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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
#164399#6116. Changing the SequencesrealIyxiangRE 3ms3888kbC++144.0kb2023-09-04 22:40:402023-09-04 22:40:41

Judging History

你现在查看的是最新测评结果

  • [2023-09-04 22:40:41]
  • 评测
  • 测评结果:RE
  • 用时:3ms
  • 内存:3888kb
  • [2023-09-04 22:40:40]
  • 提交

answer

#include <queue>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <assert.h>
#include <algorithm>

#define in read()
#define fi first
#define se second
#define pb push_back
#define rep(i, x, y) for(int i = (x); i <= (y); i++)
#define per(i, x, y) for(int i = (x); i >= (y); i--)

using namespace std;

typedef long long ll;
typedef double db;
typedef vector < int > vec;
typedef pair < int , int > pii;

int read() {
    int x = 0; bool f = 0; char ch = getchar(); while(!isdigit(ch)) f |= ch == '-',ch = getchar();
    while(isdigit(ch)) x = x * 10 + (ch ^ 48),ch = getchar(); return f ? -x : x;
}

const int N = 5010;
const int M = 50010;

int n, m, cnt;
int a[N], b[N], pr[N], pos[N], pt[N];
int p[100][100];
int ans[N];

namespace F { // max_flow_with_cost
	const int inf = 0x3f3f3f3f;
	int h[N], now[N], lst[N], cnt = 1, S, T;
	bool vis[N], ban[N];
	ll maxflow, cost, dis[N];
	struct edge { int v, w, c, nxt; } e[M << 1];
	inline void tlink(int x, int y, int w, int c) { e[++cnt] = (edge) { y, w, c, h[x]}; h[x] = cnt; }
	inline void link(int x, int y, int w, int c) { tlink(x, y, w, c); tlink(y, x, 0, -c); }
	inline void setst(int s, int t) { S = s; T = t; }
	bool bfs(int s = S) {
		queue < int > q; q.push(s); rep(i, S, T) dis[i] = -1e9;
		memset(vis, 0, sizeof vis); dis[s] = 0; now[s] = h[s]; vis[s] = 1;
		memset(lst, -1, sizeof(lst)); 
		/*if(s) {
			rep(i, 1, 2 * m) for (int j = h[i]; j; j = e[j].nxt)
			if(e[j].v > m && !e[j].w) cout << i << ' ' << e[j].v << '\n'; 
		cout << '\n'; 
		}*/
		while(!q.empty()) {
			int x = q.front(); q.pop();
			for(int i = h[x], y; i; i = e[i].nxt)
				if(e[i].w && dis[y = e[i].v] < dis[x] + e[i].c && !ban[y]) {
					dis[y] = dis[x] + e[i].c, lst[y] = (i ^ 1); now[y] = h[y];
					if(!vis[y]) vis[y] = 1, q.push(y);
				}
			vis[x] = 0;
		}
		/*if(s) {
			cout << s << '\n'; 
			rep(i, S, T) cout << i << ' ' << dis[i] << ' ' << lst[i] << '\n'; cout << '\n'; 
		}*/
		return dis[T] > -1e9; 
	}
  int dinic(int x,int flow,ll c) {
		if(x == T) return maxflow += flow, cost += c * flow, flow;
		int res = flow; vis[x] = 1;
		for(int i = now[x], y; i && res; i = e[i].nxt) {
			now[x] = i;
			if(e[i].w && dis[y = e[i].v] == dis[x] + e[i].c && !vis[y]) {
				int k = dinic(y, min(res,e[i].w), c + e[i].c); 
				e[i].w -= k; e[i ^ 1].w += k; res -= k;
			}
		} vis[x] = 0;
		return flow - res;
  }
	void runit() { 
		while(bfs()) dinic(S, inf, 0); 
		rep(i, 1, m) for (int j = h[i]; j; j = e[j].nxt)
			if(!e[j].w) ans[i] = e[j].v - m; 
		// rep(i, 1, m) cout << ans[i] << ' '; cout << '\n'; 
		rep(i, 1, m) assert(ans[i]); 
		rep(t, 1, m) { 
			int i = pr[t]; if(!i) break ; 
			rep(j, 1, ans[i] - 1) {
				bfs(j + m); 
				// cout << i << ' ' << j << ' ' << dis[i] << ' ' << p[i][j] << '\n'; 
				if(dis[i] + p[i][j] != 0) continue ;  
				// cout << i << ' ' << j << ' ' << dis[i] << ' ' << p[i][j] << '\n'; 
				// rep(k, 1, 2 * m) cout << k << ' ' << e[lst[k]].v << '\n'; cout << '\n'; 
				auto upd = [&](int x) {
					e[x].w ^= 1, e[x ^ 1].w ^= 1; 
				} ;
				int k = i; 
				for ( ; e[lst[k]].v != j + m; k = e[lst[k]].v) {
					// cout << k << ' ' << e[lst[k]].v << ' ' << e[lst[e[lst[k]].v]].v << '\n';
					upd(lst[k]); 
					k = e[lst[k]].v;
					ans[e[lst[k]].v] = k - m; 
					upd(lst[k]); 
				}
				upd(lst[k]); 
				for (k = h[i]; k; k = e[k].nxt) if(e[k].v == j + m) {
					upd(k); break ; 
				}
				ans[i] = j; 
				// rep(k, 1, m) cout << ans[k] << ' '; cout << '\n'; 
			}
			ban[i] = 1; 
		}
		// rep(i, 1, m) assert(ans[i] > 0); 
	}
}

int main() {
	n = in, m = in; rep(i, 1, n) a[i] = in; rep(i, 1, n) b[i] = in;
	rep(i, 1, n) if(!pos[a[i]]) pos[a[i]] = i, pr[++cnt] = a[i];
	rep(i, 1, n) p[a[i]][b[i]]++;
	F :: setst(0, m * 2 + 1);
	rep(i, 1, m) F :: link(0, i, 1, 0), F :: link(i + m, m * 2 + 1, 1, 0);
	rep(i, 1, m) rep(j, 1, m) F :: link(i, j + m, 1, p[i][j]);
	F :: runit();
	rep(i, 1, n) assert(ans[a[i]]), printf("%d ", ans[a[i]]); printf("\n");
	return 0; 
}

詳細信息

Test #1:

score: 100
Accepted
time: 1ms
memory: 3644kb

input:

4 3
2 2 3 3
2 2 2 2

output:

1 1 2 2 

result:

ok 4 number(s): "1 1 2 2"

Test #2:

score: 0
Accepted
time: 1ms
memory: 3660kb

input:

5 3
2 2 3 3 2
2 2 2 2 3

output:

3 3 2 2 3 

result:

ok 5 number(s): "3 3 2 2 3"

Test #3:

score: 0
Accepted
time: 1ms
memory: 3648kb

input:

1 1
1
1

output:

1 

result:

ok 1 number(s): "1"

Test #4:

score: 0
Accepted
time: 3ms
memory: 3756kb

input:

1 60
60
60

output:

60 

result:

ok 1 number(s): "60"

Test #5:

score: 0
Accepted
time: 3ms
memory: 3888kb

input:

1 60
1
60

output:

60 

result:

ok 1 number(s): "60"

Test #6:

score: 0
Accepted
time: 1ms
memory: 3760kb

input:

1 60
60
1

output:

1 

result:

ok 1 number(s): "1"

Test #7:

score: 0
Accepted
time: 1ms
memory: 3852kb

input:

1 60
1
1

output:

1 

result:

ok 1 number(s): "1"

Test #8:

score: -100
Dangerous Syscalls

input:

100000 60
18 36 47 52 31 3 43 49 2 4 60 23 22 3 4 25 11 50 25 40 51 51 59 5 11 50 47 28 29 21 46 39 46 49 23 50 1 24 15 30 45 12 5 2 4 33 23 29 35 35 47 13 10 24 20 44 23 16 27 4 25 27 47 27 57 47 35 31 24 47 27 17 45 44 29 44 43 4 23 20 22 43 2 53 41 32 56 21 28 56 21 15 44 60 11 52 36 26 33 26 4 5...

output:


result: