AtCoder Regular Contest 161 E Not Dyed by Majority (Cubic Graph)

Posted 2023-05-29 zltzlt

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给构造题提供了一种新的思路。

如果答案占总方案数的比较大，可以考虑随机化。

例如本题，考虑随机化，难点变成判断一个方案是否可行。考虑 2-SAT，如果一个点是 \$\\textB\$，那么对于这个点的边，有一条边的另一个端点是 \$\\textW\$，那么其他两个都是 \$\\textB\$。反之同理。跑一遍 2-SAT 即可知道随出来的这个解是否合法。

code

// Problem: E - Not Dyed by Majority (Cubic Graph)
// Contest: AtCoder - AtCoder Regular Contest 161
// URL: https://atcoder.jp/contests/arc161/tasks/arc161_e
// Memory Limit: 1024 MB
// Time Limit: 4000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

#include <bits/stdc++.h>
#define pb emplace_back
#define fst first
#define scd second
#define mems(a, x) memset((a), (x), sizeof(a))

using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef long double ldb;
typedef pair<int, int> pii;

const int maxn = 500100;

int n, f[maxn], head[maxn], len;
vector<int> G[maxn];
mt19937 rnd(time(NULL));
int dfn[maxn], low[maxn], times, scc[maxn], cnt, stk[maxn], top;
struct edge 
	int to, next;
 edges[maxn * 10];

inline void add_edge(int u, int v) 
	edges[++len].to = v;
	edges[len].next = head[u];
	head[u] = len;


void dfs(int u) 
	dfn[u] = low[u] = ++times;
	stk[++top] = u;
	for (int i = head[u]; i; i = edges[i].next) 
		int v = edges[i].to;
		if (!dfn[v]) 
			dfs(v);
			low[u] = min(low[u], low[v]);
		 else if (!scc[v]) 
			low[u] = min(low[u], dfn[v]);
		
	
	if (dfn[u] == low[u]) 
		++cnt;
		while (1) 
			int x = stk[top--];
			scc[x] = cnt;
			if (x == u) 
				break;
			
		
	


void solve() 
	scanf("%d", &n);
	for (int i = 1; i <= n; ++i) 
		vector<int>().swap(G[i]);
	
	for (int i = 0, u, v; i < n * 3 / 2; ++i) 
		scanf("%d%d", &u, &v);
		G[u].pb(v);
		G[v].pb(u);
	
	while (1) 
		for (int i = 1; i <= n; ++i) 
			f[i] = rnd() & 1;
		
		len = times = cnt = 0;
		for (int i = 1; i <= n * 2; ++i) 
			head[i] = dfn[i] = low[i] = scc[i] = 0;
		
		for (int i = 1; i <= n; ++i) 
			int u = G[i][0], v = G[i][1], w = G[i][2];
			if (f[i]) 
				add_edge(u, v + n);
				add_edge(u, w + n);
				add_edge(v, u + n);
				add_edge(v, w + n);
				add_edge(w, u + n);
				add_edge(w, v + n);
			 else 
				add_edge(u + n, v);
				add_edge(u + n, w);
				add_edge(v + n, u);
				add_edge(v + n, w);
				add_edge(w + n, u);
				add_edge(w + n, v);
			
		
		for (int i = 1; i <= n * 2; ++i) 
			if (!dfn[i]) 
				dfs(i);
			
		
		bool flag = 0;
		for (int i = 1; i <= n; ++i) 
			if (scc[i] == scc[i + n]) 
				for (int u = 1; u <= n; ++u) 
					putchar(f[u] ? \'B\' : \'W\');
				
				putchar(\'\\n\');
				flag = 1;
				break;
			
		
		if (flag) 
			break;
		
	


int main() 
	int T = 1;
	scanf("%d", &T);
	while (T--) 
		solve();
	
	return 0;

AtCoder Regular Contest 119 C

Problem Statement

There are $N$ buildings along the AtCoder Street, numbered $1$ through $N$ from west to east. Initially, Buildings $1, 2, \dots, N$ have the heights of $A_{1}, A_{2}, \dots, A_{N}$ , respectively.

Takahashi, the president of ARC Wrecker, Inc., plans to choose integers $l$ and $r$ $(1 \leq l < r \leq N)$ and make the heights of Buildings $l, l + 1, \dots, r$ all zero. To do so, he can use the following two kinds of operations any number of times in any order:

Set an integer $x$ $(l \leq x \leq r - 1)$ and increase the heights of Buildings $x$ and $x + 1$ by $1$ each.
Set an integer $x$ $(l \leq x \leq r - 1)$ and decrease the heights of Buildings $x$ and $x + 1$ by $1$ each. This operation can only be done when both of those buildings have heights of $1$ or greater.

Note that the range of $x$ depends on $(l, r)$ .

How many choices of $(l, r)$ are there where Takahashi can realize his plan?

Constraints

$2 \leq N \leq 300000$
$1 \leq A_{i} \leq 10^{9}$ $(1 \leq i \leq N)$
All values in input are integers.

Input

Input is given from Standard Input in the following format:

 $N$ 
 $A_{1}$   $A_{2}$   $\dots$   $A_{N}$

Output

Print the answer.

Sample Input 1 Copy

Copy

5
5 8 8 6 6

Sample Output 1 Copy

Copy

Takahashi can realize his plan for $(l, r) = (2, 3), (4, 5), (2, 5)$ .

For example, for $(l, r) = (2, 5)$ , the following sequence of operations make the heights of Buildings $2, 3, 4, 5$ all zero.

Decrease the heights of Buildings $4$ and $5$ by $1$ each, six times in a row.
Decrease the heights of Buildings $2$ and $3$ by $1$ each, eight times in a row.

For the remaining seven choices of $(l, r)$ , there is no sequence of operations that can realize his plan.

Sample Input 2 Copy

Copy

7
12 8 11 3 3 13 2

Sample Output 2 Copy

Copy

Takahashi can realize his plan for $(l, r) = (2, 4), (3, 7), (4, 5)$ .

For example, for $刷题AtCoder Regular Contest 001$

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AtCoder Regular Contest 119 C