bzoj1904: Musical Water-fence

找出最高的木块，假设在这块木块上无限加水，就会形成一些水池，然后才向两侧溢出

用并查集维护当前在某个位置使水向左/右流动，水会流向哪个水池或从某一侧溢出浪费，当某个水池满时更新并查集

#include<cstdio>
int n,m,mx=1,n2;
double ws[1000007],hs[1000007],sum[1000007],al=0,ar=0;
int f[2000007],ss[1000007],sp=0;
int get(int x){
    int a=x,c;
    while(x!=f[x])x=f[x];
    while(x!=f[a])c=f[a],f[a]=x,a=c;
    return x;
}
void cal(int w){
    if(w<n2)f[w]=w+n2;
    else f[w]=w-n2;
}
int main(){
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;++i){
        scanf("%lf%lf",ws+i,hs+i);
        if(hs[i]>hs[mx])mx=i;
    }
    ss[sp=0]=mx;
    for(int i=mx-1;i;--i){
        while(hs[ss[sp]]<hs[i])--sp;
        ss[++sp]=i;
    }
    ss[++sp]=0;
    n2=n+2;
    for(int i=sp;i;--i){
        for(int j=ss[i];j<ss[i-1];++j)f[j+1]=f[j+n2]=ss[i]+n2,sum[ss[i]]+=ws[j]*(hs[ss[i]]-hs[j]);
    }
    ss[sp=0]=mx;
    for(int i=mx+1;i<=n;++i){
        while(hs[ss[sp]]<hs[i])--sp;
        ss[++sp]=i;
    }
    ss[++sp]=n+1;
    for(int i=sp;i;--i){
        for(int j=ss[i];j>ss[i-1];--j)f[j-1+n2]=f[j]=ss[i],sum[ss[i]]+=ws[j]*(hs[ss[i]]-hs[j]);
    }
    for(int i=1;i<=m;++i){
        int x;double s,sl,sr;
        scanf("%d%lf",&x,&s);
        sl=sr=s/2;
        while(sl){
            int w=get(x);
            if(w%n2==0){
                al+=sl;
                sl=0;
            }else if(w%n2==n+1){
                ar+=sl;
                sl=0;
            }else if(sum[w%n2]>sl){
                sum[w%n2]-=sl;
                sl=0;
            }else{
                sl-=sum[w%n2];
                sum[w%n2]=0;
                cal(w);
            }
        }
        while(sr){
            int w=get(x+n2);
            if(w%n2==0){
                al+=sr;
                sr=0;
            }else if(w%n2==n+1){
                ar+=sr;
                sr=0;
            }else if(sum[w%n2]>sr){
                sum[w%n2]-=sr;
                sr=0;
            }else{
                sr-=sum[w%n2];
                sum[w%n2]=0;
                cal(w);
            }
        }
    }
    printf("%.3f
%.3f",al,ar);
    return 0;
}