寒假刷题记录,第一天(重刷题目)

2024-01-07 21:55:03

PTA甲级:

1003 Emergency

经典迪杰斯特拉算法

#include<iostream>
#include<cstring>

using namespace std;

const int N = 510 , INF = 0x3f3f3f3f;
int dist[N] , cnt[N] , g[N][N];
int helper[N] , way[N];
bool st[N];
int n , m , c1 , c2;

int main()
{
    cin >> n >> m >> c1 >> c2;

    for(int i = 0;i < n;i ++)
        cin >> cnt[i];

    memset(dist , 0x3f , sizeof dist);
    memset(g , 0x3f , sizeof g);
    while(m --)
    {
        int a , b , c;
        cin >> a >> b >> c;
        g[a][b] = g[b][a] = c;
    }

    dist[c1] = 0 , way[c1] = 1 , helper[c1] = cnt[c1];
    for(int i = 0;i < n;i ++)
    {
        int t = -1;
        for(int j = 0;j < n;j ++)
            if(!st[j] && (t == -1 || dist[t] > dist[j]))
                t = j;
        st[t] = true;
        
        for(int j = 0;j < n;j ++)
        {
            if(dist[j] > dist[t] + g[t][j])
            {
                dist[j] = dist[t] + g[t][j];
                way[j] = way[t];
                helper[j] = helper[t] + cnt[j];
            }
            else if(dist[j] == dist[t] + g[t][j])
            {
                way[j] += way[t];
                helper[j] = max(helper[t] + cnt[j] , helper[j]);
            }
        }
    }

    cout << way[c2] << " " << helper[c2] << endl;
    return 0;
}

1004 Counting Leaves

经典深度优先方法dfs

#include<iostream>
#include<vector>

using namespace std;

const int N = 200;
int n , m;
vector<int>v[N];
int cnt[N];
int dep = 1;

void dfs(int x , int depth)
{
    dep = max(dep , depth);
    if(!v[x].size()) 
    {
        cnt[depth] ++;
        return ;
    }
    for(auto i : v[x])
        dfs(i , depth + 1);
}

int main()
{
    while(cin >> n)
    {
        if(!n) break;
        cin >> m;
        for(int i = 0;i < m;i ++)
        {
            int id;
            cin >> id;
            int k;
            cin >> k;
            while(k --)
            {
                int id1;
                cin >> id1;
                v[id].push_back(id1);
            }
        }

        dfs(1 , 1);
        for(int i = 1;i <= dep;i ++)
            if(i == 1) cout << cnt[i];
            else cout << " " << cnt[i];
        cout << endl;
    }
    return 0;
}

天梯赛:

L2-001 紧急救援

同1003一样

L2-048 寻宝图

经典dfs

#include<iostream>
#include<vector>
#include<queue>

using namespace std;

const int N = 1e5 + 10;
int n , m;
int total , tre;
vector<int>nodes[N];
vector<bool>st[N];
int dx[4] = {0 , 0 , 1 , -1};
int dy[4] = {1 , -1 , 0 , 0};

void dfs(int x , int y , bool &f)
{
    if(nodes[x][y] > 1) f = true;
    st[x][y] = true;
    for(int i = 0;i < 4;i ++)
    {
        int tx = x + dx[i] , ty = y + dy[i];
        if(tx < 0 || tx >= n || ty < 0 || ty >= m) continue;
        if(st[tx][ty] || !nodes[tx][ty]) continue;
        dfs(tx , ty , f);
    }
}

int main()
{
    cin >> n >> m;
    for(int i = 0;i < n;i ++)
    {
        string s;
        cin >> s;
        for(int j = 0;j < m;j ++)
            nodes[i].push_back(s[j] - '0') , st[i].push_back(false);
    }
    
    for(int i = 0;i < n;i ++)
    {
        for(int j = 0;j < m;j ++)
        {
            if(nodes[i][j] && !st[i][j]) 
            {
                bool f = false;
                dfs(i , j , f);
                if(f) tre ++;
                total ++;
            }
        }
    }
    cout << total << " " << tre << endl;
    return 0;
}

文章来源:https://blog.csdn.net/xp_xht123/article/details/135441891
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