LuoguP1283 平板涂色(状压DP)

参考了I_AM_HelloWord的代码，\(f[i][j]\)表示转态\(i\)时最后一刷为\(j\)的最小代价，上面的块可用暴力填涂，注意边界

#include <cstdio>
#include <iostream>
#include <cmath>
#include <algorithm>
#include <cstring>
#define R(a,b,c) for(register int a = (b); a <= (c); ++a)
#define nR(a,b,c) for(register int a = (b); a >= (c); --a)
#define Fill(a,b) memset(a,b,sizeof(a))
#define Swap(a,b) ((a) ^= (b) ^= (a) ^= (b))
#define QWQ
#ifdef QWQ
#define D_e(x) cout << (#x) << " : " << x << "\n"
#define D_e_Line printf("\n----------------\n")
#define FileOpen() freopen("in.txt", "r", stdin)
#define FileSave() freopen("out.txt","w", stdout)
#define Pause() system("pause")
#define TIME() fprintf(stderr, "TIME : %.3lfms\n", clock() / CLOCKS_PER_SEC)
#endif
struct FastIO {
	template<typename ATP> inline FastIO& operator >> (ATP &x) {
		x = 0; int f = 1; char c;
		for(c = getchar(); c < '0' || c > '9'; c = getchar()) if(c == '-') f = -1;
		while(c >= '0' && c <= '9') x =x * 10 + (c ^ '0'), c = getchar();
		x = f == 1 ? x : -x;
		return *this;
	}
} io;
using namespace std;
template<typename ATP> inline ATP Max(ATP x, ATP y) {
	return x > y ? x : y;
}
template<typename ATP> inline ATP Min(ATP x, ATP y) {
	return x < y ? x : y;
}

#include <assert.h> 

const int N = 21;

int lx[N], ly[N], rx[N], ry[N], tot[N], col[N], f[1 << 17][N], sta[N][N], mp[N][N];

inline bool Check(int x, int s) {
	bool flag = true;
	R(i,1,tot[x]){
//		D_e(sta[x][i] - 1);
		assert(sta[x][i] - 1 >= 0);
		flag &= ((s >> (sta[x][i] - 1)) & 1);
		if(flag == false) return false;
	}
	return true;
}
int main() {
	int n;
	io >> n;

	R(i,1,n){
		io >> lx[i] >> ly[i] >> rx[i] >> ry[i] >> col[i];
		R(x,lx[i],rx[i] - 1){
			R(y,ly[i],ry[i] - 1){
				assert(x >= 0 && y >= 0);
				mp[x][y] = i;
			}
		}
	}
	R(i,1,n){
		if(!lx[i]) continue;
		--lx[i];
		R(j,ly[i] + 1, ry[i]){
			if(mp[lx[i]][j] != mp[lx[i]][j - 1]) sta[i][++tot[i]] = mp[lx[i]][j - 1];
		}
		if(mp[lx[i]][ry[i]] == mp[lx[i]][ry[i] - 1]) sta[i][++tot[i]] = mp[lx[i]][ry[i] - 1];
	}
	Fill(f, 0x3f);
	R(i,1,20) f[0][i] = 1;
	int maxn = (1 << n) - 1;
	R(s,1,maxn){
		R(i,1,n){
			if(((s >> (i - 1)) & 1) && Check(i, s)){
				R(j,1,20){
					if(j == col[i]) continue;
					assert((s ^ (1 << (i - 1))) >= 0);
					f[s][col[i]] = Min(f[s][col[i]], f[s ^ (1 << (i - 1))][j] + 1);
				}
				f[s][col[i]] = Min(f[s][col[i]], f[s ^ (1 << (i - 1))][col[i]]);
			}
		}
	}

	int ans = 2147483647;
	R(i,1,20) ans = Min(ans, f[maxn][i]);

	printf("%d", ans);

	return 0;
}
/*
7
0 0 2 2 1
0 2 1 6 2
2 0 4 2 1
1 2 4 4 2
1 4 3 6 1
4 0 6 4 1
3 4 6 6 2
*/