bzoj5010: [Fjoi2017]矩阵填数

#include<bits/stdc++.h>
typedef long long i64;
const int P=1e9+7;
int T,h,w,m,n;
int xs[33],ys[33],xp,yp,vs[33],vp,ts[33];
int rc[33][5],mv[33][33],as[33][33];
void mins(int&a,int b){if(a>b)a=b;}
int pw(int a,int n){
    int v=1;
    for(;n;n>>=1,a=i64(a)*a%P)if(n&1)v=i64(v)*a%P;
    return v;
}
int main(){
    for(scanf("%d",&T);T;--T){
        int ans=0;
        scanf("%d%d%d%d",&h,&w,&m,&n);
        xp=yp=vp=0;
        xs[xp++]=1;
        xs[xp++]=h+1;
        ys[yp++]=1;
        ys[yp++]=w+1;
        vs[vp++]=m;
        for(int i=0;i<n;++i){
            for(int j=0;j<5;++j)scanf("%d",rc[i]+j);
            xs[xp++]=rc[i][0];
            xs[xp++]=rc[i][2]+1;
            ys[yp++]=rc[i][1];
            ys[yp++]=rc[i][3]+1;
            vs[vp++]=rc[i][4];
            vs[vp++]=rc[i][4]-1;
        }
        std::sort(xs,xs+xp);
        xp=std::unique(xs,xs+xp)-xs-1;
        std::sort(ys,ys+yp);
        yp=std::unique(ys,ys+yp)-ys-1;
        std::sort(vs,vs+vp);
        vp=std::unique(vs,vs+vp)-vs;
        for(int i=0;i<xp;++i)
        for(int j=0;j<yp;++j)as[i][j]=(xs[i+1]-xs[i])*(ys[j+1]-ys[j]);
        for(int t=0;t<n;++t){
            rc[t][0]=std::lower_bound(xs,xs+xp,rc[t][0])-xs;
            rc[t][2]=std::lower_bound(xs,xs+xp,rc[t][2]+1)-xs;
            rc[t][1]=std::lower_bound(ys,ys+yp,rc[t][1])-ys;
            rc[t][3]=std::lower_bound(ys,ys+yp,rc[t][3]+1)-ys;
            rc[t][4]=std::lower_bound(vs,vs+vp,rc[t][4])-vs;
        }
        for(int S=0;S<(1<<n);++S){
            for(int i=0;i<xp;++i)
            for(int j=0;j<yp;++j)mv[i][j]=vp-1;
            int s=1;
            for(int t=0;t<n;++t){
                int v=rc[t][4];
                if(S>>t&1)s=-s,--v;
                for(int i=rc[t][0];i<rc[t][2];++i)
                for(int j=rc[t][1];j<rc[t][3];++j)mins(mv[i][j],v);
            }
            for(int i=0;i<vp;++i)ts[i]=0;
            for(int i=0;i<xp;++i)
            for(int j=0;j<yp;++j)ts[mv[i][j]]+=as[i][j];
            for(int i=0;i<vp;++i)s=i64(s)*pw(vs[i],ts[i])%P;
            ans=(ans+s)%P;
        }
        printf("%d\n",(ans+P)%P);
    }
    return 0;
}