2018.8.16日记

https://wenku.baidu.com/view/bc213eff700abb68a982fb22.html?from=search
https://www.cnblogs.com/AWCXV/p/9486282.html

#include 
#include
using namespace std;
struct know :map<string,int>
{
    int operator ()(const string &tmp)
    {
        if(!count(tmp))
        {
            insert(make_pair(tmp,size()));
        }
        return at(tmp);
    }
}id;
struct node
{
    node(int n):v(n){}
    struct Vex
    {
        vector<int>a;
        int s[2][521];
    };
    struct road
    {
       int from,to,op;
    };
    int fa;
    vectorv;
    vectore;
    void add(const road &st)
    {
       v[st.from].a.push_back((int)e.size());
       //这里用到了类似前向星的东西
       e.push_back(st);

    }
    void dfs(int u,int is)
    {
       v[fa].s[is][u] = 1;
       int to ;
       for(int y:v[u].a)
       {
           //
          if(to = e[y].to ,is = is&&e[y].op,!v[fa].s[is][to])
          {
              dfs(to,is);
          }
       }
    }
}exm(521);
int main()
{
    int n,m;
    cin>>n>>m;
    string t1,t2,t3;
    node::road tmp;
    for(int i=1; i<=n; i++)
    {
        cin>>t1>>t2>>t3;
        tmp.from = id(t1);
        tmp.to  =id(t3);
        tmp.op  = (t2=="is-a");
        exm.add(tmp);
    }
   for(int i=0;i<(int )id.size();i++)
   {
      exm.fa =1;
      exm.dfs(i,1);
   }
    for(int i=1; i<=m; i++)
    {
       cin>>t1>>t2>>t3;
       int j =  t3=="is-a"?1:0;
       int n = exm.v[id(t1)].s[j][id(t3)];
       if(!n)printf("Query %d: true\n",i);
       else printf("Query %d: false\n",i);
    }

    return 0;
}

#include 
using namespace std;

const int N = 1E5;

int n,m,a[N+10],_size[N+10];
int dl[N+10],h,t,step[N+10],dp[N+10][20];

int ma(int i,int j){
    if (a[i]>=a[j]) return i;
    return j;
}

int get_ma(int l,int r){
    if (l>r) return 0;
    int len  = log2(r-l+1);
    return ma(dp[l][len],dp[r-(1<1][len]);
}

int get_first_bigger(int q,int L){
    int l = L,r = n,temp = -1;
    while (l <= r){
        int mid = (l+r)>>1;
        if (a[get_ma(L,mid)]>q){
            temp = mid;
            r = mid-1;
        }else l = mid + 1;
    }
    return temp;
}
void debug(int a[],int n)
{
    cout<<"数组a为"<int top = 1;
   for(int i=0;i<=n;i++)
   {
       if(top)top =0;
       else printf(" ");
       printf("%d",a[i]);
   }
   printf("\n");
}
int main(){
    //freopen("D:\\rush.txt","r",stdin);
    //freopen("D:\\rush_out.txt","w",stdout);
    int T;
    scanf("%d",&T);
    while (T--){
        scanf("%d%d",&n,&m);
        for (int i = 1;i <= n;i++) scanf("%d",&a[i]);
        h = 1,t = 0;
        for (int i = n;i >= 1;i--){
            while (h<=t && a[i]>=a[dl[t]]) t--;
            dl[++t] = i;
            _size[i] = t-h+1;
        }
        step[1] = 1;
        int cur = 1;
        for (int i = 2;i <= n;i++)
            if (a[i]>a[cur]){
                step[i] = step[cur]+1;
                cur = i;
            }
        for (int i = 0;i <= n;i++) dp[i][0] = i;
        for (int l = 1;l <= 17;l++){
            for (int i = 1;i <= n;i++){
                if ((i+(1<1)>n) break;
                dp[i][l] = ma(dp[i][l-1],dp[i+(1<<(l-1))][l-1]);
            }
        }
      //注意这里讲的上升序列是指题目中讲的从左到右
      //遇到一个比当前值大的就进行更新。

      //step数组含义 是从1开始，到
      //当前数为止上升序列的长度，注意
      //如果step数组不为0，那么a[i]一定是确定好的上升序列中的一个
      //例如 1 2 3 4 4即使是相等，a[5]也是0；

      //_size数组含义是单调栈的大小，代表从当前标号到最后上升序列的长度


      //dl数组含义 是单调栈内元素的位置标号
      debug(step,n);
      debug(_size,n);
      debug(dl,n);
        for (int i = 1;i <= m;i++){
            int p,q;
            scanf("%d%d",&p,&q);
            int idx = get_ma(1,p-1);
            int key = q,ans = 0;
            cout<" "<" idx"<if (q<=a[idx]){
                key = a[idx];
                ans = -1;
            }
            int idx2 = get_first_bigger(key,p+1);
            cout<<"idx= "<" idx2= "<" "<<" ;"<" "<if (idx2==-1){
                ans += step[idx]+1;


            }else{
                ans += step[idx]+1+_size[idx2];
                cout<<"_size[idx2]= "<<_size[idx2]<printf("%d\n\n\n",ans);
        }
    }
    return 0;
}

#include
using namespace std;
i
int a[maxn];
int get_ma(int )
{

}
int main()
{

    return 0;
}

//5
//5 1
//1 5 2 4 3
//数组a为
//0 1 2 0 0 0
//数组a为
//0 2 1 2 1 1
//数组a为
//0 2 1 0 0 0
//3 4
//2 5 idx
//idx= 2 idx2= -1  ;2
//2
//
//
//5 1
//1 2 3  4 4
//数组a为
//0 1 2 3 4 0
//数组a为
//0 4 3 2 1 1
//数组a为
//0 4 3 2 1 0
//3 8
//2 2 idx
//idx= 2 idx2= -1  ;2
//3
//
//
//5 10
//1 2  3 4 4
//数组a为
//0 1 2 3 4 0
//数组a为
//0 4 3 2 1 1
//数组a为
//0 4 3 2 1 0
//3 2
//2 2 idx
//idx= 2 idx2= 4  ;2
//_size[idx2]= 1
//3
//