• POJ 1987 Distance Statistics


    http://poj.org/problem?id=1987

    题意:给一棵树,求树上有多少对节点满足距离<=K

    思路:点分治,我们考虑把每个距离都存起来,然后排序,一遍扫描计算一下,注意还要减掉自己加自己的方案。而且,我们还要去掉走到同一个子树的方案。复杂度:O(nlog^2n)

    #include<cstdio>
    #include<cmath>
    #include<cstring>
    #include<iostream>
    #include<algorithm>
    #define ll long long
    int tot,go[1000005],first[1000005],next[1000005];
    ll st[1000005],val[1000005];
    int sum,son[1000005],root,n,F[1000005],c[1000005];
    int pd[1000005],sz,vis[1000005];
    ll dis[1000005];
    int cnt,K,ans;
    int read(){
        int t=0,f=1;char ch=getchar();
        while (ch<'0'||ch>'9'){if (ch=='-') f=-1;ch=getchar();}
        while ('0'<=ch&&ch<='9'){t=t*10+ch-'0';ch=getchar();}
        return t*f;
    }
    void insert(int x,int y,int z){
        tot++;
        go[tot]=y;
        next[tot]=first[x];
        first[x]=tot;
        val[tot]=z;
    }
    void add(int x,int y,int z){
        insert(x,y,z);insert(y,x,z);
    }
    void findroot(int x,int fa){
        son[x]=1;F[x]=0;
        for (int i=first[x];i;i=next[i]){
            int pur=go[i];
            if (pur==fa||vis[pur]) continue;
            findroot(pur,x);
            son[x]+=son[pur];
            F[x]=std::max(F[x],son[pur]);
        }
        F[x]=std::max(F[x],sum-son[x]);
        if (F[x]<F[root]) root=x;
    }
    void bfs(int x){
        int h=1,t=1;c[1]=x;pd[x]=sz;dis[x]=0;
        while (h<=t){
            int now=c[h++];
            for (int i=first[now];i;i=next[i]){
                int pur=go[i];
                if (vis[pur]||pd[pur]==sz) continue;
                pd[pur]=sz;
                dis[pur]=dis[now]+val[i];
                c[++t]=pur;
                st[++cnt]=dis[pur];
            }
        }
        std::sort(st+1,st+1+cnt);
        int j=cnt,res=0,Cnt=0;
        for (int i=1;i<=t;i++){
            while (j>1&&st[i]+st[j]>K) j--;
            if (st[i]+st[j]<=K) res+=j;
            if (st[i]+st[i]<=K) Cnt++;
        }
        res-=Cnt;
        ans+=res/2;
    }
    int del(int x,int Dis){
        dis[x]=Dis;sz++;
        int h=1,t=1;cnt=1;st[cnt]=Dis;
        pd[x]=sz;c[1]=x;
        while (h<=t){
           int now=c[h++];
           for (int i=first[now];i;i=next[i]){
              int pur=go[i];
              if (pd[pur]==sz||vis[pur]) continue;
              dis[pur]=dis[now]+val[i];
              st[++cnt]=dis[pur];
              pd[pur]=sz;
              c[++t]=pur;
           }
        }
        int j=cnt,res=0,Cnt=0;
        std::sort(st+1,st+1+cnt);
        for (int i=1;i<=t;i++){
           while (j>1&&st[i]+st[j]>K) j--;
           if (st[i]+st[j]<=K) res+=j;
           if (st[i]+st[i]<=K) Cnt++;
        }
        res-=Cnt;
        return res/2;
    }
    void solve(int x){
        vis[x]=1;++sz;
        cnt=1;st[cnt]=0;
        bfs(x);
        for (int i=first[x];i;i=next[i]){
             int pur=go[i];
             if (vis[pur]) continue;
             ans-=del(pur,val[i]);
        }
        int Cnt=sum;
        for (int i=first[x];i;i=next[i]){
             int pur=go[i];
             if (vis[pur]) continue;
             if (son[pur]>son[x]) sum=Cnt-son[x];
             else sum=son[pur];
             root=0;    
             findroot(pur,x);
             solve(root);
        }
    }
    int main(){
        int m;
        char s[20];
        scanf("%d%d
    ",&n,&m);
        for (int i=1;i<n;i++){
            int x,y,z;
            scanf("%d%d%d",&x,&y,&z);
            add(x,y,z);
            scanf("%s",s+1);
        }
        scanf("%d
    ",&K);
        F[0]=0x7fffffff;
        root=0;sum=n;
        findroot(1,0);
        solve(root);
        printf("%d
    ",ans);
    }
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  • 原文地址:https://www.cnblogs.com/qzqzgfy/p/5671476.html
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