NOIP模拟9.23（TYVJ NOIP2017模拟赛D1）

来源：互联网发布：linux 设置字符编码编辑：程序博客网时间：2024/06/05 23:05

T1.回形遍历。（模拟）AC。
T2.排列（单调栈+暴力）只暴力了60
T3.近似排列计数（状压+矩阵快速幂）搜索本来应该30分的。。。奈何写错字母。。

T1 回形遍历

#include <bits/stdc++.h>using namespace std;#define ll long long#define inf 0x3f3f3f3f#define pa pair<ll,int>#define N 50010inline 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;}int n,m,xx,y,z,cnt=0,st=1;int main(){//  freopen("calc.in","r",stdin);//  freopen("calc.out","w",stdout);    n=read();m=read();xx=read();y=read();z=read();    if(y==0){        cnt=n-1-xx;if(cnt>=z){printf("%d %d",xx+z,0);return 0;}        if(z-cnt<=m-1){printf("%d %d",n-1,z-cnt);return 0;}cnt+=m-1;        if(z-cnt<=n-1){printf("%d %d",n-1-(z-cnt),m-1);return 0;}cnt+=n-1;        if(z-cnt<=m-2){printf("%d %d",0,m-1-(z-cnt));return 0;}cnt+=m-2;    }    else if(y==m-1){        cnt=xx;if(cnt>=z){printf("%d %d",xx-z,m-1);return 0;}        if(z-cnt<=m-2){printf("%d %d",0,m-1-(z-cnt));return 0;}cnt+=m-2;    }    else if(xx==n-1){        cnt=m-1-y;if(cnt>=z){printf("%d %d",n-1,y+z);return 0;}        if(z-cnt<=n-1){printf("%d %d",n-1-(z-cnt),m-1);return 0;}cnt+=n-1;        if(z-cnt<=m-2){printf("%d %d",0,m-1-(z-cnt));return 0;}cnt+=m-2;    }    else if(xx==0){        cnt=y-1;if(cnt>=z){printf("%d %d",0,y-z);return 0;}    }else{        for(int x=1;x<=n;++x){            if(y==x){z+=xx-(x-1);st=x;break;}            if(xx==n-1-x){z+=y-x+n-1-x-(x-1);st=x;break;}            if(y==m-1-x){z+=n-1-x-(x-1)+m-1-x-x+n-1-x-xx;st=x;break;}            if(xx==x){z+=n-1-x-(x-1)+m-1-x-x+n-1-x-x+m-1-x-y;st=x;break;}        }    }    for(int x=st;x<=n;++x){        if(x-1+z-cnt<=n-1-x){printf("%d %d",x-1+z-cnt,x);return 0;}        else cnt+=n-1-x-(x-1);        if(x+z-cnt<=m-1-x){printf("%d %d",n-1-x,x+z-cnt);return 0;}        else cnt+=(m-1-x)-x;        if(n-1-x-(z-cnt)>=x){printf("%d %d",n-1-x-(z-cnt),m-1-x);return 0;}        else cnt+=(n-1-x)-x;        if(m-1-x-(z-cnt)>=x+1){printf("%d %d",x,m-1-x-(z-cnt));return 0;}        else cnt+=m-1-x-(x+1);    }    return 0;}

T2 排列

#include <bits/stdc++.h>using namespace std;#define ll long long#define inf 0x3f3f3f3f#define N 100010#define pa pair<int,int>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;}int tst,n,a[N],pos1[N],pos2[N];//pos1[i]比i大的数第一次出现的位置，pos2比i小的 ll ans[N];stack<pa>mx1,mn1;int main(){//  freopen("a.in","r",stdin);    tst=read();    while(tst--){        n=read();for(int i=1;i<=n;++i) a[i]=read();memset(ans,0,sizeof(ans));        while(!mx1.empty()) mx1.pop();        while(!mn1.empty()) mn1.pop();        for(int i=n;i>=1;--i){            while(!mx1.empty()&&a[i]>mx1.top().first) mx1.pop();            if(mx1.empty()) pos1[a[i]]=n+1;            else pos1[a[i]]=mx1.top().second;            mx1.push(make_pair(a[i],i));            while(!mn1.empty()&&a[i]<mn1.top().first) mn1.pop();            if(mn1.empty()) pos2[a[i]]=n+1;            else pos2[a[i]]=mn1.top().second;            mn1.push(make_pair(a[i],i));        }ans[0]=n;        for(int i=1;i<=n;++i){            int mx=a[i],mn=a[i];            for(int j=i+1;j<=n;++j){                if(a[j]>mx) mx=a[j];                else mn=a[j];                int pos=min(pos1[mx],pos2[mn]);                ans[mx-mn]+=pos-j;j=pos-1;            }        }        for(int i=1;i<n;++i) ans[i]+=ans[i-1];        for(int i=0;i<=n-1;++i) printf("%lld\n",ans[i]);    }    return 0;}

T3 近似排列计数

30分的搜索。。

#include <bits/stdc++.h>using namespace std;#define ll long long#define inf 0x3f3f3f3f#define pa pair<ll,int>#define N 100010#define mod 1000000007inline 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;}int tst,n,m,k,a[N],ans=0,dx[]={-1,0,1,-2,2};bool f[N];void dfs(int x){    if(x>n) ans++;    if(a[x]) dfs(x+1);    for(int i=0;i<3;++i){        int xx=x+dx[i];        if(xx>=1&&xx<=n&&!f[xx]){            f[xx]=1;dfs(x+1);f[xx]=0;        }    }}void dfs1(int x){    if(x>n) ans++;    if(a[x]) dfs1(x+1);    for(int i=0;i<5;++i){        int xx=x+dx[i];        if(xx>=1&&xx<=n&&!f[xx]){            f[xx]=1;dfs1(x+1);f[xx]=0;        }    }}void solve0(){    ans=0;    if(k==1) dfs(1);    else dfs1(1);    printf("%d\n",ans%mod);}int main(){//  freopen("a.in","r",stdin);    tst=read();    while(tst--){        n=read();m=read();k=read();        memset(f,0,sizeof(f));memset(a,0,sizeof(a));        for(int i=1;i<=m;++i){            int x=read(),y=read();a[x]=y;f[y]=1;        }        if(k==0){puts("1");continue;}        if(n<=20) solve0();        else solve0();    }    return 0;}

按道理50分的朴素状压dp

#include <bits/stdc++.h>using namespace std;#define ll long long#define inf 0x3f3f3f3f#define pa pair<ll,int>#define N 100010#define mod 1000000007inline 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;}int tst,n,m,k,a[N],ans=0,dx[]={-1,0,1,-2,2},dp[1100000],bin[25];void solve1(){    memset(dp,0,sizeof(dp));dp[0]=1;    for(int i=1;i<=n;++i){        if(a[i]){            for(int s=bin[n]-1;s>=0;--s)                if(!(s&(1<<a[i]-1))) dp[s|(1<<a[i]-1)]=(dp[s|(1<<a[i]-1)]+dp[s])%mod;            continue;           }        for(int s=bin[n]-1;s>=0;--s)            for(int j=0;j<3;++j){                int x=i+dx[j];if(x<1||x>n) continue;                if(!(s&(1<<x-1))) dp[s|(1<<x-1)]=(dp[s|(1<<x-1)]+dp[s])%mod;            }    }    printf("%d\n",dp[bin[n]-1]);}void solve2(){    memset(dp,0,sizeof(dp));dp[0]=1;    for(int i=1;i<=n;++i){        if(a[i]){            for(int s=bin[n]-1;s>=0;--s)                if(!(s&(1<<a[i]-1))) dp[s|(1<<a[i]-1)]=(dp[s|(1<<a[i]-1)]+dp[s])%mod;            continue;           }        for(int s=bin[n]-1;s>=0;--s)            for(int j=0;j<5;++j){                int x=i+dx[j];if(x<1||x>n) continue;                if(!(s&(1<<x-1))) dp[s|(1<<x-1)]=(dp[s|(1<<x-1)]+dp[s])%mod;            }    }    printf("%d\n",dp[bin[n]-1]);}int main(){    freopen("a.in","r",stdin);    tst=read();bin[0]=1;    for(int i=1;i<=20;++i) bin[i]=bin[i-1]<<1;    while(tst--){        n=read();m=read();k=read();memset(a,0,sizeof(a));        for(int i=1;i<=m;++i){            int x=read(),y=read();a[x]=y;        }        if(k==0){puts("1");continue;}        if(k==1) solve1();        else solve2();    }    return 0;}

正解，待填坑。。。

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