);G[i].clear();cz[i] = 1;ncc[i].clear();}while(m--){scanf("%d%d", &x, &y);G[x].push_back(y);G[y].push_back(x);}memset(z, 0, sizeof z);memset(pre, 0, sizeof pre);memset(iscut, false, sizeof iscut);memset(ncc_no, 0, sizeof ncc_no);clock = ncc_cnt = 0;for(i=1, sum=0; i<=n; i++)if(!pre[i]){ncc_size[++ncc_cnt] = 1;tarjan(i, -1);for(x=ncc[ncc_cnt].size(); x--;)if(iscut[y = ncc[ncc_cnt][x]] && y!=i)z[y] = (z[y] + (long long)ncc_size[ncc_cnt] * inv((long long)cz[y] * w[y] % MOD) % MOD) % MOD;sum = (sum + ncc_size[ncc_cnt]) % MOD;}for(i=1, y=0; i<=n; i++){if(!iscut[i])z[i] = ncc[ncc_no[i]].size() > 1 ? (long long)ncc_size[ncc_no[i]] * inv(w[i]) % MOD : 0;y = ((long long)y + ((long long)z[i] + (long long)(sum - ncc_size[ncc_no[i]] + MOD) % MOD) % MOD * i % MOD) % MOD;}printf("%d\n", y);}return 0; } View Code

G.

轉(zhuǎn)載于:https://www.cnblogs.com/dgutfly/p/6037684.html

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