ABC436 题解

A. o-padding

code

#include<bits/stdc++.h>
using namespace std;
int main(){
    int n;
    string s,pre = "";
    cin >> n >> s;
    for(int i = s.size()+1; i <= n; ++i)
        pre += 'o';
    cout << pre << s;
}

B. Magic Square

code

#include<bits/stdc++.h>
using namespace std;
const int NN = 108;
int a[NN][NN];
int n;
int main(){
    ios::sync_with_stdio(false),cin.tie(0);
    cin >> n;
    int x = 0, y = (n-1)/2,cnt = 0;
    for(int i = 1; i <= n; ++i){
        for(int j = 1; j <= n; ++j){
            a[x][y] = ++cnt;
            if(j != n){
                x = (n+x-1) % n;
                y = (y+1) % n;
            }
            else{
                x = (x+1) % n;
            }
        }
    }

    for(int i = 0; i < n; ++i){
        for(int j = 0; j < n; ++j){
            cout << a[i][j] << " ";
        }
        cout << endl;
    }
}

C. 2x2 Placing

不得不说，C 题的实现是有一点困难的

至少在脑袋不够用的情况下卡了本人很久的实现 TAT

就是一个在平面内查询一个点是否被覆盖的问题

code

#include<bits/stdc++.h>
using namespace std;
const int NN = 1e6 + 8;
int n,m;
int fx[9] = {-1,-1,-1,0,0,0,1,1,1};
int fy[9] = {-1,0,1,-1,0,1,-1,0,1};
struct Node{
    int num;
    int x,y;
    bool operator < (const Node &A)const{
        if(x == A.x){
            if(y == A.y) return num < A.num;
            return y <= A.y;
        }
        else return x < A.x;
    }
    bool operator <= (const Node &A)const{
        if(x == A.x){
            if(y == A.y) return num <= A.num;
            return y <= A.y;
        }
        else return x < A.x;
    }
}a[NN],b[NN];
bool d[NN];
vector<int> now,pre;
int check(int x,int y){
    int i = lower_bound(b+1,b+1+m,(Node){0,x,y}) - b;
    if(b[i].x == x && b[i].y == y && d[b[i].num]) return b[i].num;
    else return 0x3f3f3f3f;
}
int main(){
    cin >> n >> m;
    for(int i = 1; i <= m; ++i)
        cin >> a[i].x >> a[i].y, b[i].x= a[i].x, b[i].y = a[i].y, b[i].num = i;
    sort(b+1,b+1+m);
    int ans = 0;
    for(int i = 1; i <= m; ++i){
        bool vis = 1;
        for(int j = 0; j <= 8; ++j)
            if(check(a[i].x + fx[j],a[i].y + fy[j]) < i){
                // cout << i << ":" << a[i].x + fx[j] << " " << a[i].y + fy[j] << "get\n";
                vis = 0;break;
            }
        if(vis) d[i] = 1,++ans;
    }
    cout << ans << endl;
}

D. Teleport Maze

bfs 即可

code

#include<bits/stdc++.h>
using namespace std;
const int NN = 1e3 + 8;
int H,W;
string s;
int step[NN][NN];
int vis[NN][NN];
int fx[4] = {0,0,1,-1};
int fy[4] = {1,-1,0,0};
struct Node{
    int x,y;
};
vector<Node> Ch[30];

inline int trans(char c){
    return c-'a'+1;
}
void search(){
    queue<Node> q;
    q.push({1,1});
    step[1][1] = 0;
    while(!q.empty()){
        Node u = q.front();
        q.pop();
        int ux = u.x, uy = u.y, stu = step[ux][uy];
        if(vis[ux][uy] != 0){
            int c = vis[ux][uy];
            for(auto j : Ch[c]){
                int vx = j.x, vy = j.y;
                if(step[vx][vy] == -1){
                    q.push({vx,vy});
                    step[vx][vy] = stu + 1;
                }
            }
            Ch[c].clear();
        }
        for(int j = 0; j < 4; ++j){
            int vx = ux + fx[j], vy = uy + fy[j];
            if(vx <= 0 || vx > H || vy <= 0 || vy > W) continue;
            if(step[vx][vy] == -1 && vis[vx][vy] != -1){
                q.push({vx,vy});
                step[vx][vy] = stu + 1;
            }
        }
    }
}
int main(){
    memset(step,-1,sizeof(step));
    cin >> H >> W;
    step[1][1] = 0;
    for(int i = 1; i <= H; ++i){
        cin >> s;
        s = ' ' + s;
        for(int j = 1; j <= W; ++j)
            if(s[j] == '#'){
                vis[i][j] = -1;
            }
            else if(isalpha(s[j])) Ch[trans(s[j])].push_back({i,j}), vis[i][j] = trans(s[j]);
    }
    search();
    cout << step[H][W];
}

E. Minimum Swap

这道题就是通过排列连边之后，询问最小环的大小

code

#include<bits/stdc++.h>
using namespace std;
const int NN = 3e5 + 8;
int n;
int a[NN],rev[NN];
int main(){
    cin >> n;
    for(int i = 1; i <= n; ++i)
        cin >> a[i];
    long long ans = 0;
    for(int i = 1; i <= n; ++i){
        if(i != a[i]){
            int cnt = 1;
            int t = a[i];
            while(a[i] != i) swap(a[i],a[t]), t = a[i], ++cnt;
            ans += 1ll * cnt * (cnt-1) / 2;
        }
    }
    cout << ans;
}

F. Starry Landscape Photo

可以发现，如果我们考虑一个方案里面的最大值 \(x\)

我们假设它的左边有 \(L_x\) 个小于它的值，右边有 \(R_x\) 个小于它的值

那么我们只需要让 \(b = x\) 然后调整区间，就可以得到 \((L_x+1) \times (R_x+1)\) 种方案

最后对于每一个 \(x\) 求方案数之和即可

code

#include<bits/stdc++.h>
using namespace std;
const int NN = 5e5 + 8;
int n;
int b[NN];
int rev[NN];
int a[NN];
int lowbit(int x){
	return x & (-x);
}
void add(int x,int num){
	while(x <= n){
		a[x] += num;
		x += lowbit(x);
	}
}
int ask(int x){
	int res = 0;
	while(x){
		res += a[x];
		x -= lowbit(x);
	}
	return res;
}

int main(){
	ios::sync_with_stdio(false),cin.tie(0);
	
	cin >> n;
	for(int i = 1; i <= n; ++i){
		cin >> b[i];
		rev[b[i]] = i;
	}
	
	long long ans = 0;
	for(int i = 1; i <= n; ++i){
		int numl = ask(rev[i]-1),numr = i-1-numl;
		add(rev[i],1);
		ans += 1ll * (numl + 1) * (numr + 1);
	}
	cout << ans;
} 

G. Linear Inequation