NOJ 1116 哈罗哈的大披萨 【淡蓝】 [状压dp+各种优化]

  1 #include<stdio.h>
  2 
  3 //#include<pair>
  4 
  5 #define N 20
  6 #define M 1005
  7 #define mod 1000000007
  8 //#define p 10000007
  9 #define mod2 1000000000
 10 #define ll long long
 11 #define LL long long
 12 #define eps 1e-9
 13 #define maxi(a,b) (a)>(b)? (a) : (b)
 14 #define mini(a,b) (a)<(b)? (a) : (b)
 15 
 16 int m;
 17 int x[N][N];
 18 double p[N],a[N],n[N],y[N][N];
 19 double zhe[ 32768 ][N];
 20 double ans;
 21 double dp[ 32768 ];
 22 double tta[ 32768 ];
 23 int tot;
 24 double zhezhe[N][N];
 25 int pos[ 32768 ];
 26 int son[ 32768 ];
 27 int pos2[ 32768 ][ 17 ];
 28 int cnt2[ 32768 ];
 29 int son2[ 32768 ][ 17 ];
 30 int h[16];
 31 char ss[17];
 32 int bit[32768][16];
 33 
 34 void ini1()
 35 {
 36     int i,o,j,k,ch;
 37 
 38     h[0]=1;
 39     for(i=1;i<16;i++)
 40         h[i]=2*h[i-1];
 41     for(o=0;o<32768;o++){
 42         k=o;
 43         j=0;
 44         ch=1;
 45         while(k > 0){
 46             bit[i][j] = k%2;
 47             k = k/2;
 48             j++;
 49         }
 50         for(j=0;j<15;j++)
 51         {
 52             if(bit[i][j]==0)
 53             {
 54                 pos2[o][ cnt2[o] ]=j;
 55                 cnt2[o]++;
 56             }
 57             else
 58             {
 59                 if(ch==1)
 60                 {
 61                     ch=0;
 62                     pos[o]=j;
 63                     son[o]= o - h[j];
 64                 }
 65             }
 66         }
 67     }
 68 }
 69 
 70 void ini()
 71 {
 72     ans=1000000000;
 73     int i,j;
 74     int o;
 75     tot=h[m];
 76     for(i=1;i<=m;i++){
 77         for(j=1;j<=m;j++){
 78             zhezhe[i][j]=1.0;
 79         }
 80         scanf("%lf%lf%lf",&p[i],&a[i],&n[i]);
 81         for(j=1;j<=n[i];j++){
 82             scanf("%d%lf",&x[i][j],&y[i][j]);
 83             y[i][j]=(100.0-y[i][j])/100.0;
 84             zhezhe[i][ x[i][j] ]=y[i][j];
 85         }
 86     }
 87 
 88     for(o=0;o<tot;o++){
 89         dp[o]=1000000000;
 90 
 91     }
 92     for(j=1;j<=m;j++){
 93         zhe[0][j]=1.0;
 94     }
 95     tta[0]=0.0;
 96 }
 97 
 98 void solve()
 99 {
100     int o,k,kk,j;
101     int te;
102     dp[0]=0;
103     for(o=0;o<tot;o++){
104         if(o>=1){
105             j=pos[o]+1;
106             te=son[o];
107             tta[o]=tta[te]+a[j];
108 
109             for(kk=0;kk<cnt2[o];kk++){
110                 k=pos2[o][kk]+1;
111                 zhe[o][k]=zhe[te][k]*zhezhe[j][k];
112             }
113         }
114 
115         for(kk=0;kk<cnt2[o];kk++){
116             k=pos2[o][kk]+1;
117             te=o+h[k-1];
118             dp[te]=mini(dp[te],dp[o]+p[k]*zhe[o][k] );
119         }
120     }
121 }
122 
123 void out()
124 {
125     int o;
126     for(o=1;o<tot;o++){
127         //printf(" o=%d dp=%.4f\n",o,dp[o]);
128         ans=mini(ans,dp[o]/tta[o]);
129     }
130     printf("%.4f\n",ans);
131 }
132 
133 int main()
134 {
135     ini1();
136    // freopen("data.in","r",stdin);
137     //freopen("data.out","w",stdout);
138     while(scanf("%d",&m)!=EOF)
139     {
140         if( m==0 ) break;
141 
142         ini();
143         solve();
144         out();
145     }
146 
147     return 0;
148 }