BUYING FEED

                                                          BUYING FEED

                                                                             时间限制：3000 ms  |  内存限制：65535 KB                                                                                                         难度：4描述Farmer John needs to travel to town to pick up K (1 <= K <= 100)pounds of feed. Driving D miles with K pounds of feed in his truck costs D*K cents.The county feed lot has N (1 <= N<= 100) stores (conveniently numbered 1..N) that sell feed. Each store is located on a segment of the X axis whose length is E (1 <= E <= 350). Store i is at location X_i (0 <X_i < E) on the number line and can sell John as much as F_i (1 <= F_i <= 100) pounds of feed at a cost of C_i (1 <= C_i <= 1,000,000) cents per pound.Amazingly, a given point on  the X axis might have more than one store.Farmer John  starts  at location 0 on this number line and can drive only in the positive direction, ultimately arriving at location E, with at least K pounds of feed. He can stop at any of the feed storesalong the way and buy any amount of feed up to the the store's limit.  What is the minimum amount Farmer John has to pay to buy and transport the K pounds of feed? Farmer Johnknows there is a solution. Consider a sample where Farmer John  needs two pounds of feed from three stores (locations: 1, 3, and 4) on a number line whose range is 0..5:     0   1   2  3   4   5    ---------------------------------         1       1   1                Available pounds of feed         1       2   2               Cents per poundIt is best for John to buy one pound of feed from both the second and third stores. He must pay two cents to buy each pound of feed for a total cost of 4. When John travels from 3 to 4 he is moving 1 unitof length and he has 1 pound of feed so he must pay1*1 = 1 cents.When John travels from 4 to 5 heis moving one unit and he has 2 pounds of feed so he must pay 1*2 = 2 cents. The total cost is 4+1+2 = 7 cents.输入The first line of input contains a number c giving the number of cases that followThere are multi test cases ending with EOF.Each case starts with a line containing three space-separated integers: K, E, and NThen N lines follow :every line contains three space-separated integers: Xi Fi Ci输出For each case,Output A single integer that is the minimum cost for FJ to buy and transport the feed样例输入

1

2 5 3                 

3 1 2

4 1 2

1 1 1

样例输出

7

看起来很复杂的样子，其实就是大水题。只要将何时购买东西，购买时的S=单价+路程*重量，只要将S的数从小到大排序，找够需要的数就AC了。

我的代码：

#include <iostream>#include <cstdio>#include <cstdlib>#include <algorithm>using namespace std;struct S{int x,y,z;long long tj;}a[400];bool cmp(const S& p,const S& q){return p.tj<q.tj ;}int main(){int N,n,k,i,t;long long m;scanf("%d",&N);while (N--){scanf("%d %ld %d",&n,&m,&k);for (i=0;i<k;++i){scanf("%d %d %d",&a[i].x,&a[i].y,&a[i].z);a[i].tj=m - a[i].x + a[i].z ;}sort(a,a+k,cmp);for (i=0,t=0,m=0;i<k&&t!=n;++i){while (a[i].y&&t!=n){t++;m+=a[i].tj;a[i].y--;}}printf("%d\n",m);}return 0;}