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ID题目提交者结果用时内存语言文件大小提交时间测评时间
#51664#4887. Fast Bridgeszhoukangyang#WA 4ms4824kbC++114.8kb2022-10-03 12:27:072022-10-03 12:27:09

Judging History

你现在查看的是最新测评结果

  • [2023-08-10 23:21:45]
  • System Update: QOJ starts to keep a history of the judgings of all the submissions.
  • [2022-10-03 12:27:09]
  • 评测
  • 测评结果:WA
  • 用时:4ms
  • 内存:4824kb
  • [2022-10-03 12:27:07]
  • 提交

answer

#include<bits/stdc++.h>
#define L(i, j, k) for(int i = (j); i <= (k); ++i)
#define R(i, j, k) for(int i = (j); i >= (k); --i)
#define ll long long 
#define ull unsigned long long 
#define vi vector <int>
#define sz(a) ((int) (a).size())
#define me(f, x) memset(f, x, sizeof(f)) 
#define uint unsigned int
using namespace std;
const int N = 507, mod = 998244353, iv3 = (mod + 1) / 3;

int n, k;
struct tup {
	int xl, xr, yl, yr;
} f[N];

struct vec {
	int x, y;
}; 

//int a[N];
//unsigned long long A[64]; 
//unsigned long long cur;
//inline int lst(int x) {
//    int bk = x >> 6, t = x & 63;
//    ull o = A[bk] & ((1ull << (t + 1)) - 1);
//    if(o) return 63 - __builtin_clzll(o) + (bk << 6);
//    ull G = cur & ((1ull << bk) - 1); 
//    if(!G) return 0;
//    int c = 63 - __builtin_clzll(G);
//    return (63 - __builtin_clzll(A[c])) + (c << 6);
//} 
//inline int nxt(int x) {
//    int bk = x >> 6, t = x & 63;
//    ull o = A[bk] >> t;
//    if(o) return __builtin_ctzll(o) + (bk << 6) + t;
//    ull G = cur >> (bk + 1);
//    if(!G) return 0;
//    int c = __builtin_ctzll(G) + bk + 1;
//    return __builtin_ctzll(A[c]) + (c << 6);
//}
//void init() {
//    me(a, 0x3f), me(A, 0), cur = 0, me(vis, 0);
//} 
//void ins(int x) {
//    int bk = x >> 6, t = x & 63;
//    A[bk] |= 1ull << t, cur |= 1ull << bk;
//    vis[x] = true;
//}
//void del(int x) {
//    int bk = x >> 6, t = x & 63;
//    if(A[bk] >> t & 1) A[bk] ^= 1ull << t;
//    if(!A[bk] && (cur >> bk & 1)) cur ^= 1ull << bk;
//    vis[x] = false;     
//}
//void insert(int x, int w) {
//    a[x] = min(a[x], w);
//    if(!vis[x]) {
//        int u = nxt(x);
//        if(u <= n && a[u] <= a[x]) return ;
//        ins(x);
//    }
//    while(true) {
//        int u = lst(x - 1);
//        if(!u || a[u] < a[x]) return ;
//        del(u), a[u] = 1e9;
//    }
//}

int dis[N][N];
int arrx[N], arry[N], atotx, atoty; 
bool vis[N];
int t[N]; 
vi gp[N], st[N];

int haha[N], xl[N], yl[N]; 
vector < int > id[N];
int arr[N], atot; 

int a[10][10][10][10];
int work(vector < pair < int, int > > vp) {
	L(i, 1, n) vis[i] = false, gp[i].clear();
	for(auto u : vp) vis[u.first] = true, gp[u.first].emplace_back(u.second);
	
	atotx = atoty = 0;
	L(i, 1, n) if(vis[i]) arrx[++atotx] = f[i].xl;
	sort(arrx + 1, arrx + atotx + 1), atotx = unique(arrx + 1, arrx + atotx + 1) - arrx - 1;
	L(i, 1, n) if(vis[i]) xl[i] = lower_bound(arrx + 1, arrx + atotx + 1, f[i].xl) - arrx;
	
	L(i, 1, n) if(vis[i]) arry[++atoty] = f[i].yl;
	sort(arry + 1, arry + atoty + 1), atoty = unique(arry + 1, arry + atoty + 1) - arry - 1;
	L(i, 1, n) if(vis[i]) yl[i] = lower_bound(arry + 1, arry + atoty + 1, f[i].yl) - arry;
	
	me(t, 0);
	L(i, 1, n) if(vis[i]) id[xl[i]].emplace_back(i);
	L(i, 0, n) st[i].clear();
	
	int ns = 0;
	
	R(i, atotx, 1) {
		for(const int &ic : id[i]) 
			for(const int &u : gp[ic]) t[u] = max(t[u], yl[ic]);
		
		int rns = 0, sum = 0;
		L(j, 1, n) if(t[j]) st[t[j]].emplace_back(j);
		me(haha, 0);
		R(j, atoty, 1) {
			if(sz(st[j])) {
				for(const int &a : st[j]) {
					L(t, 1, f[a].xr) 
						if(haha[t] < f[a].yr) 
							(sum += (ll) (arr[t] - arr[t - 1]) * (f[a].yr - haha[t]) % mod) %= mod, 
								haha[t] = f[a].yr;
				}
				st[j].clear();
			}
			(rns += (ll) sum * (arry[j] - arry[j - 1]) % mod) %= mod;
		}
		
		(ns += (ll) rns * (arrx[i] - arrx[i - 1]) % mod) %= mod;
	}
	L(i, 1, n) if(vis[i]) id[xl[i]].clear();
	return ns;
}

vector < pair < int, int > > vq[N];
int solve(vector < tup > T) {
	n = sz(T);
	L(i, 1, sz(T)) f[i] = T[i - 1];
	me(dis, 0x3f);
	L(i, 1, n) dis[i][i] = 0; 
	L(i, 1, n) L(j, 1, n) if(f[i].xr <= f[j].xl && f[i].yr <= f[j].yl) dis[i][j] = 1;
	L(k, 1, n) L(i, 1, n) L(j, 1, n) dis[i][j] = min(dis[i][j], dis[i][k] + dis[k][j]);
	L(i, 0, n) vq[i].clear();
	L(i, 1, n) L(j, 1, n) if(dis[i][j] < n) vq[dis[i][j]].emplace_back(make_pair(i, j)); 
	atot = 0;
	L(i, 1, n) f[i].xr = k - f[i].xr + 1, f[i].yr = k - f[i].yr + 1, arr[++atot] = f[i].xr;
	sort(arr + 1, arr + atot + 1), atot = unique(arr + 1, arr + atot + 1) - arr - 1;
	L(i, 1, n) f[i].xr = lower_bound(arr + 1, arr + atot + 1, f[i].xr) - arr;
	int qwq = 0; 
	L(i, 0, n - 1) if(sz(vq[i])) qwq += work(vq[i]), qwq %= mod;
	return qwq;
}

int main() {
	ios :: sync_with_stdio(false);
	cin.tie(0); cout.tie(0);
	int n;
	cin >> n >> k;
	vector < tup > A, B;
	while(n--) {
		tup a;
		cin >> a.xl >> a.yl >> a.xr >> a.yr;
		if(a.yl > a.yr) {
			a.yl = k - a.yl + 1;
			a.yr = k - a.yr + 1;
			A.push_back(a);
		} else {
			B.push_back(a);
		}
	}
	int ns = (ll) (k + 1) * k % mod * (k - 1) % mod * iv3 % mod * k % mod * k % mod;
	(ns += mod - solve(A)) %= mod;
	(ns += mod - solve(B)) %= mod;
	cout << ns << '\n';
	return 0;
}
/*
2 5
3 3 4 5
3 3 5 4
*/

详细

Test #1:

score: 100
Accepted
time: 0ms
memory: 4648kb

input:

2 2
1 1 2 2
1 2 2 1

output:

6

result:

ok answer is '6'

Test #2:

score: 0
Accepted
time: 3ms
memory: 4720kb

input:

0 1000000000

output:

916520226

result:

ok answer is '916520226'

Test #3:

score: 0
Accepted
time: 3ms
memory: 4824kb

input:

5 5
1 1 3 3
3 3 5 1
3 3 4 5
3 3 5 4
1 5 3 3

output:

946

result:

ok answer is '946'

Test #4:

score: -100
Wrong Answer
time: 4ms
memory: 4660kb

input:

200 5
1 1 4 2
2 5 4 4
2 3 4 2
2 4 3 5
1 4 4 2
2 5 4 2
1 2 4 4
1 2 2 4
1 4 5 1
3 4 5 1
4 2 5 1
2 2 5 4
3 2 5 1
3 1 5 2
4 2 5 3
1 3 5 1
3 4 4 5
2 2 4 3
2 3 5 4
1 4 5 3
2 2 3 1
2 5 3 3
1 1 5 3
4 4 5 5
1 3 4 4
4 3 5 1
2 3 3 4
3 4 4 2
1 4 4 5
2 1 4 4
1 3 5 2
1 1 3 3
1 5 3 1
1 1 3 5
1 4 3 5
4 5 5 4
1 1 4 ...

output:

728

result:

wrong answer expected '708', found '728'