zoukankan      html  css  js  c++  java
  • CodeForces 306B

    //#include<bits/stdc++.h>
    #include<algorithm>
    #include<stdio.h>
    #include<iostream>
    using namespace std;
    const int maxx=1000000*2+10;
    struct Node
    {
        int a,l;
        int num;
    } op[maxx];
    int use[maxx];
    bool cmp(Node a,Node b)
    {
        if(a.a!=b.a) return a.a<b.a;
        else if(a.a==b.a) return a.l>b.l;
        return a.num<b.num;
    }
    int main()
    {
        int n,m;
        scanf("%d%d",&n,&m);
        for(int i=0; i<m; i++)
        {
            scanf("%d%d",&op[i].a,&op[i].l);
            op[i].num=i+1;
        }
        sort(op,op+m,cmp);
        int  right=op[0].a+op[0].l;
        use[op[0].num]=1;
        int incl=right;
        int preid=0;
        int ans=1;
        for(int i=1; i<m; i++)
        {
            if(op[i].a>incl)
            {
                if(preid)
                {
                    use[preid]=1;
                    incl=right;
                    preid=0;
                    i--;
                    ans++;
    
                }
                else
                {
                    preid=0;
                    use[op[i].num]=1;
                    right=op[i].a+op[i].l;
                    incl=right;
                    ans++;
                }
            }
            else if(op[i].a+op[i].l>right)
            {
    
                right=op[i].a+op[i].l;
                preid=op[i].num;
            }
        }
        if(preid)
        {
            ans++;
            use[preid]=1;
        }
        printf("%d
    ",m-ans);
        if(m-ans)
            for(int i=1; i<=m; i++)
                if(!use[i]) printf("%d ",i);
        return 0;
    
    }
    /*100 6
    1 100
    2 20
    3 10
    5 10
    6 1
    2 5*/
    View Code
     H
    Time Limit:3000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

    Description

    A process RAM is a sequence of bytes that are indexed from 1 to n. Polycarpus's program contains such instructions as "memset", that is, the operations of filling memory cells on a segment with some value. The details are: the code only contains m instructions that look like "set13 a_i l_i". Instruction i fills a continuous memory segment of length li, starting from cell number ai, (that it cells with numbers ai, ai + 1, ..., ai + li - 1) with values 13.

    In Polycarpus's code, the optimizer's task is to remove the maximum number of instructions from his code in such a way that the remaining instructions set value 13 in all the memory bytes that got this value from the code before the optimization. Also, the value 13 should be set only in the memory bytes that got this value from the code before the optimization. Your task is to implement the optimizer for such program.

    Input

    The first line contains integers n and m (1 ≤ n ≤ 2·106, 1 ≤ m ≤ 2·105) — the number of bytes (memory cells) and the number of instructions in Polycarpus's code. Then m lines follow, each line contains a pair of integers aili (1 ≤ ai ≤ n, 1 ≤ li ≤ n - ai + 1).

    Output

    Print in the first line the sought maximum number of instructions that can be removed from the code. In the second line print the numbers of the instructions. The instructions are numbered from 1 to m in the order they appeared in the input. If there are multiple solutions, print any of them.

    Sample Input

    Input
    10 4
    3 3
    3 1
    4 1
    9 2
    Output
    2
    2 3
    Input
    1 1
    1 1
    Output
    0


    按照起点,路径长度排序。然后贪心,每次选择覆盖最大的一个边,知道遍历完所有的边。判断最后一个边应不应该加入。
  • 相关阅读:
    oracle创建表空间、用户 导入dmp文件
    input中只允许输入数字
    hibernate 中 Criteria 的使用介绍
    将json字符串,转换为对象实体
    unity 修改快捷键的方法 (Ubuntu 12.10)
    openjms的使用 下载、安装、启动、使用、点对点代码demo
    OGNL在页面上的常见11种用法
    jms原理简介网摘
    java中对象转换为json
    Luence初始与简单应用Document的增删改查.
  • 原文地址:https://www.cnblogs.com/superxuezhazha/p/5411138.html
Copyright © 2011-2022 走看看