POJ 2923 - Relocation(状态压缩+01背包)

Emma and Eric are moving to their new house they bought after returning from their honeymoon. Fortunately, they have a few friends helping them relocate. To move the furniture, they only have two compact cars, which complicates everything a bit. Since the furniture does not fit into the cars, Eric wants to put them on top of the cars. However, both cars only support a certain weight on their roof, so they will have to do several trips to transport everything. The schedule for the move is planed like this:

1. At their old place, they will put furniture on both cars.
2. Then, they will drive to their new place with the two cars and carry the furniture upstairs.
3. Finally, everybody will return to their old place and the process continues until everything is moved to the new place.

Note, that the group is always staying together so that they can have more fun and nobody feels lonely. Since the distance between the houses is quite large, Eric wants to make as few trips as possible.

Given the weights w_i of each individual piece of furniture and the capacitiesC₁ andC₂ of the two cars, how many trips to the new house does the party have to make to move all the furniture? If a car has capacityC, the sum of the weights of all the furniture it loads for one trip can be at mostC.

Input

The first line contains the number of scenarios. Each scenario consists of one line containing three numbersn,C₁ and C₂. C₁ andC₂ are the capacities of the cars (1 ≤C_i ≤ 100) andn is the number of pieces of furniture (1 ≤n ≤ 10). The following line will containn integers w₁, …,w_n, the weights of the furniture (1 ≤w_i ≤ 100). It is guaranteed that each piece of furniture can be loaded by at least one of the two cars.

Output

The output for every scenario begins with a line containing “Scenario #i:”, wherei is the number of the scenario starting at 1. Then print a single line with the number of trips to the new house they have to make to move all the furniture. Terminate each scenario with a blank line.

Sample Input

26 12 133 9 13 3 10 117 1 1001 2 33 50 50 67 98

Sample Output

Scenario #1:2Scenario #2:3

                                                              

题意：

n(1<=n<=10)个物品, 两辆卡车的容量c1, c2. 给出n个物品的体积，两辆卡车同时运作，要求出最少运输多少次能把货物运走。

思路：

状态压缩DP+01背包。

先枚举选择若干个的状态，使用位运算进行枚举。总的状态量为1<<n，判断这些物品能否一次性运走，若能则这一状态看成一个物品。

预处理后得到ii 种状态，在这些状态中使用01 背包，注意判断两个状态不能含有同一种物品，则转化为每一个物品的体积为state[i], 价值为1。

题目转化为求出最小价值，则dp[]初始化为inf。

CODE：

#include <iostream>#include <cstdio>#include <algorithm>#include <cmath>#include <string>#include <cstring>#include <queue>#include <stack>#include <vector>#include <set>#include <map>const int inf=0xfffffff;typedef long long ll;using namespace std;int c[15], dp[1030], state[1030], vis[1030];int n, c1, c2;bool judge(int x){    int s = 0;    memset(vis, false, sizeof(vis));    vis[0] = true;    for(int i = 0; i < n; ++i){        if((1<<i) & x){            s += c[i];            for(int j = c1; j >= c[i]; --j){                if(vis[j - c[i]]) vis[j] = true;            }        }    }    if(s > c1 + c2) return false;    for(int i = 0; i <= c1; ++i){        if(vis[i] && s - i <= c2) {            return true;        }    }    return false;}int main(){    //freopen("in", "r", stdin);    int T, ca = 0;    scanf("%d", &T);    while(T--){        ca ++;        scanf("%d %d %d", &n, &c1, &c2);        for(int i = 0;i < n; ++i){            scanf("%d", &c[i]);        }        for(int i = 0; i < (1<<n); ++i){            dp[i] = inf;        }        dp[0] = 0;        int ii = 0;        for(int i = 1; i < (1<<n); ++i){            if(judge(i)) state[ii++] = i;        }        for(int i = 0; i < ii; ++i){            for(int j = (1<<n) - 1; j >= 0; --j){                if(dp[j] == inf) continue;                if((j & state[i]) == 0){                    dp[j | state[i]] = min(dp[j | state[i]], dp[j] + 1);                }            }        }        printf("Scenario #%d:\n%d\n\n", ca, dp[(1<<n) - 1]);    }    return 0;}