Educational Codeforces Round 52

原创于 2020-11-08 01:08:00 发布 · 155 阅读

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#搜狗 #wxpython #templates #ntp #hudson

本文提供了四个不同问题的代码解决方案，包括贪心算法解决巧克力购买问题、图论问题中的孤立顶点计数、数组操作使元素相等及复杂路径规划问题。通过精心设计的数据结构和算法优化，实现了高效求解。

1065A - Vasya and Chocolate

贪心比较按照哪种买法最优然后全部按照最优买法买即可(有剩余就单独买)

solved

#include <stdio.h>
#include <iostream>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <map> 
#include <stack>
#include <sstream>
#include <set>
// #pragma GCC optimize(2)

//#define int long long
#define rep(i,a,n) for(int i=a;i<=n;i++)
#define rush() int T;scanf("%d",&T);for(int Ti=1;Ti<=T;++Ti)
#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define mm(i,v) memset(i,v,sizeof i);
#define mp(a, b) make_pair(a, b)
#define pi acos(-1)
#define fi first
#define se second
//你冷静一点，确认思路再敲！！！ 

using namespace std;
typedef long long ll;
typedef double db;
typedef pair<int, int > PII;
priority_queue< PII, vector<PII>, greater<PII> > que;
stringstream ssin; //  ssin << string   while ( ssin >> int)
const ll LINF = 0x7fffffffffffffffll;

const int N = 4e5 + 5, M = 4e5 + 5, mod = 1e9 + 7, INF = 0x3f3f3f3f;
ll _, s, a, b, c;

inline ll read() {
    char c=getchar();ll x=0,f=1;
    while(c<'0'||c>'9'){if(c=='-')f=-1; c=getchar();}
    while(c>='0'&&c<='9'){x=x*10+c-'0'; c=getchar();}
    return x*f;
}


int main()
{
    // freopen("in.txt","r",stdin);
    // freopen("out.txt","w",stdout);
    _ = read();
    while (_--) {
        s = read(); a = read(); b = read(); c = read();
        if (1.0 * a * c / (a + b) < c) {
            ll tmp = s / c;
            ll ans = tmp / a * b + tmp;
            cout << ans << '\n';
        } else {
            cout << s / c << '\n';
        }
    }
    #ifndef ONLINE_JUDGE
        system("pause");
    #endif
}

View Code

1065B - Vasya and Isolated Vertices

本题求两问，对于第一问，显然让每天边都匹配两个点是最优的，那么直接输出max(0, n - 2 * m) 即可。

第二问可以直接枚举来得到。

solved

#include <stdio.h>
#include <iostream>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <map> 
#include <stack>
#include <sstream>
#include <set>
// #pragma GCC optimize(2)

//#define int long long
#define rep(i,a,n) for(int i=a;i<=n;i++)
#define rush() int T;scanf("%d",&T);for(int Ti=1;Ti<=T;++Ti)
#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define mm(i,v) memset(i,v,sizeof i);
#define mp(a, b) make_pair(a, b)
#define pi acos(-1)
#define fi first
#define se second
//你冷静一点，确认思路再敲！！！ 

using namespace std;
typedef long long ll;
typedef double db;
typedef pair<int, int > PII;
priority_queue< PII, vector<PII>, greater<PII> > que;
stringstream ssin; //  ssin << string   while ( ssin >> int)
const ll LINF = 0x7fffffffffffffffll;

const int N = 4e5 + 5, M = 4e5 + 5, mod = 1e9 + 7, INF = 0x3f3f3f3f;
ll n, m;
ll ans1, ans2;

inline ll read() {
    char c=getchar();ll x=0,f=1;
    while(c<'0'||c>'9'){if(c=='-')f=-1; c=getchar();}
    while(c>='0'&&c<='9'){x=x*10+c-'0'; c=getchar();}
    return x*f;
}

bool ck(ll mid) {
    ll nn = n - mid;
    if (nn < 0) return false;
    if (nn == 0) return true;
    
    if ((nn * (nn - 1)) / 2 >= m && nn <= 2 * m) return true;
    return false;
}

int main()
{
    // freopen("in.txt","r",stdin);
    // freopen("out.txt","w",stdout);
    n = read(); m = read();
    ans1 = max(0ll, n - 2 * m);
    ll maxx = 0;
    for (ll i = 0; i <= n; ++i) {
        if (i == 0 && m == 0) {
            maxx = n;
            break;
        }
        if ((i * (i - 1) / 2 >= m && i <= 2 * m)) {
            maxx = n - i;
            break;
        }
    }
    ans2 = maxx;
    
    cout << ans1 << " " << ans2 << '\n';
    #ifndef ONLINE_JUDGE
        system("pause");
    #endif
}

View Code

1065C - Make It Equal

直接模拟即可

solved

#include <stdio.h>
#include <iostream>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <map> 
#include <stack>
#include <sstream>
#include <set>
// #pragma GCC optimize(2)

//#define int long long
#define rep(i,a,n) for(int i=a;i<=n;i++)
#define rush() int T;scanf("%d",&T);for(int Ti=1;Ti<=T;++Ti)
#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define mm(i,v) memset(i,v,sizeof i);
#define mp(a, b) make_pair(a, b)
#define pi acos(-1)
#define fi first
#define se second
//你冷静一点，确认思路再敲！！！ 

using namespace std;
typedef long long ll;
typedef double db;
typedef pair<int, int > PII;
priority_queue< PII, vector<PII>, greater<PII> > que;
stringstream ssin; //  ssin << string   while ( ssin >> int)
const ll LINF = 0x7fffffffffffffffll;

const int N = 4e5 + 5, M = 4e5 + 5, mod = 1e9 + 7, INF = 0x3f3f3f3f;
ll n, k;
ll a[N];
map<ll, ll>ma;

inline ll read() {
    char c=getchar();ll x=0,f=1;
    while(c<'0'||c>'9'){if(c=='-')f=-1; c=getchar();}
    while(c>='0'&&c<='9'){x=x*10+c-'0'; c=getchar();}
    return x*f;
}

int main()
{
    // freopen("in.txt","r",stdin);
    // freopen("out.txt","w",stdout);
    n = read(); k = read();
    ll Min = 1e18, Max = 0;
    for (int i = 1; i <= n; ++i) {
        a[i] = read();
        ma[a[i]]++;
        Min = min(Min, a[i]);
        Max = max(Max, a[i]);
    }
    
    bool flag = 0;
    ll ans = 1;
    ll now = Max;
    ll kk = k;
    while (now > Min) {
        if (kk >= ma[now]) {
            kk -= ma[now];
            flag = 1;
            now--;
            ma[now] += ma[now + 1];
        } else if (kk < ma[now]) {
            kk = k;
            ans++;
            continue;
        }
    }
    if (!flag) ans = 0;
    cout << ans << '\n';
    
    // #ifndef ONLINE_JUDGE
    //     system("pause");
    // #endif
}

View Code

D-Three Pieces

最短路+dp dp[x][y][rep][3][pos] 表示再点(x, y)换了rep次棋后当前使用一个棋子且已经枚举完1-pos位置的最小值

因为是双关键字，可以考虑使用加阈值转化为单关键词或者直接在dp方程式中表示

unsolved

#include <stdio.h>
#include <iostream>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <map> 
#include <stack>
#include <sstream>
#include <set>
// #pragma GCC optimize(2)

//#define int long long
#define rep(i,a,n) for(int i=a;i<=n;i++)
#define rush() int T;scanf("%d",&T);for(int Ti=1;Ti<=T;++Ti)
#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define mm(i,v) memset(i,v,sizeof i);
#define mp(a, b) make_pair(a, b)
#define pi acos(-1)
#define fi first
#define se second

using namespace std;
typedef long long ll;
typedef double db;
typedef pair<int, int > PII;
priority_queue< PII, vector<PII>, greater<PII> > que;
stringstream ssin; //  ssin << string   while ( ssin >> int)
const ll LINF = 0x7fffffffffffffffll;

const int M = 4e5 + 5, mod = 1e9 + 7, INF = 0x3f3f3f3f;
int n, N;
int MAP[15][15];
int nx[8] = {-1, -2, -2, -1, 1, 2, 2, 1};
int ny[8] = {-2, -1, 1, 2, 2, 1, -1, -2};
int dp[15][15][205][4][105];

inline ll read() {
    char c=getchar();ll x=0,f=1;
    while(c<'0'||c>'9'){if(c=='-')f=-1; c=getchar();}
    while(c>='0'&&c<='9'){x=x*10+c-'0'; c=getchar();}
    return x*f;
}

struct node {
    int x, y, cost, repl, now, num;
    bool operator<(const node &a) const {
        return cost > a.cost;
    }
//    friend bool operator < (node a,node b){
//        return a.cost > b.cost;
//    }
}s, e;

bool check(node u) {
    if (u.x >= 1 && u.x <= n && u.y >= 1 && u.y <= n && dp[u.x][u.y][u.repl][u.now][u.num] > u.cost)
        return true;
    return false;
}

bool ck(int x, int y) {
    if (x < 1 || x > n || y < 1 || y > n) return false;
    return true;
}

void solve() {
    priority_queue<node>q;        
    mm(dp, 0x3f);
    dp[s.x][s.y][0][0][1] = 0;
    dp[s.x][s.y][0][1][1] = 0;
    dp[s.x][s.y][0][2][1] = 0;
    
    q.push({s.x, s.y, 0, 0, 0, 1});
    q.push({s.x, s.y, 0, 0, 1, 1});
    q.push({s.x, s.y, 0, 0, 2, 1});
    
    while (!q.empty()) {
        node u = q.top();
        q.pop();
        if (dp[u.x][u.y][u.repl][u.now][u.num] < u.cost) continue;
        if (u.repl > 200) continue;
        node h;
        // 判断换子
        for (int i = 0; i < 3; ++i) {
            if (i == u.now) continue;
            h = u;
            h.repl++; h.now = i; h.cost++;
            if (!check(h)) continue;
            dp[h.x][h.y][h.repl][h.now][h.num] = h.cost;
            q.push(h);
        }
        
        if (u.now == 0) {  //骑士垂直走 
            for (int i = 1; i <= n; ++i) {
                if (u.x == i) continue;
                h = u;
                h.x = i; h.cost++;
                if (h.num + 1 == MAP[h.x][h.y]) h.num++;
                if (!check(h)) continue;
                dp[h.x][h.y][h.repl][h.now][h.num] = h.cost;
                q.push(h);
            }
            
            for (int i = 1; i <= n; ++i) {
                if (u.y == i) continue;
                h = u;
                h.y = i; h.cost++;
                if (h.num + 1 == MAP[h.x][h.y]) h.num++;
                if (!check(h)) continue;
                dp[h.x][h.y][h.repl][h.now][h.num] = h.cost;
                q.push(h);
            }
            
        } else if (u.now == 1) {   // 马走日字
        
            for (int i = 0; i < 8; ++i) {
                int xx = u.x + nx[i], yy = u.y + ny[i];
                if (xx < 1 || xx > n || yy < 1 || yy > n) continue;
                h = u;
                h.x = xx; h.y = yy; h.cost ++;
                if (h.num + 1 == MAP[h.x][h.y]) h.num++;
                if (!check(h)) continue;
                dp[h.x][h.y][h.repl][h.now][h.num] = h.cost;
                q.push(h);
            } 
            
        } else if (u.now == 2) {  // 象走斜线 
        
            for (int i = 1; i <= n; ++i) {   // -1, -1
                int xx = u.x + i * (-1), yy = u.y + i * (-1);
                if (!ck(xx, yy)) break;
                h = u;
                h.cost++; h.x = xx; h.y = yy;
                if (h.num + 1 == MAP[h.x][h.y]) h.num++;
                if (!check(h)) continue;
                dp[h.x][h.y][h.repl][h.now][h.num] = h.cost;
                q.push(h);
            }
            for (int i = 1; i <= n; ++i) {   //  -1, 1
                int xx = u.x + i * (-1), yy = u.y + i * 1;
                if (!ck(xx, yy)) break;
                h = u;
                h.cost++; h.x = xx; h.y = yy;
                if (h.num + 1 == MAP[h.x][h.y]) h.num++;
                if (!check(h)) continue;
                dp[h.x][h.y][h.repl][h.now][h.num] = h.cost;
                q.push(h);
            }
            for (int i = 1; i <= n; ++i) {     // 1, -1
                int xx = u.x + i * 1, yy = u.y + i * (-1);
                if (!ck(xx, yy)) break;
                h = u;
                h.cost++; h.x = xx; h.y = yy;
                if (h.num + 1 == MAP[h.x][h.y]) h.num++;
                if (!check(h)) continue;
                dp[h.x][h.y][h.repl][h.now][h.num] = h.cost;
                q.push(h);
            }
            for (int i = 1; i <= n; ++i) {   // 1, 1
                int xx = u.x + i * 1, yy = u.y + i * 1;
                if (!ck(xx, yy)) break;
                h = u;
                h.cost++; h.x = xx; h.y = yy;
                if (h.num + 1 == MAP[h.x][h.y]) h.num++;
                if (!check(h)) continue;
                dp[h.x][h.y][h.repl][h.now][h.num] = h.cost;
                q.push(h);
            }
        }
    }
    
    int Min = 1e9, u = 0;
    for (int i = 0; i <= 200; ++i) {
        for (int j = 0; j < 3; ++j) {
            if (dp[e.x][e.y][i][j][n * n] < Min) {
                Min = dp[e.x][e.y][i][j][n * n];
                u = i;
            }
        }
    }
    
    printf("%d %d\n", Min, u);
}

int main()
{
    // freopen("in.txt","r",stdin);
    // freopen("out.txt","w",stdout);
    n = read();
    N = n;
    for (int i = 1; i <= n; ++i) {
        for (int j = 1; j <= n; ++j) {
            int x;
            x = read();
            MAP[i][j] = x;
            if (x == 1) {
                s.x = i; s.y = j;
            } else if (x == n * n) {
                e.x = i; e.y = j;
            }
        }
    }
    solve();
    
//    #ifndef ONLINE_JUDGE
//        system("pause");
//    #endif
}

View Code