51nod 1351 吃点心（贪心）

http://www.51nod.com/onlineJudge/questionCode.html#!problemId=1351

题意：

思路：

要么先选low值大的，要么先选high值大的，分两种情况讨论。

每次只要选了的low值和>=x或者c-未选的high值和>=x就肯定满足了。

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<queue>
 #include<algorithm>
 using namespace std;
 const int maxn = +;

 int n,c,x,lowtot,hightot;

 struct node
 {
     int low,high;
     bool operator< (const node& rhs) const
     {
         return low > rhs.low;
     }
 }a[maxn];

 bool cmp(node a, node b)
 {
     return a.high>b.high;
 }

 int main()
 {
     //freopen("in.txt","r",stdin);
     int T;
     scanf("%d",&T);
     while(T--)
     {
         lowtot = hightot = ;
         scanf("%d%d%d",&n,&c,&x);
         for(int i=;i<=n;i++)
         {
             scanf("%d%d",&a[i].low,&a[i].high);
             lowtot += a[i].low;
             hightot += a[i].high;
         }
         int sum = ;
         int ans = ;
         sort(a+,a+n+);
         int tmp = hightot;
         for(int i=;i<=n;i++)
         {
             sum += a[i].low;
             ans++;
             hightot -= a[i].high;
             if(sum>=x || c-hightot>=x)  break;
         }

         sum = ;
         int ans2 = ;
         sort(a+,a+n+,cmp);
         hightot = tmp;
         for(int i=;i<=n;i++)
         {
             sum += a[i].low;
             ans2++;
             hightot -= a[i].high;
             if(sum>=x || c-hightot>=x)  break;
         }
         printf("%d\n",min(ans,ans2));
     }
     return ;
 }