去連結裡看題目吧

\(Link\)

解題思路

暴力 dfs 加剪枝即可。

• 先搜尋 \(0\) 的個數小的行。

• 用 \(vis_{0,i,j}\) 表示數字 \(j\) 在第 \(i\) 行是否出現過，\(vis_{1,i,j}\) 表示數字 \(j\) 在第 \(i\) 列是否出現過，\(vis_{2,i,j}\) 表示數字 \(j\) 在第 \(i\) 個九宮格里是否出現過。

• 用矩陣預處理需要的東西。

沒了。。。

AC CODE

#include <bits/stdc++.h>

using namespace std;

int read()
{
	int x = 0;
	char c = getchar();
	while(c < '0' || c > '9') c = getchar();
	while(c >= '0' && c <= '9')
	{
		x = x * 10 + c - '0';
		c = getchar();
	}
	return x;
}

void write(int x)
{
	if(x < 0)
	{
		putchar('-');
		write(-x);
		return;
	}
	if(x > 9) write(x / 10);
	putchar(x % 10 + '0');
}

int h[10][10] = {{0, 0, 0, 0, 0, 0, 0, 0, 0, 0},
			    {0, 1, 1, 1, 2, 2, 2, 3, 3, 3},
				{0, 1, 1, 1, 2, 2, 2, 3, 3, 3},
				{0, 1, 1, 1, 2, 2, 2, 3, 3, 3},
				{0, 4, 4, 4, 5, 5, 5, 6, 6, 6},
				{0, 4, 4, 4, 5, 5, 5, 6, 6, 6},
				{0, 4, 4, 4, 5, 5, 5, 6, 6, 6},
				{0, 7, 7, 7, 8, 8, 8, 9, 9, 9},
				{0, 7, 7, 7, 8, 8, 8, 9, 9, 9},
				{0, 7, 7, 7, 8, 8, 8, 9, 9, 9}};

int score[10][10] = {{0, 0, 0, 0, 0, 0, 0, 0, 0, 0},
				   {0, 6, 6, 6, 6, 6, 6, 6, 6, 6},
                   {0, 6, 7, 7, 7, 7, 7, 7, 7, 6},
                   {0, 6, 7, 8, 8, 8, 8, 8, 7, 6},
                   {0, 6, 7, 8, 9, 9, 9, 8, 7, 6},
                   {0, 6, 7, 8, 9, 10, 9, 8, 7, 6},
                   {0, 6, 7, 8, 9, 9, 9, 8, 7, 6},
                   {0, 6, 7, 8, 8, 8, 8, 8, 7, 6},
                   {0, 6, 7, 7, 7, 7, 7, 7, 7, 6},
                   {0, 6, 6, 6, 6, 6, 6, 6, 6, 6}};

int Ans;

bool flag;

int a[10][10], ans[10][10];

int num, g[107];

int vis[3][107][107];

struct abc
{
	int id, cnt;
} k[10];

bool cmp(abc a, abc b)
{
	return a.cnt < b.cnt;
}

int getans()
{
	int Ans = 0;
	for(register int i = 1; i <= 9; ++i)
		for(register int j = 1; j <= 9; ++j)
			Ans += ans[i][j] * score[i][j];
	return Ans;
}

void dfs(int u)
{
	if(u == 82)
	{
		flag = 1;
		Ans = max(Ans, getans());
		return;
	}
	int x, y;
	if(g[u] % 9) x = g[u] / 9 + 1, y = g[u] % 9;
	else x = g[u] / 9, y = 9;
//	cout << x << " " << y << endl;
	if(!a[x][y])
	{
		for(int j = 1; j <= 9; ++j)
		{
			int G = h[x][y];
//			cout << G << endl;
			if(!vis[0][x][j] && !vis[1][y][j] && !vis[2][G][j])
			{
				ans[x][y] = j;
				vis[0][x][j] = vis[1][y][j] = vis[2][G][j] = 1;
				dfs(u + 1);
				vis[0][x][j] = vis[1][y][j] = vis[2][G][j] = 0;
			}
		}
	}
	else dfs(u + 1);
}

signed main()
{
	for(register int i = 1; i <= 9; ++i)
	{
		int cnt = 0;
		for(register int j = 1; j <= 9; ++j)
		{
			a[i][j] = read();
			if(!a[i][j]) cnt++;
			else
			{
				int v = a[i][j], g = h[i][j];
				ans[i][j] = v;
				vis[0][i][v] = vis[1][j][v] = vis[2][g][v] = 1;
			}
		}
		k[i].id = i;
		k[i].cnt = cnt;
	}
	sort(k + 1, k + 9 + 1, cmp);
	for(register int i = 1; i <= 9; ++i)
		for(register int j = 1; j <= 9; ++j)
		{
			int x = k[i].id, y = j;
			g[++num] = (x - 1) * 9 + y;
		}
	dfs(1);
	if(flag) write(Ans);
	else write(-1);	
	putchar('\n');
	return 0;
}
 

本文來自部落格園，作者：蒟蒻orz，轉載請註明原文連結：https://www.cnblogs.com/orzz/p/15376320.html