1.小猫爬山
题目链接
#include<iostream>
#include<algorithm>
using namespace std;
const int N=20;
int cat[N],car[N];
int n,m;
int ans=N;
void dfs(int k,int l)
{
if(l>=ans) return;
if(k==n)
{
ans=l;
return;
}
for(int i=0;i<l;i++)
{
if(cat[k]+car[i]<=m)
{
car[i]+=cat[k];
dfs(k+1,l);
car[i]-=cat[k];
}
}
car[l]=cat[k];
dfs(k+1,l+1);
car[l]=0;
}
int main()
{
cin>>n>>m;
for(int i=0;i<n;i++)
cin>>cat[i];
sort(cat,cat+n,greater<int>());
dfs(0,0);
cout<<ans<<endl;
}
2.sudoku
#include <iostream>
using namespace std;
const int N = 9;
int ones[1 << N], map[1 << N];
int row[N], col[N], cell[5][5];
string str;
inline int lowbit(int x)
{
return x & -x;
}
void init()
{
for (int i = 0; i < N; i++)
row[i] = col[i] = (1 << N) - 1; //初始状态每一个数都可以选
for (int i = 0; i < 3; i++)
for (int j = 0; j < 3; j++)
cell[i][j] = (1 << N) - 1;
}
inline int get(int x, int y)
{
return row[x] & col[y] & cell[x / 3][y / 3]; //求交集
}
bool dfs(int cnt)
{
if (!cnt)
return true;
int minv = 10;
int x, y;
for (int i = 0; i < N; i++)
for (int j = 0; j < N; j++)
if (str[i * 9 + j] == '.')
{
int t = ones[get(i, j)];
if (t < minv)
{
minv = t;
x = i, y = j;
}
}
for (int i = get(x, y); i; i -= lowbit(i))
{
int t = map[lowbit(i)];
// 修改状态
row[x] -= 1 << t;
col[y] -= 1 << t;
cell[x / 3][y / 3] -= 1 << t;
str[x * 9 + y] = '1' + t;
if (dfs(cnt - 1))
return true;
// 恢复现场
row[x] += 1 << t;
col[y] += 1 << t;
cell[x / 3][y / 3] += 1 << t;
str[x * 9 + y] = '.';
}
return false;
}
int main()
{
for (int i = 0; i < N; i++)
map[1 << i] = i;
for (int i = 0; i < 1 << N; i++) //计算一个二进制数包含多少个1
{
int s = 0;
for (int j = i; j > 0; j -= lowbit(j))
s++;
ones[i] = s;
}
while (cin >> str, str[0] != 'e')
{
init();
int cnt = 0; //需要填的个数
for (int i = 0, k = 0; i < N; i++)
for (int j = 0; j < N; j++, k++)
{
if (str[k] != '.')
{
int t = str[k] - '1'; //1~9变成0~8
row[i] -= 1 << t;
col[j] -= 1 << t;
cell[i / 3][j / 3] -= 1 << t;
}
else
cnt++;
}
dfs(cnt);
cout << str << endl;
}
}