csp201604

折点计数

实现

#include
using namespace std;
const int N = 1010;
int a[N];
int n;
int ans;
int main()
{
    cin>>n;
    for(int i=0;i<n;i++)
    {
        scanf("%d",&a[i]);
    }
    for(int i=1;i<n-1;i++)
    {
        if((a[i]>a[i+1] && a[i]>a[i-1])||(a[i]<a[i-1] && a[i]<a[i+1]))
        ans++;
    }
    cout<<ans;
}
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俄罗斯方块

实现

#include 
#include 
#include 

using namespace std;

const int N = 17;
int g[N][N], st[N][N];
int p[4][4];

bool draw(int x, int y)
{
    memcpy(st, g, sizeof st); // 这里是对原数组的备份  因为需要多次draw  每次需要元数组的信息 所以需要拷贝
    for (int i = 0; i < 4; i ++)
        for (int j = 0; j < 4; j ++)
        {
            if (p[i][j])  // 当前点是格子
            {
                int a = x + i, b = y + j;
                st[a][b] ++;
                if (st[a][b] == 2) return true; // 为2表示重叠
            }
        }
    return false;    
}

int main()
{
    for (int i = 0; i < 15; i ++)
        for (int j = 0; j < 10; j ++)
            cin >> g[i][j];

    for (int i = 0; i < 4; i ++)
        for (int j = 0; j < 4; j ++)
            cin >> p[i][j];

    for (int j = 0; j < 10; j ++) g[15][j] = 1; //为了判断下边界  我们需要将下边届初始化为1 表示有格子

    int c;
    cin >> c;
    c --; // 从0开始做的

    for (int i = 0; ; i ++)
    {
        if (draw(i, c))
        {
            draw(i-1, c);
            break;
        }
    }

    for (int i = 0; i < 15; i ++)
    {
        for (int j = 0; j < 10; j ++)
            cout << st[i][j] << " ";
        cout << endl;    
    }
    return 0;
}

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路径解析

实现

#include
#include
#include
#include
using namespace std;
int n;
vector<string> path;
vector<string> curr;
vector<string> get(string t)
{
	vector<string> tmp;
	for (int i = 0; i < t.size(); i++)
	{
		if (t[i] == '/') continue ;
		int j = i + 1;
		while (t[j] != '/' && j < t.size())
		{
			j++;
		}
		tmp.push_back(t.substr(i, j - i));
		i = j - 1;
		
	}
// 	for(int i=0;i
// 	cout<
// 	cout<
	return tmp;
}
void walk(vector<string>& p,vector<string> cur)
{

        for(auto tt:p)
        {
            if(tt=="..")
            {
                if(cur.size())cur.pop_back();
            }
            else if(tt==".") continue ;
            else cur.push_back(tt);
        }
        if(!cur.size()) 
        {
            cout<<"/"<<endl;
             return ;
        }
        
        for(auto tt:cur)
        {
            cout<<"/";
            cout<<tt;
        }
    
    cout<<"\n";
}
int main()
{
	
	string t;
	cin >> n >> t;
	//cout << n << endl;
	curr = get(t);
	getchar();
	while (n--)
	{
		getline(cin, t);
		if(t=="")
		{
		    for(auto tt:curr)
		    {
		        cout<<"/";
		        cout<<tt;
		    }
		    cout<<endl;
		    continue;
		}
		path = get(t);
		
		vector<string> kong;
		if(t[0]!='/') walk(path,curr);
		else walk(path,kong);
	}
	return 0;
}
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83

游戏

实现

#include

using namespace std;
int n,m,t;
const int N = 110;
bool g[N][N][10010];
int dist[N][N][310];
struct node
{
    int x,y,time;
}nodes[110*110*310];

int bfs()
{
    int dx[4] = {-1,0,1,0},dy[4] = {0,1,0,-1};
    queue<node> q;
    q.push({1,1,0});
    memset(dist,0x3f,sizeof dist);
    while(q.size())
    {
        auto t = q.front();
        q.pop();
        for(int i=0;i<4;i++)
        {
            int xx = t.x+dx[i],yy = t.y+dy[i],tt = t.time+1;
            if(g[xx][yy][tt]) continue;
            if(xx<1 ||xx>n || yy<1 ||yy>m ) continue;
            if(dist[xx][yy][tt]>tt) 
            {
                dist[xx][yy][tt] = tt;
                q.push({xx,yy,tt});
                if(xx==n &&yy==m) return tt;
            }
        }
    }
    return 0;
}
int main()
{
    cin>>n>>m>>t;
    int x,y,t1,t2;
    while(t--)
    {
        cin>>x>>y>>t1>>t2;
        for(int i=t1;i<=t2;i++)
        g[x][y][i] = 1;
    }
    cout<<bfs()<<endl;
}
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