總結——關於2017 11 3測試的分析總結

NOIP 2017 模拟

11 3

T1：

题目：

难以置信这竟然会是T1。天呐，我到底参加的是NOI in Provinces还是NOI Plus 啊啊啊。。

——正解思路：

——我的乱搞：

暴力匹配，n * n * m大暴力。。。

——tips:

编程不止眼前的暴力，还有远方的暴力。

写暴力之前心中一定要有B数。。。

满载悲伤的代码：

(诶我操，正解代码怎么T了？？？？？)

(代码掺了金(#pragma)坷(GCC)垃(optimize("O3))，一秒能当两秒花)

#pragma GCC optimize("O3")#include<algorithm>#include<map>#include<fstream>using namespace std;inline int read(){int i=0;char ch;for(ch=getchar();!isdigit(ch);ch=getchar());for(;isdigit(ch);ch=getchar()) i=(i<<3)+(i<<1)+(ch^48);return i;}int buf[1024];inline void write(int x){char num = 0, c[10];do c[++num] = x % 10 | 48, x /= 10; while (x);while (num) putchar(c[num--]);putchar('\n');}int tot,nxt[8001],first[8001],goal[8001];map<int,int>id;int f1[8001],f2[8001],f[8001],sze[8001<<1],vis[8001<<1],lef,cnt,cnt1,cnt2,cnt3;int ans[1000][1000],spi[8001];int tota,pos1,pos2,x,y,now;int n,q,a[8001];void addedge(int a,int b){nxt[++tot]=first[a];first[a]=tot;goal[tot]=b;}int main(){n=read();q=read();for(register int i=1;i<=n;++i){a[i]=read();if(!id[a[i]]) id[a[i]]=++cnt;a[i]=id[a[i]];}for(register int i=n;i;--i) addedge(a[i],i);for(register int i=1;i<=cnt;++i)for(register int j=i+1;j<=cnt;++j){++tota;cnt1=0;cnt2=0;cnt3=0;for(register int p=first[i];p;p=nxt[p]) f1[++cnt1]=goal[p];for(register int p=first[j];p;p=nxt[p]) f2[++cnt2]=goal[p];cnt3=cnt1+cnt2;pos1=1;pos2=1;for(register int k=1;k<=cnt3;++k)if(pos2>cnt2||(pos1<=cnt1&&f1[pos1]<f2[pos2])) f[k]=f1[pos1++];else f[k]=f2[pos2++];now=8000;vis[now]=tota;sze[now]=f[1];ans[i][j]=f[1]*(f[1]-1)>>1;f[cnt3+1]=n+1;for(int k=1;k<=cnt3;++k){if(a[f[k]]==i) ++now;else --now;if(vis[now]<tota){sze[now]=0;vis[now]=tota;}ans[i][j]+=(f[k+1]-f[k])*sze[now]+(f[k+1]-f[k])*(f[k+1]-f[k]-1)/2;sze[now]+=f[k+1]-f[k];}ans[j][i]=ans[i][j];}for(register int i=1;i<=cnt;++i){spi[i]=0;lef=0;for(register int p=first[i];p;p=nxt[p]){spi[i]+=(goal[p]-lef)*(goal[p]-lef-1)>>1;lef=goal[p];}spi[i]+=(n-lef)*(n-lef+1)>>1;}for(register int i=1;i<=q;++i){x=read();y=read();x=id[x];y=id[y];if((!x&&!y)||(x==y)) write((n+1)*n/2);else if(!x||!y) write(spi[x^y]);else write(ans[x][y]);}return 0;}

T2:

题目：

——正解思路：

——我的乱搞：

天呐，这不是个 vector + erase 风骚走位水题嚒??关键是还真TM AK了???!!!

写线段树的大佬们一脸蒙逼。。。。。。。。。。

——tips :

那么这就很监介了，，真 - 暴力出奇迹。

#pragma GCC optimize("O3")#include <fstream>#include <vector>int n;int p[100010];std::vector <int> v;inline void read (int &x) {x = 0;register char c = getchar();while (!isdigit(c)) c = getchar();while (isdigit(c)) x = x * 10 + (c ^ 48), c = getchar();}inline void write (int x) {short num = 0;char c[10];while (x) c[++num] = x % 10 | 48, x /= 10;while (num) putchar(c[num--]);putchar(' ');}int main () {read (n);for (register int i = 1; i <= n; ++i) read (p[i]), v.push_back(i);std::vector <int> :: iterator it;for (register int i = n; i >= 1; --i) {it = v.begin() + i - p[i] - 1 + p[i - 1];p[i] = *it;v.erase(it);}for (register int i = 1; i <= n; ++i) write (p[i]);return 0;}

正解：

#include<iostream>#include<cstdio>#include<cstdlib>#include<cstring>#include<string>#include<ctime>#include<cmath>#include<algorithm>using namespace std;int n,pre,now,tmp,l,r,mid;int a[100010],tree[100010];void read(int &x){x=0;char ch=getchar();while(!isdigit(ch)) ch=getchar();while(isdigit(ch)){x=(x<<3)+(x<<1)+ch-'0';ch=getchar();}}int lowbit(int x){return x&(-x);}int get(int x){int ans=0;while(x>=1){ans+=tree[x];x-=lowbit(x);}return ans;}void add(int x,int y){while(x<=n){tree[x]+=y;x+=lowbit(x);}}int main(){//freopen("premu.in","r",stdin);read(n);pre=0;for(int i=1;i<=n;++i){read(now);a[i]=now-pre;pre=now;}for(int i=1;i<=n;++i) add(i,1);for(int i=n;i>=1;i--){tmp=i-a[i];l=0,r=n;while(l+1<r){mid=(l+r)>>1;if(get(mid)<tmp) l=mid;else r=mid;}a[i]=r;add(r,-1);}for(int i=1;i<=n;++i) cout<<a[i]<<" ";return 0;}

T3 :

题目：

点击回复你也查看不了

在USACO上学的暴力技巧派上用场了耶。。。。。

——正解思路：

——我的乱搞：

暴搜 + vector 剪枝。

——tips：

good good 暴力, day day 爆零。

满载一种难以言表的思想感情的代码：

#include <iostream>#include <fstream>#include <algorithm>#define MIN(a, b) a = min(a, b)#define SOLVE(a, b) \for (register int x = 1; x <= n; ++x) for (register int y = 1; y <= n; ++y) a[i][x][y] = a[i + b][x][y] + road[i].r;\for (register int x = 1; x <= n; ++x) {\if (a[i + b][x][road[i].y] < inf) MIN(a[i][x][road[i].x], a[i + b][x][road[i].y] + road[i].c);\if (a[i + b][x][road[i].x] < inf) MIN(a[i][x][road[i].y], a[i + b][x][road[i].x] + road[i].c);}using namespace std;struct node{int x, y, c, r;}road[30001];struct query{int x, y, l, r, id;}qry[300001], tag[300001];int n, m, q;int ans[300001];int Left[30001][31][31], Right[30001][31][31];const int inf = 1e+9;inline int read () {register int x = 0;register char c = getchar();while (!isdigit(c)) c = getchar();while (isdigit(c)) x = x * 10 + (c ^ 48), c = getchar();return x;}inline void write (int x) {char num = 0, c[10];if (x < 0) putchar('-'), x = -x;do c[++num] = x % 10 | 48, x /= 10; while (x);while (num) putchar(c[num--]);putchar('\n');}inline void solve(int lr, int rr, int lq, int rq) {if (lq > rq) return;if (lr == rr) {for (register int i = lq; i <= rq; ++i)if (qry[i].x != qry[i].y) {ans[qry[i].id] = (qry[i].x == road[lr].x and qry[i].y == road[lr].y) ? road[lr].c : -1;} else {ans[qry[i].id] = inf;(qry[i].x == road[lr].x and qry[i].y == road[lr].y) ? ans[qry[i].id] = road[lr].c : 0;MIN(ans[qry[i].id], road[lr].r);}return;}int mid = lr + rr >> 1;for (register int i = 1; i <= n; ++i)for (register int j = 1; j <= n; ++j)Left[mid + 1][i][j] = Right[mid][i][j] = inf;for (register int i = 1; i <= n; ++i)Left[mid + 1][i][i] = Right[mid][i][i] = 0;for (register int i = mid; i >= lr; --i) {SOLVE(Left, 1)}for (register int i = mid + 1; i <= rr; ++i) {SOLVE(Right, -1)}for (register int i = lq; i <= rq; ++i) if (qry[i].l <= mid and qry[i].r > mid) {int value = inf;for (register int x = 1; x <= n; ++x) MIN(value, Left[qry[i].l][x][qry[i].x] + Right[qry[i].r][x][qry[i].y]);ans[qry[i].id] = value < inf ? value : -1;}for (register int i = lq; i <= rq; ++i) tag[i] = qry[i];int wl = lq - 1, wr = rq + 1;for (register int i = lq; i <= rq; ++i) {if (tag[i].r <= mid) qry[++wl] = tag[i];if (tag[i].l >  mid) qry[--wr] = tag[i];}solve(lr, mid, lq, wl);solve(mid + 1, rr, wr, rq);}int main () {n = read(), m = read(), q = read();for (register int i = 1; i <= m; ++i) road[i] = (node){read(), read(), read(), read()};for (register int i = 1; i <= q; ++i) qry[i] = (query){read(), read(), read(), read(), i};solve(1, m, 1, q);for (register int i = 1; i <= q; ++i) write(ans[i]);return 0;}