POJ 2923 Relocation ★（状态压缩+01背包）

题目大意：有n个物品，有两辆车载重分别是c1,c2.问需要多少趟能把物品运完。好题~用到状态压缩思想的01背包。先枚举选若干个时的状态，总状态量为1<<n，判断集合里的物品能否用两辆车一次运走,如果能运走，那就把这个状态看成一个物品。预处理完找到tot个物品，再对这tot个物品进行01背包处理，每个物品的体积是state[i]，价值是1，求必选n个物品的最少价值。状态转移方程:dp[j|k] = min(dp[j|k],dp[k]+1) (k为state[i,1<=j<=(1<<n)-1])

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#define MID(x,y) ((x+y)>>1)
#define mem(a,b) memset(a,b,sizeof(a))
using namespace std;

typedef long long LL;
const int sup = 0x7fffffff;
const int inf = -0x7fffffff;

int n, c1, c2;
int w[11];
vector  state;
int dp[1 << 11];

bool ifok(int state){
    bool f[101];
    int sum = 0;
    mem(f, 0);
    f[0] = 1;
    for (int i = 0; i < n; i ++)
        if ((1 << i) & state){
            sum += w[i];
            for (int j = c1; j >= w[i]; j --){
                f[j] = f[j] || f[j-w[i]];
            }
        }
    if (sum > c1 + c2)
        return 0;
    for (int i = 0; i <= c1; i ++)
        if (f[i] && (sum - i <= c2))
            return 1;
    return 0;
}

int main(){
    int t;
    scanf("%d", &t);
    for (int caseo = 1; caseo <= t; caseo ++){
        scanf("%d %d %d", &n, &c1, &c2);
        for (int i = 1; i <= n; i ++)
            scanf("%d", &w[i]);
        for (int i = 0; i < (1 << n); i ++){
            dp[i] = sup;
            if (ifok(i)){
                state.push_back(i);
            }
        }
        dp[0] = 0;
        for (int i = 0; i < (int)state.size(); i ++){
            for (int j = 0; j < (1 << n); j ++){
                if (dp[j] != sup){
                    int p = state[i];
                    if (j & p == 0){
                        dp[j | p] == min(dp[j | p], dp[j] + 1);
                    }
                }
            }
        }
        printf("Scenario #%d:\n", caseo);
        printf("%d\n", dp[(1<

举杯独醉，饮罢飞雪，茫然又一年岁。 ------AbandonZHANG