补题

10.8 c

https://vjudge.net/contest/332874#problem/C

#include<iostream >
using namespace std;
const int maxn=2e5+10;
char a[2][maxn];
int len,n;
bool flag;
void dfs(int x,int y,int d)//x,y为坐标点，d=1向上，d=2向下，d=3向右（由于可以旋转，3，4，5，6可以当成一个） 
{
    if(y>=len)
      return;
    if(x==1&&y==len-1&&d==3)//到达"终点" 
    {
         flag=1;
         return;
    } 
    if(x==0)//在第一行，向右或向下 
    {
        if(d==3)
        {
            if(a[x][y+1]=='1'||a[x][y+1]=='2')
              dfs(x,y+1,3);
            else
              dfs(x,y+1,2);
        } 
        else if(d==2)
        {
            if(a[x+1][y]!='1'&&a[x+1][y]!='2')
               dfs(x+1,y,3);
        }
    }
    else if(x==1)//在第二行，向右或向上
    {
        if(d==3)
        {
            if(a[x][y+1]!='1'&&a[x][y+1]!='2')
              dfs(x,y+1,1);
            else
              dfs(x,y+1,3);
        }
        else if(d==1)
        {
            if(a[x-1][y]!='1'&&a[x-1][y]!='2')
               dfs(x-1,y,3);
        }
    } 
    
}
int main()
{
    cin>>n;
    while(n--)
    {
        flag=0;
        cin>>len;
        cin>>a[0];
        cin>>a[1];
        if(a[0][0]=='1'||a[0][0]=='2')
           dfs(0,0,3);
        else
           dfs(0,0,2);
        if(flag)
           cout<<"YES"<<endl;
        else 
           cout<<"NO"<<endl;
    }
    return 0;
}

 10.9

https://vjudge.net/contest/333223#problem/B

这道题考查的是最短路，floyd算法的应用

由于对于任意一条至少包含两条边的路径i--j，一定存在一个中间点，使得i--k+k--j=i--j，当然，对于不同的点k，i--k和k--j的长度值可能不同，

所以需要取一个最小值才是最短路径

e[i][j]=min(e[i][j],max(e[i][k],e[k][j]));
max是取从i--j中，以k为节点，能承受的最大噪音分贝值，然后从这些最大中再选出最小的那个路径，就是最短路径

#include<iostream>
using namespace std;
int e[1010][1010];
int inf=0x3f3f3f3f;
int c,s,q;
int main()
{
    int c1,c2,d,x=1,a,b;
    while(cin>>c>>s>>q)
    { 
        if(c==0&&s==0&&q==0)break;
        if(x!=1)cout<<endl; 
        //初始化 
        for(int i=0;i<=c;i++)
          for(int j=0;j<=c;j++)
            if(i==j)e[i][j]=0;
            else  e[i][j]=inf;
        
        for(int i=1;i<=s;i++)
        {
            cin>>c1>>c2>>d;
            e[c1][c2]=d;
            e[c2][c1]=d;
        }
        cout<<"Case #"<<x<<endl;
        
        for(int k=0;k<=c;k++)
          for(int i=0;i<=c;i++)
            for(int j=0;j<=c;j++)
                e[i][j]=min(e[i][j],max(e[i][k],e[k][j]));
                
        for(int i=0;i<q;i++)
        {
            cin>>a>>b;
            if(e[a][b]==inf)
               cout<<"no path"<<endl;
            else
               cout<<e[a][b]<<endl;
        }
        x++;
    } 
    return 0;
}

 10.11

A - Slim Span

 UVA - 1395 

给你一个图，点以及权值，找出查找的两点间最小差值

kruskal算法，

#include<iostream>
#include<algorithm>
using namespace std;
#define inf 0x3f3f3f3f
int n,m,flag,f[5500];
struct node
{
    int u,v,w;
}s[5500];
bool cmp(node x,node y)
{
    return x.w<y.w;
}
int getf(int v)//并查集找祖先 
{
    if(f[v]==v)
      return v;
    else
    {
        f[v]=getf(f[v]);
        return f[v];
    }
}
void kruskal()
{
    for(int i=0;i<m;i++)
    {
        int cnt=0;
        for(int j=0;j<=n;j++)
          f[j]=j;
        for(int j=i;j<m;j++)//并查集合并 
        {
            int t1=getf(s[j].u);
            int t2=getf(s[j].v);
            if(t1!=t2)//两个点是否在一个集合里 
            {
                cnt++;
                f[t2]=t1;
                 if(cnt==n-1)
                 {
                     flag=min(flag,s[j].w-s[i].w);
                     break;
                 }
            } 
        }
    }
} 
int main()
{
    while(cin>>n>>m&&(n||m))
    {
        for(int i=0;i<m;i++)
           cin>>s[i].u>>s[i].v>>s[i].w;
        sort(s,s+m,cmp);
        flag=inf;
        kruskal();
        if(flag==inf)
           cout<<"-1"<<endl;
        else
          cout<<flag<<endl;
        
    }
    return 0;
}

B - Resort

 CodeForces - 350B 

Valera's finally decided to go on holiday! He packed up and headed for a ski resort.

Valera's fancied a ski trip but he soon realized that he could get lost in this new place. Somebody gave him a useful hint: the resort has n objects (we will consider the objects indexed in some way by integers from 1 to n), each object is either a hotel or a mountain.

Valera has also found out that the ski resort had multiple ski tracks. Specifically, for each object v, the resort has at most one object u, such that there is a ski track built from object u to object v. We also know that no hotel has got a ski track leading from the hotel to some object.

Valera is afraid of getting lost on the resort. So he wants you to come up with a path he would walk along. The path must consist of objects v₁, v₂, ..., v_k (k ≥ 1) and meet the following conditions:

1. Objects with numbers v₁, v₂, ..., v_k - 1 are mountains and the object with number v_k is the hotel.
2. For any integer i (1 ≤ i < k), there is exactly one ski track leading from object v_i. This track goes to object v_i + 1.
3. The path contains as many objects as possible (k is maximal).

Help Valera. Find such path that meets all the criteria of our hero!

Input

The first line contains integer n (1 ≤ n ≤ 10⁵) — the number of objects.

The second line contains n space-separated integers type₁, type₂, ..., type_n — the types of the objects. If type_i equals zero, then the i-th object is the mountain. If type_i equals one, then the i-th object is the hotel. It is guaranteed that at least one object is a hotel.

The third line of the input contains n space-separated integers a₁, a₂, ..., a_n (0 ≤ a_i ≤ n) — the description of the ski tracks. If number a_i equals zero, then there is no such object v, that has a ski track built from v to i. If number a_i doesn't equal zero, that means that there is a track built from object a_i to object i.

Output

In the first line print k — the maximum possible path length for Valera. In the second line print k integers v₁, v₂, ..., v_k — the path. If there are multiple solutions, you can print any of them.

Examples

Input

5
0 0 0 0 1
0 1 2 3 4

Output

5
1 2 3 4 5

Input

5
0 0 1 0 1
0 1 2 2 4

Output

2
4 5

Input

4
1 0 0 0
2 3 4 2

Output

1
1

题意：n为所给数长度
第一排数0是表示山，1表示酒店
第二排数是表示可走的滑道
如样例2
5

序号  1 2 3 4 5
      0 0 1 0 1
二排   0 1 2 2 4
0--1
1--2
2--3
2--4
4--5（只有3和5是酒店，2既可以到3又可以到4，不满足题目条件，所以选择包含尽可能多的节点）
答案为4--5
两个代码，思路异曲同工

#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn=1e5+10;
int n;
int a[maxn],b[maxn],c[maxn],p[maxn];
bool vis[maxn];
int main()
{
    cin>>n;
    memset(c,0,sizeof(c));
    memset(vis,false,sizeof(vis));
    for(int i=0;i<maxn;i++)
       p[i]=i;
       
    for(int i=1;i<=n;i++)
       cin>>a[i];
    for(int i=1;i<=n;i++)
    {
       cin>>b[i];
       c[b[i]]++;//记录出度
    }
    for(int i=1;i<=n;i++)
    {
         if(c[b[i]]<=1)
         {
             p[b[i]]=i;//记录满足的节点 
        }
    } 
    int dx=0,dc=1;
    for(int i=1;i<=n;i++)
    {
        if(vis[i]) continue;
        if(c[i]<=1)
        {
            int t=i,z=1;//z控制可执行的点，从一个跳到可执行的点
            while(p[t]!=t&& !vis[i])
            {
                if(c[t]>1)break;
                z++,t=p[t];
                vis[t]=true;
            }
            if(a[t]==1&&z>dx)
               dx=max(dx,z),dc=i;
        }
        vis[i]=true;
    } 
    cout<<dx<<endl;
    while(p[dc]!=dc)
    {
        cout<<dc<<" ";
        dc=p[dc];
    }
    cout<<dc<<endl;
      return 0;
}

将可执行的点存数组，倒序查找

#include <cstdio>
#include <vector>
#include <algorithm>
using namespace std;
const int maxn=100000;
int n,a[maxn],b[maxn],c[maxn];
int main()
{
    scanf("%d",&n);
    for (int i=0;i<n;i++)
        scanf("%d",&a[i]);
    for (int i=0;i<n;i++){
        scanf("%d",&b[i]);
        b[i]--;
    }
    for (int i=0;i<n;i++)
        c[i]=0;
    for (int i=0;i<n;i++)
        if (b[i]!=-1)
            c[b[i]]++;//可走的路 
    int ret=0,reti=-1;
    for (int i=0;i<n;i++)
        if (a[i]==1){
            int x=1;
            int j=i;
            while (b[j]!=-1 && c[b[j]]==1){//查到酒店跳转查看前一个 
                j=b[j];
                x++;//记录次数 
            }
            if (x>ret){//最长 
                ret=x;
                reti=i;
            }
        }
    printf("%d
",ret);
    vector<int> output;
    for (int i=0;i<ret;i++){
        output.push_back(reti);//倒序放入 
        reti=b[reti];
    }
    for (int i=0;i<ret;i++){
        if (i)
            printf(" ");
        printf("%d",output[ret-1-i]+1);
    }
    puts("");
    return 0;
}