Crixalis's Equipment
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3957 Accepted Submission(s): 1625
Problem Description
Crixalis
- Sand King used to be a giant scorpion(蝎子) in the deserts of
Kalimdor. Though he's a guardian of Lich King now, he keeps the living
habit of a scorpion like living underground and digging holes.Someday Crixalis decides to move to another nice place and build a new house for himself (Actually it's just a new hole). As he collected a lot of equipment, he needs to dig a hole beside his new house to store them. This hole has a volume of V units, and Crixalis has N equipment, each of them needs Ai units of space. When dragging his equipment into the hole, Crixalis finds that he needs more space to ensure everything is placed well. Actually, the ith equipment needs Bi units of space during the moving. More precisely Crixalis can not move equipment into the hole unless there are Bi units of space left. After it moved in, the volume of the hole will decrease by Ai. Crixalis wonders if he can move all his equipment into the new hole and he turns to you for help.
Input
The
first line contains an integer T, indicating the number of test cases.
Then follows T cases, each one contains N + 1 lines. The first line
contains 2 integers: V, volume of a hole and N, number of equipment
respectively. The next N lines contain N pairs of integers: Ai and Bi.
0<T<= 10, 0<V<10000, 0<N<1000, 0 <Ai< V, Ai <= Bi < 1000.
0<T<= 10, 0<V<10000, 0<N<1000, 0 <Ai< V, Ai <= Bi < 1000.
Output
For each case output "Yes" if Crixalis can move all his equipment into the new hole or else output "No".
Sample Input
2
20 3
10 20
3 10
1 7
10 2
1 10
2 11
Sample Output
Yes
No
Source
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代码：
1 /* 2 应该先放实际体积小而需要的体积大的，所以按照差值由大到小排序 3 */ 4 #include<iostream> 5 #include<string> 6 #include<cstdio> 7 #include<cmath> 8 #include<cstring> 9 #include<algorithm> 10 #include<vector> 11 #include<iomanip> 12 #include<queue> 13 #include<stack> 14 using namespace std; 15 int t,v,n; 16 struct eq 17 { 18 int a,b; 19 }e[1005]; 20 bool cmp(eq x,eq y) 21 { 22 return x.b-x.a>y.b-y.a; 23 } 24 int main() 25 { 26 scanf("%d",&t); 27 while(t--) 28 { 29 scanf("%d%d",&v,&n); 30 for(int i=0;i<n;i++) 31 { 32 scanf("%d%d",&e[i].a,&e[i].b); 33 } 34 sort(e,e+n,cmp); 35 for(int i=0;i<n;i++) 36 { 37 if(v<e[i].b) 38 { 39 v=-1; 40 break; 41 } 42 v-=e[i].a; 43 } 44 if(v==-1) 45 printf("No "); 46 else printf("Yes "); 47 } 48 return 0; 49 }