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+10;
const int INF=0x3f3f3f3f;
map<int,int> mp;
int32_t main()
{
    int n,m;
    cin>>n>>m;
    for(int i=1;i<=n;i++)
    {
        int a,b; cin>>a>>b;
        for(int j=a;j<=b;j++)
            mp[j]=1;
    }
    int t=0;
    for(int i=1;i<=m;i++)
    {
        if(mp[i]==0)
        {
            t++;
        }
    }cout<<t<<endl;
    for(int i=1;i<=m;i++)
    {
        if(mp[i]==0)
        {
            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+10;
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=0;
    for(int i=0;i<n;i++)
    {
        if(ss[j]==tt[i])
        {
            j++; continue;
        }
        int t=0;
        for(int k=j+1;k<n;k++)
        {

            if(ss[k]==tt[i])
            {
                for(int x=k-1;x>=j;x--)
                {
                        swap(ss[x],ss[x+1]);
                        t=1;
                        pp.push_back(x+1);
                }
                if(t==1) { j++;break;}
            }
        }
        if(t==0) { cout<<-1<<endl; return 0;}
    }
    cout<<pp.size()<<endl;
    for(int i=0;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+10;
const int INF=0x3f3f3f3f;
int a[maxn];
int b[maxn];
int d[maxn];
int32_t main()
{
    int n,m; cin>>n>>m;  int ans1=0; int ans2=0;
    for(int i=1;i<=n;i++)
    {
        cin>>a[i]>>b[i];
        d[i]=a[i]-b[i];
        ans1+=a[i];
        ans2+=b[i];
    }
    if(ans2>m) { cout<<-1<<endl;return 0;}
    //if(ans1<=m) {  cout<<0<<endl;return 0;}
    sort(d+1,d+1+n); int t=0;
    for(int i=n;i>=1;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+10;
const int INF=0x3f3f3f3f;
int32_t main()
{
    int n,k,s; cin>>n>>k>>s;
    if(s>(n-1)*k||s<k) { cout<<"NO"<<endl; return 0;}
    cout<<"YES"<<endl;
    int t=0;
    while(s)
    {
        if(s-n+1>=k-1)
        {
            if(t%2==0) cout<<n<<" ";
            else       cout<<1<<" ";
            t++;
            k--;
            s=s-(n-1); //cout<<s<<endl;
        }
        else
        {
            int d=s-(k-1);// cout<<d<<endl;
            int pos=0;
            if(t%2==0) { cout<<1+d<<" "; pos=1+d; }
            else {        cout<<n-d<<" "; pos=n-d;}
            t++;
            k--;
            int x=0;
            while(k)
            {
                if(x%2==0)
                {
                    if(pos-1>=1) cout<<pos-1<<" ";
                    else cout<<pos+1<<" ";
                }
                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[4]={ {0,1},{0,-1},{1,0},{-1,0} };
bool tf[105][105];
char a[105][105];
int x2[10005];
int x1[10005];
int x3[10005];
int32_t main()
{
    int n,m; cin>>n>>m; getchar();
    for(int i=0;i<n;i++)
       gets(a[i]);
        int t=0;
    for(int i=0;i<n;i++)
    {
        for(int j=0;j<m;j++)
        {
            if(a[i][j]=='*')
            {
                //cout<<i<<"   "<<j<<endl;
                 int z=1;
                 while(1)
                 {
                     int x=0;
                     for(int k=0;k<4;k++)
                     {
                         int x1=i+z*pos[k].first;
                         int y1=j+z*pos[k].second;
                         if(x1<0||x1>=n||y1<0||y1>=m)
                         {
                             continue;
                         }
                         if(a[x1][y1]=='*')
                         {
                              x++;
                         }

                     }
                     //cout<<x<<endl;
                     if(x==4)
                     {
                        tf[i][j]=1;
                       for(int k=0;k<4;k++)
                       {
                         int x1=i+z*pos[k].first;
                         int y1=j+z*pos[k].second;
                         if(x1<0||x1>=n||y1<0||y1>=m)
                         {
                              continue;
                         }
                         if(a[x1][y1]=='*')
                         {
                             tf[x1][y1]=1;
                         }

                       }
                     }
                     else break;
                     z++;
                 }
                 if(z==1) continue;
                 else
                 {
                     x1[t]=i;
                     x2[t]=j;
                     x3[t]=z-1; t++;
                 }
            }
        }
    } // cout<<t<<endl;
    int k1=0;
    int k2=0;
    for(int i=0;i<n;i++)
    {
        for(int j=0;j<m;j++)
        {
            if(a[i][j]=='*') k1++;
            if(tf[i][j]==1) k2++;
        }
    }
    if(k1!=k2) cout<<-1<<endl;
    else
    {
        cout<<t<<endl;
        for(int i=0;i<t;i++)
        {
            cout<<x1[i]+1<<" "<<x2[i]+1<<" "<<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+5;
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 = 1; i <= n; i++)
    for (j = 1; j <= m; j++) cin >> a[i][j];
    for (i = 1; i <= n; i++) for (j = 1; j <= m; j++)
    if (a[i][j] == '*') l[i][j] = l[i][j-1] + 1; else l[i][j] = 0;

    for (i = 1; i <= n; i++) for (j = m; j >= 1; j--)
    if (a[i][j] == '*') r[i][j] = r[i][j+1] + 1; else r[i][j] = 0;

    for (j = 1; j <= m; j++) for (i = 1; i <= n; i++)
    if (a[i][j] == '*') u[i][j] = u[i-1][j] + 1; else u[i][j] = 0;

    for (j = m; j >= 1; j--) for (i = n; i >= 1; i--)
    if (a[i][j] == '*') d[i][j] = d[i+1][j] + 1; else d[i][j] = 0;
    
    int t=1;
    for (i = 1; i <= n; i++)
    for (j = 1; j <= m; j++) if (a[i][j] == '*') {
        s = min(min(r[i][j+1],l[i][j-1]),min(u[i-1][j],d[i+1][j]));
        if (s > 0) {
            ans1[t]=i; ans2[t]=j; ans3[t]=s;  t++;

            for (y = j-s; y <= j+s; y++) f[i][y] = 1;
            for (y = i-s; y <= i+s; y++) f[y][j] = 1;
        }
    }
    for (i = 1; i <= n; i++) for (j = 1; j <= m; j++)
    if (a[i][j] == '*' && f[i][j] == 0) {
        cout << -1 << endl;
        return 0;
    }
    cout<<t-1<<endl;
    for(int i=1;i<=t-1;i++)
        cout<<ans1[i]<<" "<<ans2[i]<<" "<<ans3[i]<<endl;
}
E2.cpp