写了一下午，终于AC了。
由于n<=5, 所以不需要太多的剪枝和技巧也能过。可以将操作后的消方块和下落和剪枝函数写到一个结构体中，这样会减少调试难度，更加简洁。
可以采用如下剪枝：
1. 如果当前有一种颜色少于两个，则一定无解。
2. 如果相邻颜色相同则不交换。
3. 优先考虑右移
4. 其实可以开一个unordered_map来避免重复搜索，由于笔者已经筋疲力尽，就没有进行这部优化（但其实n的规模太小，不开也无所谓）。
代码：
#include<cstdio>
#include<vector>
#include<algorithm>
#include<cstring>
#include<iostream>
#include<deque>
using namespace std;;
const int n = 7;
const int m = 5;
int board[10][10], spare[10][10], target;
struct Ans
{
int x,y,g;
Ans(int x = 0, int y = 0,int g = 0):x(x),y(y),g(g){}
};
deque<Ans>Q;
struct Operation
{
inline int get_num()
{
int cnt = 0;
for(int i = 1;i <= n; ++i)
for(int j = 1;j <= m; ++j)
if(spare[i][j])++cnt;
return cnt;
}
inline int cut()
{
int col[16];
memset(col,0,sizeof(col));
for(int i = 1;i <= n;++i)
for(int j = 1;j <= m;++j) ++ col[spare[i][j]];
for(int i = 1;i <= 15;++i)
if(col[i] && col[i] <= 2)return 1;
return 0;
}
inline int clear_block()
{
int flag = 0;
int mark[10][10];
memset(mark,0,sizeof(mark));
for(int i = 1;i <= n;++i)
{
for(int j = 1;j <= m;++j)
{
if(!spare[i][j])continue;
if(i + 2 <= n)
{
int cur = i;
while(spare[cur + 1][j] == spare[cur][j]) ++cur;
if(cur - i + 1 >= 3)
{
flag = 1;
int y = i;
while(spare[y][j] == spare[i][j]) mark[y][j] = 1, y += 1;
}
}
if(j + 2 <= m)
{
int cur = j;
while(spare[i][cur + 1] == spare[i][cur])++cur;
if(cur - j + 1 >= 3)
{
flag = 1;
int y = j;
while(spare[i][y] == spare[i][j]) mark[i][y] = 1, y += 1;
}
}
}
}
for(int i = 1;i <= n; ++i)
for(int j = 1;j <= m; ++j)
if(mark[i][j])spare[i][j] = 0;
return flag;
}
inline void down()
{
for(int col = 1; col <= m; ++col)
{
int bottom = n, cur = n;
while(bottom >= 1 && cur >= 1)
{
while(spare[bottom][col] && bottom >= 1) --bottom;
cur = bottom;
while(cur >= 1 && !spare[cur][col]) --cur;
if(cur >= 1)
{
int i = bottom ,j = cur;
while(spare[j][col] && j >= 1 && i >= 1)
{
spare[i][col] = spare[j][col];
spare[j][col] = 0;
--i;
--j;
}
}
}
}
}
inline void update()
{
down();
while(clear_block()) down();
}
}T;
void dfs(int nums,int arr[10][10])
{
for(int i = 1;i <= n;++i)
for(int j = 1;j <= m;++j)spare[i][j] = arr[i][j];
T.update();
for(int i = 1;i <= n;++i)
for(int j = 1;j <= m;++j)arr[i][j] = spare[i][j];
if(nums == target)
{
if(!T.get_num())
{
while(!Q.empty())
{
Ans p = Q.front();
printf("%d %d %d
",p.x-1, 7-p.y, p.g);
Q.pop_front();
}
exit(0);
}
return;
}
if(!T.get_num() || T.cut())return;
int h[10][10];
memset(h,0,sizeof(h));
for(int i = 1;i <= n;++i)
for(int j = 1;j <= m;++j)h[i][j] = arr[i][j];
for(int j = 1;j <= m;++j)
for(int i = 7; i >= 1;--i)
{
if(!h[i][j])continue;
if(j < m && arr[i][j] != arr[i][j+1])
{
swap(h[i][j],h[i][j+1]);
Q.push_back(Ans(j,i,1));
dfs(nums+1,h);
Q.pop_back();
for(int i = 1;i <= n;++i)
for(int j = 1;j <= m;++j)h[i][j] = arr[i][j];
}
if(j > 1 && !arr[i][j-1])
{
swap(h[i][j],h[i][j-1]);
Q.push_back(Ans(j,i,-1));
dfs(nums+1,h);
Q.pop_back();
swap(h[i][j],h[i][j-1]);
for(int i = 1;i <= n;++i)
for(int j = 1;j <= m;++j)h[i][j] = arr[i][j];
}
}
}
int main()
{
//freopen("in.txt","r",stdin);
scanf("%d",&target);
for(int i = 1;i <= 5; ++i)
{
for(int j = 7; ;--j)
{
int a;
scanf("%d",&a);
if(!a)break;
board[j][i] = a;
}
}
dfs(0,board);
printf("-1");
return 0;
}