POJ 2923 Relocation（状态压缩 + 两次DP）

题意：

有n个家具要搬运，两辆搬运车的容量分别是c1, c2，搬运的过程要求两辆搬运车一起来回。问最少需要多少趟能把家具搬完。

思路：

1. 因为n的范围是 n <= 10, 所以可以把家具压缩成状态 s

2. 判断状态 s 所表达的家具是否能有两辆车一次搬运完成，如果能则把状态 s 在 state[] 数组记录下来为 state[] = s

3. 对 state[] 状态数组再进行一次背包，状态的价值为 1，求最少需要多少趟能完成搬运 ：dp[s|state[i]] = min(dp[s|state[i]], dp[s] + 1);

 

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <algorithm>

using namespace std;



int w[20];

int n, C1, C2;

int dp[1100], state[1100];

bool pack[1100];



bool judge(int s)

{

    int sum = 0;

    

    memset(pack, false, sizeof(pack));

    pack[0] = true;



    for (int i = 0; i < n; ++i)

        if (s & (1 << i))

        {

            sum += w[i];

            for (int v = C1; v >= w[i]; --v)

                if (pack[v-w[i]])

                    pack[v] = true;

        }

    for (int v = 0; v <= sum; ++v)

        if (pack[v] && sum - v <= C2)

            return true;



    return false;

}



int main()

{

    int cases;

    int cnt = 0;

    scanf("%d", &cases);

    while (cases--)

    {

        scanf("%d %d %d", &n, &C1, &C2);

        for (int i = 0; i < n; ++i)

            scanf("%d", &w[i]);



        int m = 0;

        const int inf = (1 << n) - 1;



        memset(dp, 0x3f, sizeof(dp));

        memset(state, 0, sizeof(state));



        for (int s = 0; s <= inf; ++s)

            if (judge(s))

                state[m++] = s;



        dp[0] = 0;

        for (int i = 0; i < m; ++i)

            for (int s = inf - state[i]; s >= 0; --s)

                if (!(s & state[i]))

                    dp[s|state[i]] = min(dp[s|state[i]], dp[s] + 1);



        printf("Scenario #%d:\n%d\n\n", ++cnt, dp[inf]);

    }

    return 0;

}