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。
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<queue> 5 #include<algorithm> 6 using namespace std; 7 const int maxn = 50+5; 8 9 int n,c,x,lowtot,hightot; 10 11 struct node 12 { 13 int low,high; 14 bool operator< (const node& rhs) const 15 { 16 return low > rhs.low; 17 } 18 }a[maxn]; 19 20 bool cmp(node a, node b) 21 { 22 return a.high>b.high; 23 } 24 25 int main() 26 { 27 //freopen("in.txt","r",stdin); 28 int T; 29 scanf("%d",&T); 30 while(T--) 31 { 32 lowtot = hightot = 0; 33 scanf("%d%d%d",&n,&c,&x); 34 for(int i=1;i<=n;i++) 35 { 36 scanf("%d%d",&a[i].low,&a[i].high); 37 lowtot += a[i].low; 38 hightot += a[i].high; 39 } 40 int sum = 0; 41 int ans = 0; 42 sort(a+1,a+n+1); 43 int tmp = hightot; 44 for(int i=1;i<=n;i++) 45 { 46 sum += a[i].low; 47 ans++; 48 hightot -= a[i].high; 49 if(sum>=x || c-hightot>=x) break; 50 } 51 52 sum = 0; 53 int ans2 = 0; 54 sort(a+1,a+n+1,cmp); 55 hightot = tmp; 56 for(int i=1;i<=n;i++) 57 { 58 sum += a[i].low; 59 ans2++; 60 hightot -= a[i].high; 61 if(sum>=x || c-hightot>=x) break; 62 } 63 printf("%d ",min(ans,ans2)); 64 } 65 return 0; 66 }