JZOJsenior5619.【NOI2018模拟4.1】重围

problem

Description

Input

Output

Sample Input

Sample Output

Data Constraint

---

analysis

• 正解贪心+排序

• 当$a[i]<b[i]$时，可以看成都是盈，即每个都一定能杀死

• 这样的话把$a[i]$为第一关键字$b[i]$为第二关键字从小到大排序，$c<=d$就是NO

• 当$a[i]>b[u]$，可以看成都是亏，即一定先杀死$a[i]$大的

• 为什么？如果一直都是亏的话，那么$a[i]$大的可能一开始能杀死而后面杀不死了

• 如此同理分类讨论即可

---

code

#include<stdio.h>
#include<algorithm>
#define MAXN 100001

using namespace std;

long long c,d;
int n,l;

struct node
{
    int x,y,z,num;
}a[MAXN];

int read()
{
    int x=0,f=1;
    char ch=getchar();
    while (ch<'0' || '9'<ch)
    {
        if (ch=='-')f=-1;
        ch=getchar();   
    }
    while ('0'<=ch && ch<='9')
    {
        x=x*10+ch-'0';
        ch=getchar();
    }
    return x*f;
}

long long readll()
{
    long long x=0,f=1;
    char ch=getchar();
    while (ch<'0' || '9'<ch)
    {
        if (ch=='-')f=-1;
        ch=getchar();   
    }
    while ('0'<=ch && ch<='9')
    {
        x=x*10+ch-'0';
        ch=getchar();
    }
    return x*f;
}

bool cmp(node a,node b)
{
    return (a.z>b.z)||(a.z==b.z && a.x<b.x && a.z==1)||(a.z==b.z && a.y>b.y && a.z==-1);
}

int judge(int x)
{
    return x>0?1:-1;
}

int main()
{
    //freopen("readin.txt","r",stdin);
    freopen("resist.in","r",stdin);
    freopen("resist.out","w",stdout);

    n=read(),c=readll(),d=readll(),read();
    for (int i=1;i<=n;i++)
    {
        a[i].x=read(),a[i].y=read(),a[i].z=judge(a[i].y-a[i].x),a[i].num=i;
    }
    sort(a+1,a+n+1,cmp);
    for (int i=1;i<=n;i++)
    {
        c-=a[i].x;
        if (c<=d)
        {
            printf("NO");
            return 0;
        }
        c+=a[i].y;
    }
    printf("YES\n");
    for (int i=1;i<=n;i++)printf("%d ",a[i].num);
    return 0;
}