hdu 5691 Sitting in Line 状压dp

题目链接：

http://acm.hdu.edu.cn/showproblem.php?pid=5691

题解：

和tsp用的状压差不多，就是固定了一些访问顺序。

dp[i][j]表示前cnt个点中布满状态i且最后一个为j的状态的最大乘积和。

则有dp[i|(1<<k)][k]=max(dp[i|(1<<k)][k],dp[i][j]+a[j]*a[k])。

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
using namespace std;

const int maxn = 22;
const int INF = 2e9;
int dp[1 << 16][22];
int cnt[1 << 16];
int a[maxn], p[maxn],f[maxn];
int n;

void pre() {
    for (int i = 0; i < (1 << 16); i++) {
        cnt[i] = 0;
        for (int j = 0; j < 16; j++) {
            if (i&(1 << j)) cnt[i]++;
        }
    }
}

void init() {
    for (int i = 0; i < (1 << n); i++) {
        for (int j = 0; j <= n; j++) {
            dp[i][j] = -INF;
        }
    }
    memset(f, -1, sizeof(f));
}

int main() {
    pre();
    int tc,kase=0;
    scanf("%d", &tc);
    while (tc--) {
        scanf("%d", &n);
        init();
        for (int i = 0; i < n; i++) {
            scanf("%d%d", a + i, p + i);
            if (p[i] != -1) f[p[i]] = i;
        }
        a[n] = 0; p[n] = n;
        dp[0][n] = 0; 
        for (int i = 0; i < (1 << n); i++) {
            int sum = cnt[i];
            for (int j = 0; j <= n; j++) {
                if ((i&(1 << j)) == 0&&j!=n) continue;
                //被限制的点：
                if (f[sum] != -1) {
                    if ((i&(1 << f[sum])) == 0) {
                        dp[i | (1 << f[sum])][f[sum]] =
                            max(dp[i | (1 << f[sum])][f[sum]], dp[i][j]+a[j]*a[f[sum]]);
                    }
                }
                else {
                    //可以自由移动的点
                    for (int k = 0; k < n; k++) {
                        if (i&(1 << k)) continue;
                        if (p[k] ==-1) {
                            dp[i | (1 << k)][k] = max(dp[i | (1 << k)][k], dp[i][j] + a[j] * a[k]);
                        }
                    }
                }
            }
        }
        int ans = -INF;
        for (int j = 0; j < n; j++) ans = max(ans, dp[(1 << n) - 1][j]);
        printf("Case #%d:
", ++kase);
        printf("%d
", ans);
    }
    return 0;
}

再一发：

#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<stack>
#include<ctime>
#include<vector>
#include<cstdio>
#include<string>
#include<bitset>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
using namespace std;
#define X first
#define Y second
#define mkp make_pair
#define lson (o<<1)
#define rson ((o<<1)|1)
#define mid (l+(r-l)/2)
#define sz() size()
#define pb(v) push_back(v)
#define all(o) (o).begin(),(o).end()
#define clr(a,v) memset(a,v,sizeof(a))
#define bug(a) cout<<#a<<" = "<<a<<endl
#define rep(i,a,b) for(int i=a;i<(b);i++)
#define scf scanf
#define prf printf

typedef long long LL;
typedef vector<int> VI;
typedef pair<int,int> PII;
typedef vector<pair<int,int> > VPII;

const int INF=0x3f3f3f3f;
const LL INFL=10000000000000000LL;
const double eps=1e-9;

const double PI = acos(-1.0);

//start----------------------------------------------------------------------

const int maxn=17;

LL dp[1<<maxn][maxn];
LL arr[maxn];
int pos[maxn],mp[maxn];
int sumv[1<<maxn];
int n;

void pre(){
    clr(sumv,0);
    for(int i=0;i<(1<<maxn);i++){
        for(int j=0;j<maxn;j++){
            if(i&(1<<j)){
                sumv[i]++;
            }
        }
    }
}

void init(){
    clr(mp,-1);
}

int main() {
    pre();
    int tc,kase=0;
    scf("%d",&tc);
    while(tc--){
        scf("%d",&n);
        init();
        rep(i,0,n){
            scf("%lld%d",&arr[i],&pos[i]);
            if(pos[i]>=0) mp[pos[i]]=i;
        }

        rep(i,0,(1<<maxn)) rep(j,0,maxn) dp[i][j]=-INFL;
        if(mp[0]>=0){
            dp[1<<mp[0]][mp[0]]=0;
        }else{
            for(int i=0;i<n;i++){
                if(pos[i]>=0) continue;
                dp[1<<i][i]=0;
            }
        }

        rep(i,1,(1<<n)){
            rep(j,0,n){
                if(!(i&(1<<j))) continue;
                if(mp[sumv[i]-1]>=0&&mp[sumv[i]-1]!=j) continue;
                rep(k,0,n){
                    if(k==j||!(i&(1<<k))) continue;
                    if(mp[sumv[i^(1<<j)]-1]>=0&&mp[sumv[i^(1<<j)]-1]!=k) continue;
                    dp[i][j]=max(dp[i][j],dp[i^(1<<j)][k]+arr[k]*arr[j]);
                }
            }
        }

        LL ans=-INFL;
        rep(i,0,n) ans=max(ans,dp[(1<<n)-1][i]);
        prf("Case #%d:
",++kase);
        prf("%lld
",ans);

    }
    return 0;
}

//end-----------------------------------------------------------------------