题意：N(1 <= N <= 15)个矩形，每个矩形要涂上指定的颜色C(1 <= C <= 20)，如果给一个矩形涂色，那么与它相邻的上方矩形必须已经涂色，问最少要取几次画笔。

题目链接：http://poj.org/problem?id=1691

——>>状态：dp[S][color] 表示达到状态 S 且最后一次涂色为 color 时的最小取画笔数

状态转移方程：dp[S][color] = min(dp[S][color], dp[sub][i]); 或者 dp[S][color] = min(dp[S][color], dp[sub][i] + 1);

时间复杂度：O(N * C * 2 ^ N)

#include <cstdio>
#include <algorithm>
#include <cstring>

using std::min;

const int MAX_COLOR = 20;
const int MAX_RECTANGLE = 15;
const int INF = 0x3f3f3f3f;

struct RECTANGLE
{
    int ly;
    int lx;
    int ry;
    int rx;
    int color;
    int fa[MAX_RECTANGLE];
    int cnt;
} r[MAX_RECTANGLE];

int M, N;
int dp[1 << MAX_RECTANGLE][MAX_COLOR + 1];

void Read()
{
    scanf("%d", &N);
    for (int i = 0; i < N; ++i)
    {
        scanf("%d%d%d%d%d", &r[i].ly, &r[i].lx, &r[i].ry, &r[i].rx, &r[i].color);
        r[i].cnt = 0;
    }
}

void Init()
{
    for (int i = 0; i < N; ++i)
    {
        for (int j = 0; j < N; ++j)
        {
            if (j != i && r[j].ry == r[i].ly && r[j].rx > r[i].lx && r[j].lx < r[i].rx)
            {
                r[i].fa[r[i].cnt++] = (1 << j);
            }
        }
    }
}

bool IsReady(int x, int S)
{
    for (int i = 0; i < r[x].cnt; ++i)
    {
        if (!(r[x].fa[i] & S))
        {
            return false;
        }
    }

    return true;
}

int Idx(int x)
{
    int i = 0;

    while ((1 << i) != x)
    {
        ++i;
    }

    return i;
}

void Dp()
{
    memset(dp, 0x3f, sizeof(dp));
    for (int i = 1; i <= MAX_COLOR; ++i)
    {
        dp[0][i] = 1;
    }

    for (int S = 1; S <= (1 << N) - 1; ++S)
    {
        for (int t = S; t; t -= t & -t)
        {
            int cur = Idx(t & -t);
            int sub = S & ~(t & -t);
            int color = r[cur].color;

            if (IsReady(cur, sub))
            {
                for (int i = 1; i <= MAX_COLOR; ++i)
                {
                    if (color == i)
                    {
                        dp[S][color] = min(dp[S][color], dp[sub][i]);
                    }
                    else
                    {
                        dp[S][color] = min(dp[S][color], dp[sub][i] + 1);
                    }
                }
            }
        }
    }
}

void Output()
{
    int ret = INF;

    for (int i = 1; i <= 20; ++i)
    {
        ret = min(ret, dp[(1 << N) - 1][i]);
    }

    printf("%d\n", ret);
}

int main()
{
    scanf("%d", &M);
    while (M--)
    {
        Read();
        Init();
        Dp();
        Output();
    }

    return 0;
}

poj - 1691 - Painting A Board（状态压缩dp）