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IDProblemSubmitterResultTimeMemoryLanguageFile sizeSubmit timeJudge time
#338773#4805. Grammy Sortingchenxinyang2006WA 1ms4072kbC++145.4kb2024-02-26 11:40:442024-02-26 11:40:46

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你现在查看的是最新测评结果

  • [2024-02-26 11:40:46]
  • 评测
  • 测评结果:WA
  • 用时:1ms
  • 内存:4072kb
  • [2024-02-26 11:40:44]
  • 提交

answer

#include <bits/stdc++.h>
#define rep(i,j,k) for(int i=(j);i<=(k);i++)
#define per(i,j,k) for(int i=(j);i>=(k);i--)
#define uint unsigned int
#define ll long long
#define ull unsigned long long
#define db double
#define ldb long double
#define pii pair<int,int>
#define pll pair<ll,ll>
#define mkp make_pair
#define eb emplace_back
#define SZ(S) (int)S.size()
//#define mod 998244353
//#define mod 1000000007
#define inf 0x3f3f3f3f
#define linf 0x3f3f3f3f3f3f3f3f
using namespace std;

template <class T>
void chkmax(T &x,T y){
	if(x < y) x = y;
}

template <class T>
void chkmin(T &x,T y){
	if(x > y) x = y;
}

inline int popcnt(int x){
	return __builtin_popcount(x);
}

inline int ctz(int x){
	return __builtin_ctz(x);
}


/*ll power(ll p,int k = mod - 2){
	ll ans = 1;
	while(k){
		if(k % 2 == 1) ans = ans * p % mod;
		p = p * p % mod;
		k /= 2;	
	}
	return ans;
}*/
int n,m,A,B;
int a[1005];
int _u[2005],_v[2005];

int cnt;
int head[2005];
struct eg{
	int to,nxt;
}edge[4005];

void make(int u,int v){
	edge[++cnt].to = v;
	edge[cnt].nxt = head[u];
	head[u] = cnt;
}

int dfn[1005],low[1005],N;
vector <int> sta,vcc[1005],eid[1005];
void tarjan(int u){
	dfn[u] = low[u] = ++cnt;
	sta.eb(u);
	for(int i = head[u];i;i = edge[i].nxt){
		int v = edge[i].to;
//		printf("%d %d\n",u,v);
		if(!dfn[v]){
			tarjan(v);
			chkmin(low[u],low[v]);
			if(low[v] >= dfn[u]){
				++N;
				while(1){
					int p = sta.back();
					sta.pop_back();
					vcc[N].eb(p);
					if(p == v) break;
				}
				vcc[N].eb(u);
			}
		}else{
			chkmin(low[u],dfn[v]);
		}
	}
}

int fa[2005],dep[2005];
void dfs(int u,int f){
	fa[u] = f;
	dep[u] = dep[f] + 1;
	for(int i = head[u];i;i = edge[i].nxt){
		int v = edge[i].to;
		if(v == f) continue;
		dfs(v,u);
	}
}

vector <int> _S,temp;
void test(){
	int p = A,q = B;
	while(p != q){
		if(dep[p] > dep[q]){
			_S.eb(p);
			p = fa[p];
		}else{
			temp.eb(q);
			q = fa[q];
		}
	}
	_S.eb(p);
	while(!temp.empty()){
		_S.eb(temp.back());
		temp.pop_back();
	}
}

vector <int> L[1005],ans;
int ord[1005],sum[1005],vis[1005];
void starjan(int u){
	dfn[u] = low[u] = ++cnt;
	ord[cnt] = u;
	for(int i = head[u];i;i = edge[i].nxt){
		int v = edge[i].to;
		if(!dfn[v]){
			fa[v] = u;
			starjan(v);
			sum[u] += sum[v];
			chkmin(low[u],low[v]);
		}else{
			chkmin(low[u],dfn[v]);
		}
	}
//	printf("%d %d %d %d\n",u,fa[u],ord[low[u]],sum[u]);
	if(fa[u] && !sum[u]){
		L[fa[u]].eb(u);
		L[ord[low[u]]].eb(u);
	}
}

void cover(int u){
	if(vis[u]) return;
	vis[u] = 1;
	ans.eb(u);
	for(int v:L[u]) cover(v);
}

void clr(int u){
	head[u] = dfn[u] = low[u] = fa[u] = sum[u] = vis[u] = 0;
	L[u].clear();
}
int rk[1005];

void solve(int c,int S,int T){
//	printf("solve vcc %d S=%d T=%d\n",c,S,T);
	cnt = 0;
	for(int e:eid[c]){
		clr(_u[e]);
		clr(_v[e]);
//		printf("%d %d\n",_u[e],_v[e]);
	}
	for(int e:eid[c]){
		make(_u[e],_v[e]);
		make(_v[e],_u[e]);
	}
	sum[T] = 1;
	cnt = 0;
	starjan(S);

	temp.clear();
	int p = T;
	while(p){
		temp.eb(p);
		p = fa[p];
	}
	reverse(temp.begin(),temp.end());
	for(int q:temp) cover(q);
}
int pst[1005];
vector <int> answer[1005];

int main(){
//	freopen("test.in","r",stdin);
	scanf("%d%d%d%d",&n,&m,&A,&B);
	rep(i,1,n) scanf("%d",&a[i]);
	rep(i,1,m){
		scanf("%d%d",&_u[i],&_v[i]);
		make(_u[i],_v[i]);
		make(_v[i],_u[i]);
	}
/*	if(n == 2){
		if(a[A] > a[B]){
			printf("1\n");
			printf("2 %d %d\n",A,B);
		}else{
			printf("0\n");
		}
		return 0;
	}*/
	cnt = 0;
	tarjan(1);
	cnt = 0;
	memset(head,0,sizeof(head));
	rep(c,1,N){
//		printf("vcc %d\n",c);
		for(int u:vcc[c]){
			make(u,c + n);
			make(c + n,u);
//			printf("%d ",u);
		}
//		printf("\n");
	}
//	int rt = -1;
//	rep(u,1,n) if(u != A && u != B) rt = u;
//	assert(rt != -1);
//	dfs(rt,0);
	dfs(1,0);
	test();
//	for(int u:_S) printf("%d ",u);
//	printf("\n");
	if(SZ(_S) / 2 < N){
		printf("-1\n");
		return 0;
	}
	rep(i,1,m){
		if(fa[_u[i]] == fa[_v[i]]) eid[fa[_u[i]] - n].eb(i);
		else if(fa[fa[_u[i]]] == _v[i]) eid[fa[_u[i]] - n].eb(i);
		else if(fa[fa[_v[i]]] == _u[i]) eid[fa[_v[i]] - n].eb(i);
	}
//	for(int k = 0;k + 2 < SZ(_S);k += 2) printf("%d ",_S[k]);
//	printf("\n"); 
	for(int k = 0;k + 2 < SZ(_S);k += 2){
		if(k) ans.pop_back();
		solve(_S[k + 1] - n,_S[k],_S[k + 2]);
	}
	for(int u:ans) printf("%d ",u);
	printf("\n");
	//ans 是一组合法双极定向

	cnt = 0;
	fill(head + 1,head + n + 1,0);
	rep(k,0,n - 1) rk[ans[k]] = k + 1;
	rep(i,1,m){
		if(rk[_u[i]] > rk[_v[i]]) swap(_u[i],_v[i]);
		make(_u[i],_v[i]);
		pst[_v[i]] = _u[i];
	}
//	rep(u,1,n) printf("%d ",rk[u]);
//	printf("\n");
	per(k,n - 1,0){
		int p = ans[k];
		while(p != A){
			answer[k].eb(p);
			p = pst[p];
		}
		answer[k].eb(A);
		reverse(answer[k].begin(),answer[k].end());
		p = ans[k];
		while(1){
			int q = -1;
			for(int i = head[p];i;i = edge[i].nxt){
				int v = edge[i].to;
				if(a[v] <= a[A] && (q == -1 || a[v] < a[q])) q = v;
			}
			if(q == -1) break;
			answer[k].eb(q);
			p = q;
		}
		rep(i,0,SZ(answer[k]) - 2) swap(a[answer[k][i]],a[answer[k][i + 1]]);
	}
	int ssum = 0;
	rep(k,0,n - 1) ssum += !answer[k].empty();
	printf("%d\n",ssum);
	per(k,n - 1,0){
		if(answer[k].empty()) continue;
		printf("%d ",SZ(answer[k]));
		for(int p:answer[k]) printf("%d ",p);
		printf("\n");
	}
	return 0;
}

Details

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Test #1:

score: 0
Wrong Answer
time: 1ms
memory: 4072kb

input:

5 6 1 2
1 2 3 4 5
1 3
2 3
1 4
2 4
1 5
3 5

output:

1 4 5 3 2 
5
3 1 4 2 
4 1 5 3 2 
4 1 5 3 2 
2 1 4 
2 1 5 

result:

wrong answer a_1 != A at operation #1