Codeforces 938 正方形方格最多0/1 足球赛dijkstra建图

A

#include <bits/stdc++.h>
#define PI acos(-1.0)
#define mem(a,b) memset((a),b,sizeof(a))
#define TS printf("!!!
")
#define pb push_back
#define inf 1e9
//std::ios::sync_with_stdio(false);
using namespace std;
//priority_queue<int,vector<int>,greater<int>> que; get min
const double eps = 1.0e-8;
typedef pair<int, int> pairint;
typedef long long ll;
typedef unsigned long long ull;
const int maxn = 3e7 + 10;
const int  maxm = 300;
const int turn[4][2] = {{1, 0}, { -1, 0}, {0, 1}, {0, -1}};
//priority_queue<int, vector<int>, less<int>> que;
//next_permutation
ll mod = 1e9 + 7;
string ans;
int n;
string a;
int flag;
bool ok(char x)
{
        if (x == 'a' || x == 'e' || x == 'i' || x == 'o' || x == 'u' || x == 'y')
        {
                return true;
        }
        return false;
}
int main()
{
        cin >> n >> a;
        flag = 0;
        for (int i = 0; i < n; i++)
        {
                if (flag)
                {
                        if (ok(a[i]))
                        {
                                continue;
                        }
                        else
                        {
                                ans += a[i];
                                flag = 0;
                        }
                }
                else
                {
                        if (ok(a[i]))
                        {
                                flag = 1;
                                ans += a[i];
                        }
                        else
                        {
                                ans += a[i];
                                flag = 0;
                        }
                }
        }
        cout << ans << endl;
        return 0;
}

B

#include <bits/stdc++.h>
#define PI acos(-1.0)
#define mem(a,b) memset((a),b,sizeof(a))
#define TS printf("!!!
")
#define pb push_back
#define inf 1e9
//std::ios::sync_with_stdio(false);
using namespace std;
//priority_queue<int,vector<int>,greater<int>> que; get min
const double eps = 1.0e-8;
typedef pair<int, int> pairint;
typedef long long ll;
typedef unsigned long long ull;
const int maxn = 3e7 + 10;
const int  maxm = 300;
const int turn[4][2] = {{1, 0}, { -1, 0}, {0, 1}, {0, -1}};
//priority_queue<int, vector<int>, less<int>> que;
//next_permutation
ll mod = 1e9 + 7;
int number[1000005];
int anser = 0;
int main()
{
        int n;
        cin >> n;
        for (int i = 1; i <= n; i++)
        {
                int cur;
                cin >> cur;
                number[cur] = 1;
        }
        for (int i = 2; i <= 500000; i++)
        {
                if (number[i])
                {
                        anser = max(anser, i - 1);
                }
        }
        for (int i = 999999; i >= 500001; i--)
        {
                if (number[i])
                {
                        anser = max(anser, 1000000 - i);
                }
        }
        cout << anser << endl;
        return 0;
}

C

N*N的方格里 每个方格可以填0或者1 要求每个M*M的小方格里面必须要有一个0

题目给你一个数 问你有没有一对 N,M使之成立

总共有N^2个方格 因为每个M*M的小方格里面需要有一个0 所以每行和每列最少都需要N/M个0 总共就是（N/M）*（N/M）个0

所以答案就是N^2-（N/M）^2个 利用立方差公式可以分解为 (N+(N/M)*(N-(N/M)=ANS 也就是分解质因数 复杂度T*SQRT X

每次遇到一个N M都要检测是否符合条件

#include <bits/stdc++.h>
#define PI acos(-1.0)
#define mem(a,b) memset((a),b,sizeof(a))
#define TS printf("!!!
")
#define pb push_back
#define inf 1e9
//std::ios::sync_with_stdio(false);
using namespace std;
//priority_queue<int,vector<int>,greater<int>> que; get min
const double eps = 1.0e-10;
const double EPS = 1.0e-4;
typedef pair<int, int> pairint;
typedef long long ll;
typedef unsigned long long ull;
//const int maxn = 3e5 + 10;
const int  maxm = 300;
const int turn[4][2] = {{1, 0}, { -1, 0}, {0, 1}, {0, -1}};
//priority_queue<int, vector<int>, less<int>> que;
//next_permutation
int main()
{
        int n;
        cin >> n;
        while (n--)
        {
                int flag = 0;
                int x;
                cin >> x;
                if (x == 0)
                {
                        cout << 1 << " " << 1 << endl;
                        continue;
                }
                for (int i = 1; i <= ((int)sqrt(x) + 1) && (!flag); i++)
                {
                        if (x % i == 0)
                        {
                                int chu;
                                int beichu;
                                int a;
                                a = x / i;
                                chu = abs(i - a) / 2;
                                if (chu == 0)
                                {
                                        continue;
                                }
                                beichu = min(i, a) + chu;
                                int m;
                                m = beichu / chu;
                                if (beichu * beichu - (beichu / m) * (beichu / m) == x)
                                {
                                        cout << beichu << " " << m << endl;
                                        flag = 1;
                                }
                                //break;
                        }
                }
                if (!flag)
                {
                        cout << -1 << endl;
                }
        }
}

D

给你N个点 每个点可以都办足球比赛 票价为Ai 同时有M条路（不保证图联通）每条路有COST

问你在城市i的球迷想看球需要的最少的钱（往返加球票钱）

用floyd肯定不行 需要建一个超级源点 然后上堆优化的迪杰斯特拉跑一边 超级源点与原来每个点之间连的边的COST为其票价

#include <bits/stdc++.h>
#define PI acos(-1.0)
#define mem(a,b) memset((a),b,sizeof(a))
#define TS printf("!!!
")
#define pb push_back
#define inf 1e9
//std::ios::sync_with_stdio(false);
using namespace std;
//priority_queue<int,vector<int>,greater<int>> que; get min
const double eps = 1.0e-10;
const double EPS = 1.0e-4;
typedef pair<int, int> pairint;
typedef long long ll;
typedef unsigned long long ull;
//const int maxn = 3e5 + 10;
const int  maxm = 300;
const int turn[4][2] = {{1, 0}, { -1, 0}, {0, 1}, {0, -1}};
//priority_queue<int, vector<int>, less<int>> que;
//next_permutation
int n, m;
priority_queue<pair<ll, int> > que;
vector<pair<int, ll> > gra[200005];
ll sdis[200005];
int visit[200005];
int main()
{
        for (int i = 1; i <= 200000; i++)
        {
                sdis[i] = 1e18;
        }
        cin >> n >> m;
        int from, to;
        ll cost;
        for (int i = 1; i <= m; i++)
        {
                scanf("%d %d %lld", &from, &to, &cost);
                cost *= 2LL;
                gra[from].pb(make_pair(to, cost));
                gra[to].pb(make_pair(from, cost));
        }
        for (int i = 1; i <= n; i++)
        {
                scanf("%lld", &cost);
                gra[0].pb(make_pair(i, cost));
        }
        que.push({0, 0});
        while (!que.empty())
        {
                int now = que.top().second;
                que.pop();
                if (visit[now])
                {
                        continue;
                }
                visit[now] = 1;
                int len = gra[now].size();
                for (int i = 0; i < len; i++)
                {
                        to = gra[now][i].first;
                        ll value = gra[now][i].second;
                        if (!visit[to] && sdis[to] > sdis[now] + value)
                        {
                                sdis[to] = sdis[now] + value;
                                que.push({ -sdis[to], to});
                        }
                }
        }
        for (int i = 1; i <= n; i++)
        {
                cout << sdis[i] << " ";
        }
        cout << endl;
}