Codeforces Div3 #501 A-E(2) F以后补

感觉自己有点强迫症  不都写出来就找理由不写题解

http://codeforces.com/contest/1015   题目链接

A. Points in Segments

题目意思  n个线段 去覆盖1-m 中的点 问你没有覆盖的点的个数和位置

这个数据很小，可以直接暴力查找

思考：如果n<1e6, m<=1e8 呢？

#include<bits/stdc++.h>
#define int long long
#define MAX(a,b,c) max(a,max(b,c))
#define MIN(a,b,c) min(a,min(b,c))
#define pb push_back
#define fi first
#define se second
typedef long long ll;
typedef long long LL;
typedef unsigned long long ull;
typedef unsigned long long uLL;
using namespace std;
const int maxn=1e5+;
const int INF=0x3f3f3f3f;
map<int,int> mp;
int32_t main()
{
    int n,m;
    cin>>n>>m;
    for(int i=;i<=n;i++)
    {
        int a,b; cin>>a>>b;
        for(int j=a;j<=b;j++)
            mp[j]=;
    }
    int t=;
    for(int i=;i<=m;i++)
    {
        if(mp[i]==)
        {
            t++;
        }
    }cout<<t<<endl;
    for(int i=;i<=m;i++)
    {
        if(mp[i]==)
        {
            cout<<i<<" ";
        }
    }
}

A.cpp

如果 m<=1e8的 话，可以标记 线段的起点 重点后一位     左右搜一遍

如1-5   a[1]=1; a[6]=-1;

B. Obtaining the String

一个n,两个字符串(ss,tt)，左右移动前一个字符串 使两个字符串相等  求最小移动次数；

对比 ss ,tt 当ss[i] !=tt[i]; 在ss[i]后面找和tt[i]一样的， 移动到ss[i]；没有就输出-1；

#include<bits/stdc++.h>
#define int long long
#define MAX(a,b,c) max(a,max(b,c))
#define MIN(a,b,c) min(a,min(b,c))
#define pb push_back
#define fi first
#define se second
typedef long long ll;
typedef long long LL;
typedef unsigned long long ull;
typedef unsigned long long uLL;
using namespace std;
const int maxn=1e5+;
const int INF=0x3f3f3f3f;
map<int,int> mp;
vector<int> pp;
int32_t main()
{
    int n; cin>>n;
    string ss,tt; cin>>ss; cin>>tt;
    //if(ss==tt) { cout<<-1<<endl; return 0;}
    int j=;
    for(int i=;i<n;i++)
    {
        if(ss[j]==tt[i])
        {
            j++; continue;
        }
        int t=;
        for(int k=j+;k<n;k++)
        {
 
            if(ss[k]==tt[i])
            {
                for(int x=k-;x>=j;x--)
                {
                        swap(ss[x],ss[x+]);
                        t=;
                        pp.push_back(x+);
                }
                if(t==) { j++;break;}
            }
        }
        if(t==) { cout<<-<<endl; return ;}
    }
    cout<<pp.size()<<endl;
    for(int i=;i<pp.size();i++)
    {
        cout<<pp[i]<<" ";
    }
}

B.cpp

C. Songs Compression

n 组数据 一开始的歌曲数据大小  压缩后的数据大小， 要是数据小于等于m;

直接计算一开始的大小，每次减去 差值最大的；看多少次后数据小于等于m; 压缩不到m就输出-1；

#include<bits/stdc++.h>
#define int long long
#define MAX(a,b,c) max(a,max(b,c))
#define MIN(a,b,c) min(a,min(b,c))
#define pb push_back
#define fi first
#define se second
typedef long long ll;
typedef long long LL;
typedef unsigned long long ull;
typedef unsigned long long uLL;
using namespace std;
const int maxn=1e5+;
const int INF=0x3f3f3f3f;
int a[maxn];
int b[maxn];
int d[maxn];
int32_t main()
{
    int n,m; cin>>n>>m;  int ans1=; int ans2=;
    for(int i=;i<=n;i++)
    {
        cin>>a[i]>>b[i];
        d[i]=a[i]-b[i];
        ans1+=a[i];
        ans2+=b[i];
    }
    if(ans2>m) { cout<<-<<endl;return ;}
    //if(ans1<=m) {  cout<<0<<endl;return 0;}
    sort(d+,d++n); int t=;
    for(int i=n;i>=;i--)
    {
        if(ans1<=m)  break;
        ans1=ans1-d[i];
        t++;
    }
    cout<<t<<endl;
}

C.cpp

D. Walking Between Houses

在1-n 中走 k次（随你走到哪，不能不走） 使走的路程为 s;

在保证每次都至少走一步的情况下 先走最大的 1 - n- 1 - n -1 -n....；

最后再每次走一步；

10 9 45 来说 每次走的距离为 9 9 9 9 5 1 1 1 1；

#include<bits/stdc++.h>
#define int long long
#define MAX(a,b,c) max(a,max(b,c))
#define MIN(a,b,c) min(a,min(b,c))
#define pb push_back
#define fi first
#define se second
typedef long long ll;
typedef long long LL;
typedef unsigned long long ull;
typedef unsigned long long uLL;
using namespace std;
const int maxn=1e5+;
const int INF=0x3f3f3f3f;
int32_t main()
{
    int n,k,s; cin>>n>>k>>s;
    if(s>(n-)*k||s<k) { cout<<"NO"<<endl; return ;}
    cout<<"YES"<<endl;
    int t=;
    while(s)
    {
        if(s-n+>=k-)
        {
            if(t%==) cout<<n<<" ";
            else       cout<<<<" ";
            t++;
            k--;
            s=s-(n-); //cout<<s<<endl;
        }
        else
        {
            int d=s-(k-);// cout<<d<<endl;
            int pos=;
            if(t%==) { cout<<+d<<" "; pos=+d; }
            else {        cout<<n-d<<" "; pos=n-d;}
            t++;
            k--;
            int x=;
            while(k)
            {
                if(x%==)
                {
                    if(pos->=) cout<<pos-<<" ";
                    else cout<<pos+<<" ";
                }
                else       cout<<pos<<" ";
                k--;
                x++;
            }
            break;
        }
    }
}

D.cpp

E1. Stars Drawing (Easy Edition)

可以直接暴力搜  找到一个点  直接往上下左右搜  看照射的距离  大于1就全部标记

#include<bits/stdc++.h>
#define int long long
#define MAX(a,b,c) max(a,max(b,c))
#define MIN(a,b,c) min(a,min(b,c))
#define pb push_back
#define fi first
#define se second
typedef long long ll;
typedef long long LL;
typedef unsigned long long ull;
typedef unsigned long long uLL;
using namespace std;
pair<int,int> pos[]={ {,},{,-},{,},{-,} };
bool tf[][];
char a[][];
int x2[];
int x1[];
int x3[];
int32_t main()
{
    int n,m; cin>>n>>m; getchar();
    for(int i=;i<n;i++)
       gets(a[i]);
        int t=;
    for(int i=;i<n;i++)
    {
        for(int j=;j<m;j++)
        {
            if(a[i][j]=='*')
            {
                //cout<<i<<"   "<<j<<endl;
                 int z=;
                 while()
                 {
                     int x=;
                     for(int k=;k<;k++)
                     {
                         int x1=i+z*pos[k].first;
                         int y1=j+z*pos[k].second;
                         if(x1<||x1>=n||y1<||y1>=m)
                         {
                             continue;
                         }
                         if(a[x1][y1]=='*')
                         {
                              x++;
                         }
 
                     }
                     //cout<<x<<endl;
                     if(x==)
                     {
                        tf[i][j]=;
                       for(int k=;k<;k++)
                       {
                         int x1=i+z*pos[k].first;
                         int y1=j+z*pos[k].second;
                         if(x1<||x1>=n||y1<||y1>=m)
                         {
                              continue;
                         }
                         if(a[x1][y1]=='*')
                         {
                             tf[x1][y1]=;
                         }
 
                       }
                     }
                     else break;
                     z++;
                 }
                 if(z==) continue;
                 else
                 {
                     x1[t]=i;
                     x2[t]=j;
                     x3[t]=z-; t++;
                 }
            }
        }
    } // cout<<t<<endl;
    int k1=;
    int k2=;
    for(int i=;i<n;i++)
    {
        for(int j=;j<m;j++)
        {
            if(a[i][j]=='*') k1++;
            if(tf[i][j]==) k2++;
        }
    }
    if(k1!=k2) cout<<-<<endl;
    else
    {
        cout<<t<<endl;
        for(int i=;i<t;i++)
        {
            cout<<x1[i]+<<" "<<x2[i]+<<" "<<x3[i]<<endl;
        }
    } //cout<<k1<<"  "<<k2<<endl;
}

E1.cpp

E2. Stars Drawing (Hard Edition)

先预处理 每个点 上下左右 可以照射的距离   再标记

预处理有些技巧 ，不能暴力搜  要找相邻两个点的关系

由于预处理 和 标记 分开  不会超时;

#include<bits/stdc++.h>
using namespace std;
const int maxn=1e3+;
int l[maxn][maxn];
int r[maxn][maxn];
int u[maxn][maxn];
int d[maxn][maxn];
int ans1[maxn*maxn];
int ans2[maxn*maxn];
int ans3[maxn*maxn];
char a[maxn][maxn];
bool f[maxn][maxn];
int main()
{
    int n,m;
    cin >> n >> m;
    int i,j,s,y;
    for (i = ; i <= n; i++)
    for (j = ; j <= m; j++) cin >> a[i][j];
    for (i = ; i <= n; i++) for (j = ; j <= m; j++)
    if (a[i][j] == '*') l[i][j] = l[i][j-] + ; else l[i][j] = ;
 
    for (i = ; i <= n; i++) for (j = m; j >= ; j--)
    if (a[i][j] == '*') r[i][j] = r[i][j+] + ; else r[i][j] = ;
 
    for (j = ; j <= m; j++) for (i = ; i <= n; i++)
    if (a[i][j] == '*') u[i][j] = u[i-][j] + ; else u[i][j] = ;
 
    for (j = m; j >= ; j--) for (i = n; i >= ; i--)
    if (a[i][j] == '*') d[i][j] = d[i+][j] + ; else d[i][j] = ;
 
    int t=;
    for (i = ; i <= n; i++)
    for (j = ; j <= m; j++) if (a[i][j] == '*') {
        s = min(min(r[i][j+],l[i][j-]),min(u[i-][j],d[i+][j]));
        if (s > ) {
            ans1[t]=i; ans2[t]=j; ans3[t]=s;  t++;
 
            for (y = j-s; y <= j+s; y++) f[i][y] = ;
            for (y = i-s; y <= i+s; y++) f[y][j] = ;
        }
    }
    for (i = ; i <= n; i++) for (j = ; j <= m; j++)
    if (a[i][j] == '*' && f[i][j] == ) {
        cout << - << endl;
        return ;
    }
    cout<<t-<<endl;
    for(int i=;i<=t-;i++)
        cout<<ans1[i]<<" "<<ans2[i]<<" "<<ans3[i]<<endl;
}

E2.cpp