ICPC2019银川区域赛 L. Xian Xiang

链接

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题解

注意到非空格子数 $\le 18$，所以我们可以进行状压 $dp$

代码

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#define iinf 0x3f3f3f3f
#define linf (1ll<<60)
#define eps 1e-8
#define maxn 1000010
#define maxe 1000010
#define cl(x) memset(x,0,sizeof(x))
#define rep(i,a,b) for(i=a;i<=b;i++)
#define drep(i,a,b) for(i=a;i>=b;i--)
#define em(x) emplace(x)
#define emb(x) emplace_back(x)
#define emf(x) emplace_front(x)
#define fi first
#define se second
#define de(x) cerr<<#x<<" = "<<x<<endl
using namespace std;
using namespace __gnu_pbds;
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
ll read(ll x=0)
{
    ll c, f(1);
    for(c=getchar();!isdigit(c);c=getchar())if(c=='-')f=-f;
    for(;isdigit(c);c=getchar())x=x*10+c-0x30;
    return f*x;
}
ll w[20][20], need[20][20][2], num[10][10], tot, f[(1<<18)+10], s[20], n, m, k;
string str[20], mp[10][10];
pll pos[20];
ll match(string &s1, string &s2)
{
    ll cnt=0, i;
    rep(i,0,k-1)cnt+=(s1[i]==s2[i]);
    return s[cnt];
}
void calc_need(ll a, ll b)
{
    ll i1, j1, i2, j2, i, j, ret1=0, ret2=0;
    i1=pos[a].fi, j1=pos[a].se;
    i2=pos[b].fi, j2=pos[b].se;
    if(i1>i2)swap(i1,i2),swap(j1,j2);
    if(j1<j2)
    {
        /*
        #...
        ....
        ...#
        */
        rep(j,j1,j2)
            if(num[i1][j]!=-1)
                ret1 |= 1<<num[i1][j];
        rep(i,i1,i2)
            if(num[i][j2]!=-1)
                ret1 |= 1<<num[i][j2];
        
        rep(i,i1,i2)
            if(num[i][j1]!=-1)
                ret2 |= 1<<num[i][j1];
        rep(j,j1,j2)
            if(num[i2][j]!=-1)
                ret2 |= 1<<num[i2][j];
    }
    else
    {
        /*
        ...#
        ....
        #...
        */
        rep(j,j2,j1)
            if(~num[i1][j])
                ret1 |= 1<<num[i1][j];
        rep(i,i1,i2)
            if(~num[i][j2])
                ret1 |= 1<<num[i][j2];
        
        rep(j,j2,j1)
            if(~num[i2][j])
                ret2 |= 1<<num[i2][j];
        rep(i,i1,i2)
            if(~num[i][j1])
                ret2 |= 1<<num[i][j1];
    }
    need[min(a,b)][max(a,b)][0]=ret1^(1<<a)^(1<<b);
    need[min(a,b)][max(a,b)][1]=ret2^(1<<a)^(1<<b);
}
void pre()
{
    ll i, j;
    cin >> n >> m >> k;
    tot=0;
    memset(num,-1,sizeof(num));
    rep(i,1,n)rep(j,1,m)
    {
        cin >> mp[i][j];
        if(mp[i][j][0]!='-')
        {
            num[i][j]=tot;
            pos[tot]=pll(i,j);
            str[tot]=mp[i][j];
            tot++;
        }
    }
    rep(i,0,k)cin >> s[i];
    rep(i,0,tot-1)rep(j,i+1,tot-1)
    {
        w[i][j] = match(str[i],str[j]);
        calc_need(i,j);
    }
}
void dp()
{
    ll s, i, j;
    rep(s,1,(1<<tot)-1)f[s]=-linf; f[0]=0;
    rep(s,0,(1<<tot)-1)
    {
        rep(i,0,tot-1)rep(j,i+1,tot-1)
        {
            if( (s&(1<<i)) == 0 )continue;
            if( (s&(1<<j)) == 0 )continue;
            if((s&need[i][j][0])==need[i][j][0] or (s&need[i][j][1])==need[i][j][1])
                f[s] = max(f[s], f[s^(1<<i)^(1<<j)] + w[i][j] );
        }
    }
    cout << f[(1<<tot)-1] << endl;
}
int main()
{
    ll T=read();
    while(T--)
    {
        pre();
        dp();
    }
    return 0;
}