第六章-数据结构入门

6.1 forest

#include<iostream>
#include<stdio.h>
#include<string>
#include<algorithm>
using namespace std;  

struct person
{
    int number;
    int streng;//体力
    int seat;//站位
}per[10086];  

bool compare(const person &p1,const person &p2)
{
    return p1.seat < p2.seat;
}  

int main()
{  

    int i,j;
    int n;
    cin >> n;
    for(i = 1; i <= n; i++)
    {
        cin >> per[i].streng;
        per[i].seat = i;
        per[i].number = i;
    }  

    int head = 1;
    int tail = n;
    for(i = 1; i <= n; i++)
    {
        if(per[i].streng > per[head].streng)
        {
            per[i].seat = per[head].seat - 1;
            head = i;
        }
        else
        {
            per[i].seat = per[tail].seat + 1;
            tail = i;
        }
    }  

    sort(per+1,per+n+1,compare);  

    for(i = 1; i <= n; i++)
        cout << per[i].number <<" ";  

    return 0;
}   

6.2 queue1

#include<iostream>
#include<stdio.h>
#include<string>
#include<algorithm>
using namespace std;  

int arr[10086];  

int main()
{  

    int i,j;
    int n;
    cin >> n;  

    int head= 0;
    int len = 0;
    string s;
    int num;
    for(i = 0; i < n; i++)
    {
        cin >> s;
        if(s == "INSERT")
        {
            cin >> num;
            arr[len++] = num;
        }  

        else
        {
            if(head > len-1)
                cout << "ERROR" <<endl;
            else
            {
                cout << arr[head] <<endl;
                head++;
            }
        }
    }  

    return 0;
}  

6.3 queue2

#include<iostream>
#include<stdio.h>
#include<string>
#include<algorithm>
using namespace std;  

struct person
{
    bool in;
    int left;
    int right;
}per[100086];  

int main()
{
    int i,j;
    int n;
    cin >> n;  

    per[0].right = 1;
    per[0].in = false;  

    per[1].left = 0;
    per[1].right = n+1;
    per[1].in = true;  

    per[n+1].in = false;  

    int num,op;
    for(i = 2; i <= n; i++)
    {
        cin >> num >> op;   

        if(op == 0)
        {
            per[per[num].left].right = i;
            per[i].left = per[num].left;
            per[num].left = i;
            per[i].right = num;
            per[i].in = true;
        }
        else
        {
            per[per[num].right].left = i;
            per[i].right = per[num].right;
            per[num].right = i;
            per[i].left = num;
            per[i].in = true;
        }
    }  

    int m;
    cin >> m;
    for(i = 0; i < m; i++)
    {
        cin >> num;
        per[num].in = false;
    }  

    i = 0;
    while(per[i].right < n+1)
    {
        if(per[per[i].right].in)
            cout << per[i].right << " ";
        i = per[i].right;
    }  

    return 0;
}  

6.4 power

#include<iostream>
#include<cstdio>
#include<string>
#include<algorithm>
#include<stack>
using namespace std;  

stack<int> s;
int main()
{  

    int i,j;
    int n,size;
    cin >> n >> size;
    int now= 0;  

    int num;
    for(i = 0; i < n; i++)
    {
        cin >> num;  

        while(now < num)
            s.push(++now);  

        if(s.size() > size)
            break;  

        if(s.top() != num)
            break;  

        else
            s.pop();
    }  

    if(i < n)
        cout <<"ERROR " << num;
    else
        cout <<"OK";  

    return 0;
}  

6.5 maze1

#include<iostream>
#include<cstdio>
#include<string>
#include<algorithm>
#include<queue>
using namespace std;  

struct maze
{
    int x;
    int y;
}a,b;  

int vis[1010][1010];
int sum[1010][1010];
int step[4][2]={{0,1},{0,-1},{1,0},{-1,0}};
queue<maze> que;  

int main()
{  

    int i,j;  

    int n,m,t;
    cin >> n >> m >> t;   //n行m列 t障碍物
    int sx,sy;   //起始坐标
    int fx,fy;   //终点坐标
    cin >> sx >> sy >> fx >> fy;  

    int tx,ty;
    for(i = 0; i < t; i++)
    {
        cin >> tx >> ty;
        vis[tx][ty] = 1;
    } //障碍物标记1  

    a.x = sx;
    a.y = sy;
    que.push(a);  

    sum[sx][sy] = 1;  

    while(!que.empty())
    {
        a = que.front();
        que.pop();  

        if(a.x == fx && a.y == fy)
        {
            cout << sum[a.x][a.y];
            return 0;
        }  

        else
        {
            for(i = 0; i < 4; i++)
            {
                int X = a.x + step[i][0];
                int Y = a.y + step[i][1];
                //在图内且未访问
                if(!vis[X][Y] && X > 0 && Y > 0 && X <= n && Y <= m)
                {
                    vis[X][Y] = 1;
                    sum[X][Y] = sum[a.x][a.y] + 1;  

                    b.x = X;
                    b.y = Y;
                    que.push(b);
                }
            }
        }  

    }  

    cout <<"-1";
    return 0;
}