Principle of Bzoj 3812 melody-like pressure dp+ tolerance

Topic: Given an undirected graph, find out how many strong connected graphs are in the generated sub-graph of this undirected graph

We think about the principle of repulsion.

If a graph is not connected, then this image will be indented to form a dag with a >=2 point.

There must be some 0 points in a DAG, and we enumerate the point sets of these points for the repulsion

Specific DP equations and details see code comments written or more detailed I won't say more = =

#include <cstdio> #include <cstring> #include <iostream> #include <algorithm> #define M 16# Define MOD 1000000007using namespace Std;int n,m,digit[1<<8];int Into[1<<15],out_of[1<<15];long Long F[1<<15],g[1<<15],h[1<<15];long long Power_2[m*m];/*f[s] indicates that the generation of the point set S is strongly connected to the scheme number G[s] represents the generation of the point set S in the sub-figure g, If all the connecting blocks of G are strong, then G has a contribution to g[s] If G has an odd number of connected blocks, then the contribution to G[s] is +1, otherwise -1h[s] indicates how many edges in the induction sub-map of the point set S f[s]=2^h[s]-σ[t is the non-empty set of S]2^ (h[s]-h[t ]) *g[t] (note that at this time G[s] does not include the entire S-strong unicom case) */int Count (int x) {return digit[x>>8] + digit[x&255];} int main () {int i,j,x,y;cin>>n>>m;for (i=1;i<1<<8;i++) digit[i]=digit[i>>1]+ (i&1); for (power_2[0]=1,i=1;i<=m;i++) power_2[i]= (power_2[i-1]<<1)%mod;for (i=1;i<=m;i++) {scanf ("%d%d", &x,&y); out_of[1<<x-1]|=1<<y-1;into[1<<y-1]|=1<<x-1;} for (i=1;i<1<<n;i++) {int One=i&-i,sta=i^one;//one is any point in the S collection//sta for the S collection to remove the remaining points set outside one h[i]=h[sta]+count ( Into[one]&sta) +count (Out_of[one]&sta);(J=sta;j; (--j) &=sta)//Enumerate point sets that are not connected to one (g[i]+=mod-f[i^j]*g[j]%mod)%=mod;static int w[1<<15];//w[t] The number of edges that represent points in the collection T to the point in the collection s-t f[i]=power_2[h[i]];for (j=i;j; (--j) &=i)//enumeration T set {if (j==i) W[j]=0;else{int temp= (i^j) & -(I^J);//Select a point in the S-t collection W[j]=w[j^temp]-count ((I^J) &out_of[temp]) +count (J&into[temp]);} (F[i]+=mod-power_2[h[i^j]+w[j]]*g[j]%mod)%=mod;} (G[i]+=f[i])%=mod;} cout<<f[(1<<n) -1]<<endl;return 0;}

Principle of Bzoj 3812 melody-like pressure dp+ tolerance