BZOJ 1226 学校食堂(状压DP)

状压DP 

f(i,j,k)表示前i−1个人已经吃了饭，且在i之后的状态为j的人也吃了饭（用二进制表示后面的状态），最后吃的那个人是i之后的第k个 
（注意k可以是负数） 
然后 
如果j&1=1那么就表明第i个人也是吃了的，所以可以转移到f(i+1,j>>1,k−1) 
否则就枚举下一个吃饭的人，转移到f(i,j+1<<l,l) 
这么看也不是很难吧哈。。

# include <cstdio>
# include <cstring>
# include <cstdlib>
# include <iostream>
# include <vector>
# include <queue>
# include <stack>
# include <map>
# include <set>
# include <cmath>
# include <algorithm>
using namespace std;
# define lowbit(x) ((x)&(-x))
# define pi acos(-1.0)
# define eps 1e-9
# define MOD 1000000000
# define INF 1000000000
# define mem(a,b) memset(a,b,sizeof(a))
# define FOR(i,a,n) for(int i=a; i<=n; ++i)
# define FO(i,a,n) for(int i=a; i<n; ++i)
# define bug puts("H");
# define lch p<<1,l,mid
# define rch p<<1|1,mid+1,r
//# define mp make_pair
# define pb push_back
typedef pair<int,int> PII;
typedef vector<int> VI;
# pragma comment(linker, "/STACK:1024000000,1024000000")
typedef long long LL;
int Scan() {
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
void Out(int a) {
    if(a<0) {putchar('-'); a=-a;}
    if(a>=10) Out(a/10);
    putchar(a%10+'0');
}
const int N=1005;
//Code begin...

struct Node{int t, b;}node[N];
int dp[N][1<<8][20];

int main ()
{
    int T, n;
    scanf("%d",&T);
    while (T--) {
        scanf("%d",&n); FO(i,0,N) FO(j,0,1<<8) FO(k,0,20) dp[i][j][k]=INF;
        FOR(i,1,n) scanf("%d%d",&node[i].t,&node[i].b);
        int mi=INF;
        FOR(i,1,n) {
            --mi;
            if (mi<0) break;
            mi=min(mi,node[i].b);
            dp[1][1<<(i-1)][(i-1)+8]=0;
        }
        FOR(i,1,n) FO(j,0,1<<8) FOR(k,0,15) {
            if (dp[i][j][k]==INF) continue;
            if (j&1) dp[i+1][j>>1][k-1]=min(dp[i+1][j>>1][k-1],dp[i][j][k]);
            else {
                mi=INF;
                FOR(l,i,n) {
                    --mi;
                    if (mi<0) break;
                    if ((j&(1<<(l-i)))==0) {
                        dp[i][j|(1<<(l-i))][l-i+8]=min(dp[i][j|(1<<(l-i))][l-i+8],dp[i][j][k]+(node[l].t^node[i+k-8].t));
                        mi=min(mi,node[l].b);
                    }
                }
            }
        }
        int ans=INF;
        FOR(i,0,15) ans=min(ans,dp[n+1][0][i]);
        printf("%d
",ans);
    }
    return 0;
}