Codeforces Round #568 (Div. 2)G1. Playlist for Polycarp (easy version)（状压dp）

https://codeforc.es/contest/1185/problem/G1

不难想到用二进制表示第i首歌选还是不选，d[s][t]表示当状态为s且最后一首歌的类型为t的方案数。

#define bug(x) cout<<#x<<" is "<<x<<endl
#define IO std::ios::sync_with_stdio(0)
#include <bits/stdc++.h>
#define iter ::iterator
//#define pa pair<int,int>
using namespace  std;
#define ll long long
#define mk make_pair
#define pb push_back
#define se second
#define fi first
#define ls o<<1
#define rs o<<1|1
ll mod=1e9+7;
const int inf=2e9+10;
const int N=(1<<15)+5;
int n,T;
ll d[N][4];
int t[20],g[20];
ll cal(ll x,ll y){
    ll res=1;
    while(y){
        if(y&1)res=res*x%mod;
        x=x*x%mod;
        y>>=1;
    }
    return res;
}
int main(){
    IO;
    cin>>n>>T;
    for(int i=0;i<n;i++){
        cin>>t[i]>>g[i];
    }
    int mx=(1<<n);
    //d[0][1]=d[0][2]=d[0][3]=1;
    d[0][0]=1;
    for(int s=0;s<mx;s++){
        for(int i=0;i<=3;i++){
            for(int j=0;j<n;j++){
                if(!((s>>j)&1)){
                    if(g[j]!=i){
                        int ns=s^(1<<j);
                        d[ns][g[j]]=(d[ns][g[j]]+d[s][i])%mod;
                        //printf("ns=%d d=%d
",ns,d[ns][g[j]]);
                    }
                }
            }
        }
    }
    ll ans=0;
    for(int s=0;s<mx;s++){
        int x=s;
        int h=0;
        int res=0;
        while(x){
            if(x&1)res+=t[h];
            x>>=1;
            h++;
        }
        if(res==T){
            for(int j=1;j<=3;j++){
                //printf("s=%d d=%d
",s,d[s][j]);
                ans=(ans+d[s][j])%mod;
            }
        }
    }
    cout<<ans<<endl;
}