Codeforces Round #336 (Div. 2) 题解

原创

mb62d7cd05038ca 2015-12-25 09:45:58 博主文章分类：其他 ©著作权

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A.
暴力

#include<bits/stdc++.h>
using namespace std;
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
typedef long long ll;
typedef unsigned long long ull;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return (a-b+llabs(a-b)/F*F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
int read()
{
    int x=0,f=1; char ch=getchar();
    while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}
    while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}
    return x*f;
} 
int main()
{
//  freopen("A.in","r",stdin);
//  freopen(".out","w",stdout);

    int n=read(),s=read();
    int ans=s;
    For(i,n) {
        int a=read(),b=read();
        ans=max(ans,max(s-a,b)+a);

    } 
    cout<<ans<<endl;
    return 0;
}

B.
注意前后部分重叠的情况

#include<bits/stdc++.h>
using namespace std;
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
typedef long long ll;
typedef unsigned long long ull;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return (a-b+llabs(a-b)/F*F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
int read()
{
    int x=0,f=1; char ch=getchar();
    while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}
    while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}
    return x*f;
} 
string a,b;
int c[200000+10];
int d[300000];
int main()
{
//  freopen("B.in","r",stdin);
//  freopen(".out","w",stdout);

    cin>>a>>b;
    int n=a.length(),m=b.length();
    int t=0;
    Rep(i,n) if (a[i]=='0') c[i]=++t; else c[i]=t;
    Fork(i,n,m) c[i]=c[i-1];
    t=0;
    RepD(i,n+1) if (a[i]=='0') d[i]=++t; else d[i]=t;

    ll ans=0;
    Rep(i,m) {
        int r=min(n-1,i),l=max(0,n-m+i);
        int len=r-l+1;

        int cost=c[r];
        if (l) cost-=c[l-1];

        if (b[i]=='0') ans+=len-cost; else ans+=cost;

    }



    cout<<ans<<endl;
    return 0;
}

#include<bits/stdc++.h>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define ForkD(i,k,n) for(int i=n;i>=k;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=Pre[x];p;p=Next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=Next[p])  
#define Lson (o<<1)
#define Rson ((o<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (100000007)
#define pb push_back
#define mp make_pair 
#define fi first
#define se second
#define vi vector<int> 
#define pi pair<int,int>
#define SI(a) ((a).size())
typedef long long ll;
typedef unsigned long long ull;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return (a-b+llabs(a-b)/F*F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
int read()
{
    int x=0,f=1; char ch=getchar();
    while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}
    while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}
    return x*f;
} 
#define MAXN (100000+10)
#define MAXM (100000+10)

int Edge[MAXM],Next[MAXM],Pre[MAXN],weight[MAXM],size;      
void addedge(int u,int v,int w)        
{        
    Edge[++size]=v;        
    weight[size]=w;        
    Next[size]=Pre[u];        
    Pre[u]=size;        
}       
int a[MAXN],b[MAXN]; 
set<pi> S;
set<pi>::iterator it;
int c[MAXN];
int dfs(int x) {
    int dis=1;
    Forp(x) {
        dis=max(dis,1+dfs(Edge[p]));
    }
    return dis;
}
int main()
{
//  freopen("C.in","r",stdin);
//  freopen(".out","w",stdout);

    int n=read();
    For(i,n) {
        cin>>a[i]>>b[i];
        S.insert(mp(a[i],i));
    }

    For(i,n) {
        int t=a[i]-b[i]-1;
        it=S.upper_bound(mp(t,INF));
        if (it==S.begin()) c[i]=0;
        else {
            it--;
            int p = it->se;
            c[i]=p; 
        }
        addedge(c[i],i,1);
    }
    cout<<n+1-dfs(0);
    return 0;
}

D.
dp

#include<bits/stdc++.h>
using namespace std;
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
typedef long long ll;
typedef unsigned long long ull;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return (a-b+llabs(a-b)/F*F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
int read()
{
    int x=0,f=1; char ch=getchar();
    while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}
    while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}
    return x*f;
} 
#define
int f[MAXN][MAXN]={0};
int a[MAXN];
int main()
{
//  freopen("D.in","r",stdin);
//  freopen(".out","w",stdout);

    int n=read();
    For(i,n) cin>>a[i];

    For(i,n) f[i][i]=1;

    Fork(len,2,n) {
        For(i,n-len+1) {
            int j=i+len-1; 
            f[i][j]=n;
            Fork(k,i,j-1) {
                f[i][j]=min(f[i][j],f[i][k]+f[k+1][j]);
            }
            if (a[i]==a[j]&&j-i==1) f[i][j]=min(f[i][j],1);
            else if  (a[i]==a[j]) f[i][j]=min(f[i+1][j-1],f[i][j]);
        }
    }
    cout<<f[1][n]<<endl;

    return 0;
}

E.
题意：有2条只有东南西北的路径，2人分别站在2条路线的起点，如果其中一个人向某个方向走了一步，那么另一个人也会往这个方向走一步（如果有路），问这2人可能最后都走到终点吗？
题解：可能当且仅当｛存在后缀，使第一条路径的后缀的翻转和第二条路径的后缀完全相同｝，然后kmp

#include<bits/stdc++.h>
using namespace std;
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
typedef long long ll;
typedef unsigned long long ull;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return (a-b+llabs(a-b)/F*F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
int read()
{
    int x=0,f=1; char ch=getchar();
    while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}
    while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}
    return x*f;
} 
#define
char s[MAXN];
int dir[4][2]={{0,-1},{0,1},{-1,0},{1,0}};
int h[10000];
int n;
int q[MAXN];
int x[MAXN],y[MAXN];
void calc(int i,int d,int &i2,char *s ) {
    if (i>0 && d==(1^h[s[i]])) i2=i-1;
    else if (i<n && d==h[s[i+1]]) i2=i+1;
    else i2=i;
}
class KMP
{
public:
    int f2[MAXN]; //字符串从0开始，但是f[i]表示匹配第i个字符，前面留一个 f[0]--a-->f[1]--...这样的 
    char T2[MAXN],P2[MAXN]; //T is long,P is model str
    void mem(){MEM(f2) MEM(T2) MEM(P2)  }
    int getFail(char *P=0,int* f=0)
    {
        if (P==0) P=P2;if (f==0) f=f2;
        int m=strlen(P);
        f[0]=f[1]=0; 
        For(i,m-1)
        {
            int j=f[i];
            while(j&&P[i]!=P[j]) j=f[j];
            f[i+1]= P[i] == P[j] ? j+1 : 0;
        }
    }
    int find(char* T=0,char* P=0,int* f=0)
    {
        if (T==0) T=T2;if (P==0) P=P2;if (f==0) f=f2;
        int n=strlen(T),m=strlen(P);
        getFail(P,f);
        int j=0;
        Rep(i,n)
        {
            while(j&&T[i]!=P[j]) j=f[j];
            if (T[i]==P[j]) j++;
        //  if (j==m) return i-m+1;
        }
    }
}S;
int main()
{
//  freopen("E.in","r",stdin);
//  freopen(".out","w",stdout);

    cin >>n; n--;
    scanf("%s%s",s+1,s+1+n);

    h['N']='S'; h['S']='N'; h['E']='W'; h['W']='E'; 

    For(i,n) s[i]=h[s[i]];
    reverse(s+1,s+1+n);

    S.getFail(s+1);
    if (!S.f2[2*n]) cout<<"YES"<<endl;
    else cout<<"NO"<<endl;
    return 0;
}