• HDU3870- intervals(差分约束)


    You are given n closed, integer intervals [ai, bi] and n integers c1, ..., cn. 

    Write a program that: 

    > reads the number of intervals, their endpoints and integers c1, ..., cn from the standard input, 

    > computes the minimal size of a set Z of integers which has at least ci common elements with interval [ai, bi], for each i = 1, 2, ..., n, 

    > writes the answer to the standard output 

    Input

    The first line of the input contains an integer n (1 <= n <= 50 000) - the number of intervals. The following n lines describe the intervals. The i+1-th line of the input contains three integers ai, bi and ci separated by single spaces and such that 0 <= ai <= bi <= 50 000 and 1 <= ci <= bi - ai + 1. 

    Process to the end of file. 
     

    Output

    The output contains exactly one integer equal to the minimal size of set Z sharing at least ci elements with interval [ai, bi], for each i = 1, 2, ..., n. 

    Sample Input

    5
    3 7 3
    8 10 3
    6 8 1
    1 3 1
    10 11 1

    Sample Output

    6

    这是一个差分约束的基础题;

    求最长路即可;

    AC代码为:

    //scanf,printf过了。cin cout T了Orz 
    #include<bits/stdc++.h>
    using namespace std;
    const int INF=0x3f3f3f3f;
    const int maxn=1e5+10;
    int n,tot,S,T,u,v,w;
    int dis[maxn],first[maxn],vis[maxn];
    struct Node{
        int v,w,net;
    } edge[maxn*10];
    
    void addedge(int u,int v,int w)
    {
        edge[tot].v=v;
        edge[tot].w=w;
        edge[tot].net=first[u];
        first[u]=tot++;
    }
    
    void SPFA(int s)
    {
        queue<int> q;
        memset(dis,-INF,sizeof dis);
        memset(vis,0,sizeof vis);
        q.push(s); dis[s]=0;
        while(!q.empty())
        {
            int u=q.front(); q.pop();
            vis[u]=0;
            for(int i=first[u];~i;i=edge[i].net)
            {
                if(dis[edge[i].v]<dis[u]+edge[i].w) 
                {
                    dis[edge[i].v]=dis[u]+edge[i].w;
                    if(!vis[edge[i].v])
                    {
                        vis[edge[i].v]=1;
                        q.push(edge[i].v);  
                    }   
                }
            }
        }
    }
    int main()
    {
        while(~scanf("%d",&n))
        {
            tot=1, S=INF,T=-INF;
            memset(first,-1,sizeof first);
            for(int i=1;i<=n;i++)
            {
                scanf("%d%d%d",&u,&v,&w);
                S=min(S,u-1),T=max(T,v);
                addedge(u-1,v,w);
            }
            for(int i=S;i<T;i++)
            {
                addedge(i,i+1,0);
                addedge(i+1,i,-1);
            }
            SPFA(S);
            printf("%d
    ",dis[T]);
        }
        
        
        return 0;   
    } 
    View Code
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  • 原文地址:https://www.cnblogs.com/csushl/p/9386777.html
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