• pb_ds(平板电视)整理


      有人说BZOJ3040用普通的<queue>中priority_queue搞dijkstra过不了。

    我只想说你们的djk可能写的太丑了。

    先上代码

    #include<iostream>
    #include<cstdio>
    #include<cstring>
    #include<algorithm>
    #include<queue>
    #include<ext/pb_ds/priority_queue.hpp>
    #define pir pair<ll,int>
    using namespace std;
    typedef long long ll;
    //typedef std::priority_queue<pir,vector<pir>,greater<pir> > heap;
    //3840 ms
    typedef __gnu_pbds::priority_queue<pir,greater<pir> > heap;//默认是pairing_heap_tag 
    //3304 ms
    const int N=1e6+5,M=1e7+5;
    const ll INF=1e15;
    inline int read(){
        char c=getchar();int x=0,f=1;
        while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
        while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}
        return x*f;
    }
    int n,m,T,rxa,rxc,rya,ryc,rp,a,b;
    int x,y,z;
    struct node{
        int v,w,next;
    }e[M];
    int cnt,head[N];ll dis[N];bool vis[N];
    inline void add(int u,int v,int w){
        e[++cnt].v=v;e[cnt].w=w;e[cnt].next=head[u];head[u]=cnt;
    }
    heap q;
    inline void dijkstra(){
        int S;
        for(int i=1;i<=n;i++) dis[i]=INF;
        dis[S=1]=0;
        q.push(make_pair(dis[S],S));
        while(!q.empty()){
            pir t=q.top();q.pop();
            int x=t.second;
            if(vis[x]) continue;
            vis[x]=1;
            for(int i=head[x];i;i=e[i].next){
                int v=e[i].v;
                if(!vis[v]&&dis[v]>dis[x]+e[i].w){
                    dis[v]=dis[x]+e[i].w;
                    q.push(make_pair(dis[v],v)); 
                }
            }
        }
    }
    int main(){
        n=read();m=read();
        T=read();rxa=read();rxc=read();rya=read();ryc=read();rp=read();
        m=m-T;
        x=rxc%rp;y=ryc%rp;
        a=min(x%n+1,y%n+1);
        b=max(y%n+1,y%n+1);
        while(T--) add(a,b,100000000-100*a);
        while(m--) x=read(),y=read(),z=read(),add(x,y,z);
        dijkstra();
        printf("%lld",dis[n]);
        return 0;
    }

    对比:

    平衡二叉树(Balanced Binary Tree)

    SPOJ3273

    In this problem, you have to maintain a dynamic set of numbers which support the two fundamental operations

    • INSERT(S,x): if x is not in S, insert x into S
    • DELETE(S,x): if x is in S, delete x from S

    and the two type of queries

    • K-TH(S) : return the k-th smallest element of S
    • COUNT(S,x): return the number of elements of S smaller than x

    Input

    • Line 1: Q (1 ≤ Q ≤ 200000), the number of operations
    • In the next Q lines, the first token of each line is a character I, D, K or C meaning that the corresponding operation is INSERT, DELETE, K-TH or COUNT, respectively, following by a whitespace and an integer which is the parameter for that operation.

    If the parameter is a value x, it is guaranteed that 0 ≤ |x| ≤ 109. If the parameter is an index k, it is guaranteed that 1 ≤ k ≤ 109.

    Output

    For each query, print the corresponding result in a single line. In particular, for the queries K-TH, if k is larger than the number of elements in S, print the word 'invalid'.

    Example

    Input
    8
    I -1
    I -1
    I 2
    C 0
    K 2
    D -1
    K 1
    K 2
    
    Output
    1
    2
    2
    invalid
    #include<cstdio>
    #include<ext/pb_ds/assoc_container.hpp>
    //pb_ds库这次内置了红黑树(red-black tree)、伸展树(splay tree)和排序向量树(ordered-vector tree,没找到通用译名,故自行翻译)。
    //这些封装好的树都支持插入(insert)、删除(erase)、求kth(find_by_order)、求rank(order_of_key)操作,O(logn)内完成
    using namespace std;
    using namespace __gnu_pbds;
    int read(){
        int x=0,f=1;char ch=getchar();
        while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
        while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
        return x*f;
    }
    tree<int,null_mapped_type,less<int>,rb_tree_tag,tree_order_statistics_node_update>bbt;
    //SPOJG++版本稍旧(4.3.2),需要写成null_mapped_type才可以(高级版本可以写null_type)
    char in(){
        for(char ch=getchar();;ch=getchar()) if(ch>='A'&&ch<='Z') return ch;
    }
    int main(){
        char c;int x;
        for(int T=read();T--;){
            c=in();x=read();
            if(c=='I'){
                bbt.insert(x);
            }
            else if(c=='D'){
                bbt.erase(x);
            }
            else if(c=='K'){
                if(x<=bbt.size())
                    printf("%d
    ",*bbt.find_by_order(x-1));
                else
                    puts("invalid");
            }
            else{
                printf("%d
    ",bbt.order_of_key(x));
            }
        }
        return 0;
    }

     BZOJ3224: Tyvj 1728 普通平衡树

    #include<cstdio>
    #include<ext/pb_ds/assoc_container.hpp>
    #include<ext/pb_ds/tree_policy.hpp>
    #pragma GCC optimize("02")
    using namespace std;
    using namespace __gnu_pbds;
    typedef long long ll;
    ll read(){
        ll x=0,f=1;char ch=getchar();
        while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
        while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
        return x*f;
    } 
    typedef __gnu_pbds::tree<ll,null_mapped_type,less<ll>,rb_tree_tag,tree_order_statistics_node_update> T;
    T bbt;ll ans;
    int main(){
        ll n=read();
        for(ll i=1,o,k;i<=n;i++){
            o=read();k=read();
            if(o==1) bbt.insert((k<<20)+i);else
            if(o==2) bbt.erase(bbt.lower_bound(k<<20));else
            if(o==3) printf("%d
    ",bbt.order_of_key(k<<20)+1);else
            {
                if(o==4) ans=*bbt.find_by_order(k-1);else
                if(o==5) ans=*--bbt.lower_bound(k<<20);else
                if(o==6) ans=*bbt.lower_bound((k+1)<<20);
                printf("%lld
    ",ans>>20);
            }
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/shenben/p/6291080.html
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