题解【洛谷P1074】[NOIP2009]靶形数独
题解
一开始写了一个朴素的数独，无任何剪枝优化，得到了\(55\)分的好成绩。
就是这道题加一个计算分数。
代码如下（\(\mathrm{55\ pts}\)）：
/********************************
Author: csxsl
Date: 2019/10/28
Language: C++
Problem: P1074
********************************/
#include <bits/stdc++.h>
#define itn int
#define gI gi
using namespace std;
inline int gi()
{
int f = 1, x = 0; char c = getchar();
while (c < '0' || c > '9') {if (c == '-') f = -1; c = getchar();}
while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
return f * x;
}
inline long long gl()
{
long long f = 1, x = 0; char c = getchar();
while (c < '0' || c > '9') {if (c == '-') f = -1; c = getchar();}
while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
return f * x;
}
int n, m, ans, a[10][10];
bool h[10][10], l[10][10], fz[10][10];
const int fenshu[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}
};
inline int getfz(int x, int y) {return (x - 1) / 3 * 3 + (y - 1) / 3 + 1;}
inline void getans()
{
int sum = 0;
for (int i = 1; i <= 9; i+=1)
{
for (int j = 1; j <= 9; j+=1)
{
sum = sum + a[i][j] * fenshu[i][j];
}
}
if (sum > ans) ans = sum;
}
void dfs(int x, int y)
{
if (a[x][y])
{
if (x == 9 && y == 9) {getans(); return;}
else if (y == 9) dfs(x + 1, 1);
else dfs(x, y + 1);
}
else
{
for (int i = 1; i <= 9; i+=1)
{
if (!a[x][y] && !h[x][i] && !l[y][i] && !fz[getfz(x, y)][i])
{
a[x][y] = i;
h[x][i] = l[y][i] = fz[getfz(x, y)][i] = 1;
if (x == 9 && y == 9) {getans(); return;}
else if (y == 9) dfs(x + 1, 1);
else dfs(x, y + 1);
a[x][y] = 0;
h[x][i] = l[y][i] = fz[getfz(x, y)][i] = 0;
}
}
}
}
int main()
{
//freopen(".in", "r", stdin);
//freopen(".out", "w", stdout);
for (int i = 1; i <= 9; i+=1)
{
for (int j = 1; j <= 9; j+=1)
{
a[i][j] = gi();
if (a[i][j]) h[i][a[i][j]] = l[j][a[i][j]] = fz[getfz(i, j)][a[i][j]] = 1;//标记数字
}
}
dfs(1, 1);//搜索
printf("%d\n", (ans == 0) ? (-1) : (ans));//输出
return 0;
}
常见的搜索优化方式有：
-
调换搜索顺序，让方案数少的先搜。
-
剪枝，又分为可行性剪枝和最优性剪枝。
这里可以思考如何调换搜索顺序：
不难发现，只要一个位置上填了数字，我们就直接递归下一个数字即可，这一行枚举的数的个数就与这一行\(0\)的个数有关。
因此，我们可以预处理处每一行\(0\)的个数，从小到大排序后再进行搜索。
用不同的顺序进行搜索，效率也会大不一样！
注意此处需要开一个三维数组\(\mathrm{vis[0/1/2][i][j]}\)。
\(\mathrm{vis[0/1/2][i][j]}\)分别表示第\(i\)行、第\(i\)列、第\(i\)个方阵有没有\(j\)这个数。
这样设的原因留给读者作为练习。
代码中的\(\mathrm{b[i]}\)表示搜索到了第几个数，顺序与每行\(0\)的个数有关。
调换搜索顺序后发现就可以\(\mathrm{AC}\)了。
代码
/********************************
Author: csxsl
Date: 2019/10/28
Language: C++
Problem: P1074
********************************/
#include <bits/stdc++.h>
#define itn int
#define gI gi
using namespace std;
inline int gi()
{
int f = 1, x = 0; char c = getchar();
while (c < '0' || c > '9') {if (c == '-') f = -1; c = getchar();}
while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
return f * x;
}
inline long long gl()
{
long long f = 1, x = 0; char c = getchar();
while (c < '0' || c > '9') {if (c == '-') f = -1; c = getchar();}
while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
return f * x;
}
int n, m, ans, a[10][10], b[85];
bool vis[3][10][10];
const int fenshu[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}
};//每个点对应的分值
struct Node
{
int Zero_geshu/*每行0的个数*/, h/*是哪一行*/;
} zz[10];
inline int getfz(int x, int y) {return (x - 1) / 3 * 3 + (y - 1) / 3 + 1;}//(i,j)对应的方阵
inline void getans()//计算分值
{
int sum = 0;
for (int i = 1; i <= 9; i+=1)
{
for (int j = 1; j <= 9; j+=1)
{
sum = sum + a[i][j] * fenshu[i][j];
}
}
if (sum > ans) ans = sum;//更新答案
}
void dfs(int bh)//搜索
{
if (bh == 82) {getans(); return;}//搜完了
int x = b[bh] / 9 + 1, y = b[bh] % 9;//求出当前的行和列
if (!y) y = 9, x = b[bh] / 9;//特判y=0
int g = getfz(x, y);//求出当前所在的方阵编号
if (a[x][y]) dfs(bh + 1);//当前位置已经有数
else
{
for (int i = 1; i <= 9; i+=1)//枚举
{
if (!vis[0][x][i] && !vis[1][y][i] && !vis[2][g][i])
{
vis[0][x][i] = vis[1][y][i] = vis[2][g][i] = 1;
a[x][y] = i;
//填数
dfs(bh + 1);
//递归
a[x][y] = 0;
vis[0][x][i] = vis[1][y][i] = vis[2][g][i] = 0;
//回溯
}
}
}
}
inline bool cmp(Node x, Node y) {return x.Zero_geshu < y.Zero_geshu;}//按每一行0的个数排序
int main()
{
//freopen(".in", "r", stdin);
//freopen(".out", "w", stdout);
for (int i = 1; i <= 9; i+=1)
{
zz[i].h = i;
int cnt = 0;
for (int j = 1; j <= 9; j+=1)
{
a[i][j] = gi();
int g = getfz(i, j);
if (a[i][j])
{
vis[0][i][a[i][j]] = 1;
vis[1][j][a[i][j]] = 1;
vis[2][g][a[i][j]] = 1;//标记有数
}
else ++cnt;//增加这一行0的个数
}
zz[i].Zero_geshu = cnt;
}
sort(zz + 1, zz + 1 + 9, cmp);//排序
int num = 0;
for (int i = 1; i <= 9; i+=1)
{
for (int j = 1; j <= 9; j+=1)
{
b[++num] = (zz[i].h - 1) * 9 + j;//编号
}
}
dfs(1);
printf("%d\n", (ans == 0) ? (-1) : (ans));//输出,注意判断-1
return 0;//结束
}
