Codeforces Round #436 (Div. 2)解题报告

这可能是我人生第一场计rating的cf啊。。怎么就炸的这么惨啊。

不得不说，题目确实很水啊。。然而我还是被艹翻了啊。

似乎前四题都太水了。于是扔代码跑。

Fair Game

#include <bits/stdc++.h>#define gc getchar()#define ll long long#define N 109using namespace std;int a[N],vis[N],b[N];int read(){    int x=1;    char ch;    while (ch=gc,ch<'0'||ch>'9') if (ch=='-') x=-1;    int s=ch-'0';    while (ch=gc,ch>='0'&&ch<='9') s=s*10+ch-'0';    return x*s;}int main(){    int n=read(),num=0;    for (int i=1;i<=n;i++)        if (!vis[a[i]=read()]) vis[a[i]]=1,b[++num]=a[i];        else vis[a[i]]++;    if (num==2&&vis[b[1]]==vis[b[2]])    {        puts("YES");        printf("%d %d\n",b[1],b[2]);    }    else        puts("NO");    return 0;}

Polycarp and Letters

#include <bits/stdc++.h>#define gc getchar()#define ll long long#define N 2009using namespace std;int read(){    int x=1;    char ch;    while (ch=gc,ch<'0'||ch>'9') if (ch=='-') x=-1;    int s=ch-'0';    while (ch=gc,ch>='0'&&ch<='9') s=s*10+ch-'0';    return x*s;}char a[N],ch;int ans=0,now=0,vis[N];int n;int main(){    n=read();    scanf("%s",a+1);    memset(vis,0,sizeof(vis));    for (int i=1;i<=n;i++)    {        if (a[i]>='A'&&a[i]<='Z')        {            ans=max(ans,now);            now=0;            memset(vis,0,sizeof(vis));        }        else        {            if (!vis[a[i]]) vis[a[i]]=1,now++;        }    }    ans=max(ans,now);    printf("%d\n",ans);    return 0;}

Bus

#include <bits/stdc++.h>#define gc getchar()#define ll long longusing namespace std;int read(){    int x=1;    char ch;    while (ch=gc,ch<'0'||ch>'9') if (ch=='-') x=-1;    int s=ch-'0';    while (ch=gc,ch>='0'&&ch<='9') s=s*10+ch-'0';    return x*s;}int main(){    int a=read(),b=read(),f=read(),k=read();    int rest=b,ans=0;    if (k==1)    {        if (b<f||b<a-f)        {            puts("-1");            return 0;        }        if (b>=a)        {            puts("0");            return 0;        }        if (b<a)        {            puts("1");            return 0;        }    }    if (k==2)    {        if (b<f||b<2*(a-f))        {            puts("-1");            return 0;        }        if (b>=2*a)        {            puts("0");            return 0;        }        if (b>=2*a-f)        {            puts("1");            return 0;        }        if (b<2*a-f)        {            puts("2");            return 0;        }    }    if (b<2*f||b<2*(a-f))    {        puts("-1");        return 0;    }    for (int i=1;i<=k;i++)    {        if (rest>=a) rest-=a;        else        {            ans++;            if (i&1)            {                rest=b-(a-f);            }            else            {                rest=b-f;            }        }        if (i==k) break;        if (i&1)        {            if (rest<a-f)            {                ans++;                rest=b-(a-f);            }        }        else        {            if (rest<f)            {                ans++;                rest=b-f;            }        }    }    printf("%d\n",ans);    return 0;}

Make a Permutation!

#include <bits/stdc++.h>#define gc getchar()#define ll long long#define N 200009using namespace std;int n,a[N],last[N],b[N],vis[N];priority_queue<int,vector<int>,greater<int> > q;int read(){    int x=1;    char ch;    while (ch=gc,ch<'0'||ch>'9') if (ch=='-') x=-1;    int s=ch-'0';    while (ch=gc,ch>='0'&&ch<='9') s=s*10+ch-'0';    return x*s;}int main(){    n=read();    for (int i=1;i<=n;i++) a[i]=read();    for (int i=1;i<=n;i++) last[a[i]]=i;    for (int i=1;i<=n;i++)        if (!last[i]) q.push(i);    int ans=0;    memset(vis,0,sizeof(vis));    for (int i=1;i<=n;i++)    {        if (!vis[a[i]]&&last[a[i]]==i) b[i]=a[i],vis[a[i]]=1;        else        {            while (vis[q.top()]) q.pop();            int x=q.top();            if (x<a[i]||vis[a[i]])            {                b[i]=x;                vis[x]=1;                ans++;                q.pop();            }            else            {                b[i]=a[i];                vis[a[i]]=1;            }        }    }    printf("%d\n",ans);    for (int i=1;i<=n;i++) printf("%d%s",b[i],i==n?"\n":" ");    return 0;}

Fire

按照di从小到大排个序，然后01背包即可。
然而我记录方案居然是一维的，果然还是菜啊。

#include <bits/stdc++.h>#define gc getchar()#define ll long long#define N 2009#define M 20009#define inf 0x3f3f3f3fusing namespace std;int n,dp[M],vis[N][M],ans[N];struct node{    int t,d,p,pos;    bool operator <(const node &rhs) const    {        return d<rhs.d;    }}q[N];int read(){    int x=1;    char ch;    while (ch=gc,ch<'0'||ch>'9') if (ch=='-') x=-1;    int s=ch-'0';    while (ch=gc,ch>='0'&&ch<='9') s=s*10+ch-'0';    return x*s;}int main(){    n=read();    for (int i=1;i<=n;i++)    {        q[i].t=read();        q[i].d=read();        q[i].p=read();        q[i].pos=i;    }    sort(q+1,q+n+1);    for (int i=1;i<=n;i++)    {        for (int j=q[i].d-1;j>=q[i].t;j--)        {            if (dp[j-q[i].t]+q[i].p>dp[j])            {                dp[j]=dp[j-q[i].t]+q[i].p;                vis[i][j]=1;            }        }    }    int Ans=0,pos=0;    for (int i=0;i<20000;i++)        if (dp[i]>Ans)        {            Ans=dp[i];            pos=i;        }    printf("%d\n",Ans);    int m=0;    for (int i=n;i;i--)    {        if (vis[i][pos])        {            ans[++m]=q[i].pos;            pos-=q[i].t;        }    }    printf("%d\n",m);    for (int i=m;i;i--)        printf("%d%s",ans[i],i==1?"\n":" ");    if (!m) puts("");    return 0;}

Cities Excursions

大致思路是每个点跑出一棵树，然后倍增，树剖什么的跳一下就好了。
然而难点就是建树，要搞清楚环的特殊性（环上点后来出去的边都是到不了的）。最重要的是要时刻考虑到环可能扩大影响区域。。

#include <bits/stdc++.h>#define gc getchar()#define ll long long#define N 3002using namespace std;int n,m,q,first[N],number,vis[N];int now_cir;struct edge{    int to,next,vis;    void add(int x,int y)    {        to=y,next=first[x],first[x]=number;    }}e[N<<1];bool cmp(int x,int y){    return e[x].to<e[y].to;}struct Tree{    int size[N],Mson[N],fa[N],top[N],deep[N],dfn[N],id[N],cnt;    void init()    {        memset(fa,0,sizeof(fa));        memset(deep,0,sizeof(deep));        deep[0]=size[0]=cnt=0;    }    void dfs1(int x)    {        Mson[x]=0,size[x]=1;        for (int i=first[x];i;i=e[i].next)            if (e[i].vis)            {                dfs1(e[i].to),size[x]+=size[e[i].to];                if (size[e[i].to]>size[Mson[x]]) Mson[x]=e[i].to;            }    }    void dfs2(int x,int y)    {        id[dfn[x]=++cnt]=x,top[x]=y;        if (Mson[x]) dfs2(Mson[x],y);        for (int i=first[x];i;i=e[i].next)            if (e[i].vis&&e[i].to!=Mson[x]) dfs2(e[i].to,e[i].to);    }    int qry(int y,int z)    {        z=deep[y]-z+1;        if (z<=0) return -1;        while (z>deep[y]-deep[top[y]]+1&&y)        {            z-=deep[y]-deep[top[y]]+1;            y=fa[top[y]];        }        return id[dfn[y]-z+1];    }    void dfs(int x,bool flag)    {        if (!flag) deep[x]=deep[fa[x]]+1;        vector<int> to;        for (int i=first[x];i;i=e[i].next)            if (vis[e[i].to]!=2&&vis[e[i].to]!=1)                to.push_back(i);        sort(to.begin(),to.end(),cmp);        vis[x]=flag?2:3;        for (int i=0;i<(int)to.size();i++)            if (vis[e[to[i]].to]==3)            {                if (!now_cir||deep[e[to[i]].to]<deep[now_cir])                    now_cir=e[to[i]].to;            }            else                if (!vis[e[to[i]].to])                {                    if (!now_cir) e[to[i]].vis=1,fa[e[to[i]].to]=x;                    dfs(e[to[i]].to,(now_cir>0));                }        vis[x]=1;        if (x==now_cir) now_cir=0;    }}t[N];int read(){    int x=1;    char ch;    while (ch=gc,ch<'0'||ch>'9') if (ch=='-') x=-1;    int s=ch-'0';    while (ch=gc,ch>='0'&&ch<='9') s=s*10+ch-'0';    return x*s;}int main(){    n=read(),m=read(),q=read();    for (int i=1;i<=m;i++)    {        int x=read(),y=read();        e[++number].add(x,y);    }    for (int i=1;i<=n;i++)    {        now_cir=0,t[i].init();        for (int i=1;i<=number;i++) e[i].vis=0;        memset(vis,0,sizeof(vis));        t[i].deep[i]=1;        t[i].dfs(i,0);        t[i].dfs1(i),t[i].dfs2(i,i);    }    while (q--)    {        int x=read(),y=read(),k=read();        if (!t[x].deep[y]) puts("-1");        else printf("%d\n",t[x].qry(y,k));    }    return 0;}