POJ 2923Relocation（状态压缩入门）
Description
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:
- At their old place, they will put furniture on both cars.
- Then, they will drive to their new place with the two cars and carry the furniture upstairs.
- 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 wi of each individual piece of furniture and the capacities C1 and C2 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 capacity C, the sum of the weights of all the furniture it loads for one trip can be at most C.
Input
The first line contains the number of scenarios. Each scenario consists of one line containing three numbers n, C1 and C2. C1 and C2 are the capacities of the cars (1 ≤ Ci ≤ 100) and n is the number of pieces of furniture (1 ≤ n ≤ 10). The following line will contain n integers w1, …, wn, the weights of the furniture (1 ≤ wi ≤ 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:”, where i 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
2 6 12 13 3 9 13 3 10 11 7 1 100 1 2 33 50 50 67 98
Sample Output
Scenario #1: 2 Scenario #2: 3
Source
TUD Programming Contest 2006, Darmstadt, Germany
年代久远的状压题，这是我的第一道状压题。。。。
大致讲下状态压缩的思想与操作吧。。。
题意：
两辆车n个物品，每个物品有体积，两辆车也有体积，
要求把物品全部运走最少需要多少次
每次每辆车运送的物体总体积不得大于车的体积
分析：
物品个数最多是10个，可以用0和1分别表示物品是否被选中运送
假设有3个物品，那么可以用001表示当前这一次选择将第3个物品运走
那么，所有的状态可以用0~2^n-1对应的二进制数表示出来
对于上述每一种状态，选择其中可以一次运走的状态进行01背包
其中，所有位全部为1的二进制数表示背包总体积
每个物品的体积是该状态对应二进制数中1的个数
为了把物品全部运走，要选择一些状态把背包的1全部被填满
值得注意的是，因为一件物品不会被运2次，所以所有选取的状态应该是没有交集的
比如1001101和1110010是不行的，第一件物品被运了2次
然后为了使次数最小，可以将次数抽象成01背包中的物品价值，每种状态对应运送一次，价值是1
如何判断某种状态是否可以一次运走？
因为有2辆车，所以将要判断该状态是否能分成满足体积分别小于等于车体积的2份，
如何从十进制中取出该状态被选择的物体？
也就是将二进制中的1的位置取出来
利用 ：>> :右移运算符 << 左移运算符 & 按位与运算符
假设有一个二进制数： x = 01101
那么将它右移2位 : x>>=2,他会变成 00011（移出位被丢弃，左边移出的空位或者一律补0，或者补符号位）
我们知道1的二进制数是最后一位为1其他全为0，如果某数的二进制从又往左数第2位是1，移位之后这个1
变成最后一位，和二进制只有最后一位是1的数字1&之后还是1，而如果某数的二进制从又往左数第2位是0
的话，和二进制只有最后一位是1的数字1&之后将是0（因为1除了最后一位其他位全为0，而该数移位后最后一位是0）
综上所述 (x>>i)&1可以判断x从右往左第i位是0还是1
知道怎么找到二进制数中1的位置基本状压的代码也不难懂了
直接上代码：
#include<iostream>
#include<algorithm>
#include<string>
#include<cstring>
#include<cstdio>
#include<cmath>
#define mem(a,x) memset(a,x,sizeof(a))
#define inf 1<<30
using namespace std;
const double PI = acos(-1.0);
typedef long long LL;
int state[1030];
int dp[1030];
bool vis[1005];
int n,v1,v2,tot;
int c[12];
bool ok(int x)//判断一种状态是否可行（可以一次运走）
{
int sun = 0;
mem(vis,0);
vis[0] = 1;
for (int i = 0;i < n;i++)
{
if ((x>>i)&1)
{
sun += c[i];
for (int j = v1;j >= c[i];j--) //这个真的非常巧妙 开始看半天都不懂，自己模拟一遍才懂
{ //比如说此状态有c1、c2、c3,3个体积，第一次操作把体积c1标记为1，
if (vis[j-c[i]]) //第二次操作把c2和c1+c2两种体积标记为1，第三次把c3和前面的组合标记为1，
vis[j] = 1; //最后这些体积能组合成的所有体积就都被标记成了1
}
}
}
if (sun > v1+v2)//装不下
return 0;
//总体积小不代表一定装得下，拆分成2份要2份都装得下
for (int i = 0;i <= v1;i++)
{
if (vis[i]&&sun-i <= v2)//如果存在（i,sun-i）这样的组合
return 1; //满足i可以被v1装下(前面for循环是对于v1的，vis[i]表示体积i可以被v1装下)，sun-i可以被v2装下
}
return 0;
}
void init()//初始化找到满足条件的状态
{
tot = 0;
for (int i = 1;i < (1<<n);i++)
{
dp[i] = inf;
if (ok(i))
state[tot++] = i;
}
}
int main()
{
int T;
cin>>T;
int oo = 0;
while (T--)
{
cin>>n>>v1>>v2;
for (int i = 0;i < n;i++)
scanf("%d",&c[i]);
init();
int V = (1<<n) - 1;//V是n个1的二进制数
dp[0] = 0;//没有物品当然是0次运走
for (int i = 0;i < tot;i++)
{
for (int j = V;j >= 0;j--)
{
if (dp[j] == inf)
continue; //原版的背包是dp[j] = min(dp[j],dp[j-c[i]]+w[i])
//但是显然二进制不好表示减，但是可以用|抽象加
//这就相当于背包改版成dp[j+c[i]] = min(dp[j+c[i]],dp[j] + w[i])
if ((j&state[i])==0) //当然2种状态不能有交集
{
dp[j|state[i]] = min(dp[j|state[i]] ,dp[j] + 1);
}
}
}
printf("Scenario #%d:\n%d\n",++oo,dp[V]);
if (T) puts("");
}
return 0;
}
参考大神题解：http://www.cnblogs.com/kuangbin/archive/2012/09/14/2685430.html