hdu5575

题解：

每一次最短的那块板合并

先装水到溢出

然后合并

代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
using namespace std;
typedef pair<int,int> mp;
const int N=;
vector<mp> vec;
int LH[N],RH[N],L[N],R[N],O[N],X[N],heap[N],l[N],r[N],dist[N];
int T,kase,v[N],f[N],tot,n,m,x,y,z;
int find(int x){return f[x]==x?x:f[x]=find(f[x]);}
int merge(int x,int y)
{
    if (!x||!y)return x+y;
    if (v[x]>v[y])swap(x,y);
    r[x]=merge(r[x],y);
    if (dist[l[x]]<dist[r[x]])swap(l[x],r[x]);
    dist[x]=dist[r[x]]+;
    return x;
}
void Union(int x,int y)
{
    x=find(x);y=find(y);
    if (x==y)return;
    f[y]=x;
    if (x<y)
     {
        RH[x]=RH[y];
        L[R[x]]=x;
        R[x]=R[y];
     }
    else
     {
        LH[x]=LH[y];
        R[L[x]]=x;
        L[x]=L[y];
     }
    heap[x]=merge(heap[x],heap[y]);
    X[x]+=X[y];
    O[x]+=O[y];
}
int pop(int x){return merge(l[x],r[x]);}
int main()
{
    scanf("%d",&T);
    while(T--)
     {
        scanf("%d%d",&n,&m);
        int ans=;LH[]=RH[n]=1e9;L[n]=n-;
        for (int i=;i<n;i++)
         {
            scanf("%d",&RH[i]);
            LH[i+]=RH[i];
            L[i]=i-;R[i]=i+;
         }
        vec.clear();
        memset(heap,,sizeof(heap));
        tot=;
        while(m--)
         {
            scanf("%d%d%d",&x,&y,&z);
            if (z)vec.push_back(mp(y+,x));
            else
             {
                ++ans;
                v[++tot]=y;
                l[tot]=r[tot]=dist[tot]=;
                heap[x]=heap[x]?merge(heap[x],tot):tot;
             }
         }
        for (int i=;i<=n;i++) f[i]=i;
        sort(vec.begin(),vec.end());
        for (int i=;i<=n;i++) O[i]=X[i]=;
        for (int i=;i<vec.size();i++)
         {
            int x=find(vec[i].second),y=vec[i].first;
            while(y>LH[x])Union(x,L[x]),x=find(x);
            while(y>RH[x])Union(x,R[x]),x=find(x);
            while(heap[x]!=&&v[heap[x]]<y)
             {
                heap[x]=pop(heap[x]);
                ++X[x];
             }
            if (++O[x]>=X[x])
             {
                ans+=(O[x]-X[x]);
                O[x]=X[x]=;
             }
         }
        printf("Case #%d: %d\n",++kase,ans);
     }
}