POJ - 1691 Painting A Board (状态压缩 ＋ 暴力)

题目大意：要将一个大矩形內的所有小矩形涂色，涂色要求，只有该矩形上面的所有矩形都涂色了才可以涂该颜色，换一种填涂的颜色就要花费一点体力值，问填涂完需要花费的最小体力值

解题思路：先处理一下填涂该矩形的前提条件，我用了一个can数组表示填涂该矩形时要满足的状态量

用dp[color][state]表示当前用的颜色是color，填涂的状态为state时所用的最少体力值

然后暴力得出转换和结果

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
using namespace std;
#define maxn 15
#define maxc 22
#define maxs (1 << 15)
#define INF 0x3f3f3f3f

struct Rectangles{
    int x1, x2, y1, y2, color;
}Rec[maxn];

struct paiting{
    int s, color;
}start;

int n;
int can[maxn];
int dp[maxc][maxs];

bool judge(int i, int j) {
    if (Rec[j].y2 <= Rec[i].y1)
        if (((Rec[j].x1 <= Rec[i].x1 && Rec[j].x2 > Rec[i].x1 && Rec[j].x2 <= Rec[i].x2) || (Rec[j].x2 >= Rec[i].x2 && Rec[j].x1 >= Rec[i].x1 && Rec[j].x1 < Rec[i].x2)))
            return true;
    return false;
}

void init() {
    scanf("%d", &n);
    for (int i = 0; i < n; i++)
        scanf("%d%d%d%d%d", &Rec[i].y1, &Rec[i].x1, &Rec[i].y2, &Rec[i].x2, &Rec[i].color);
    memset(can, 0, sizeof(can));
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            if (i == j)
                continue;
            if (judge(i, j))
                can[i] |= (1 << j);
        }
    }
    memset(dp, 0x3f, sizeof(dp));
    dp[0][0] = 0;
}

void solve() {
    queue<paiting> q;
    start.s = 0;
    start.color = 0;
    q.push(start);
    while (!q.empty()) {
        paiting t = q.front();
        q.pop();
        int state = t.s;
        int color = t.color;
        for (int i = 0; i < n; i++)
            if ((state & (1 << i)) == 0 && (state & can[i]) == can[i]) {
                if (Rec[i].color == color) {
                    dp[color][state | (1 << i)] = min(dp[color][state | (1 << i)], dp[color][state]);
                }
                else {
                    dp[Rec[i].color][state | (1 << i)] = min(dp[Rec[i].color][state | (1 << i)], dp[color][state] + 1);
                }
                paiting tt;
                tt.s = state | (1 << i);
                tt.color = Rec[i].color;
                q.push(tt);
            }
    }

    int ans = INF;
    for (int i = 0; i < maxc; i++)
        ans = min(ans, dp[i][(1 << n) - 1]);
    printf("%d\n", ans);
}

int main() {
    int test;
    scanf("%d", &test);
    while (test--) {
        init();
        solve();

    }
    return 0;
}

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