题意：n个计算机零件（类型，名称，价格，质量）和m元钱，零件种类相同的只买一种，计算机质量取决于最差的零件质量，要求输出计算机最好的质量的大小在能支付的情况下。
题解：先把同种类零件排序放在一起并统计个数，然后把所有零件质量按从大到小排序，然后从最大的质量开始一一判断是否符合。
#include <stdio.h>
#include <iostream>
#include <string>
#include <algorithm>
using namespace std;
const int N = 1005;
const int INF = 0x3f3f3f3f;
struct P {
string type;
int price, quality;
}p[N];
int num[N], t, n, money, p1[N], temp1, k, minn, sum;
int cmp(P a, P b) {
return a.type > b.type;
}
int cmp1(int a, int b) {
return a > b;
}
int solve() {
int count = 0;
for (int i = 0; i < n; i += num[count - 1]) {
minn = INF;
for (int j = i; j < i + num[count]; j++) {
if (p[j].quality >= temp1 && minn > p[j].price) {
minn = p[j].price;
}
}
count++;
if (minn == INF)
return -1;
sum += minn;
}
if (sum <= money)
return temp1;
return -1;
}
int main(){
string temp;
scanf("%d", &t);
while (t--) {
scanf("%d%d", &n, &money);
for (int i = 0; i < n; i++) {
cin >> p[i].type >> temp >> p[i].price >> p[i].quality;
p1[i] = p[i].quality;
}
sort(p, p + n, cmp);
sort(p1, p1 + n, cmp1);
k = 0;
num[k] = 1;
temp = p[0].type;
for (int i = 1; i < n; i++) {
if (p[i].type == temp) {
num[k]++;
}
else {
k++;
temp = p[i].type;
num[k] = 1;
}
}
k++;
temp1 = p1[0];
sum = 0;
int ans = solve();
if (ans != -1) {
printf("%d\n", ans);
continue;
}
for (int q = 1; q < n; q++) {
if (temp1 != p1[q]) {
temp1 = p1[q];
sum = 0;
ans = solve();
if (ans != -1) {
printf("%d\n", ans);
break;
}
}
}
}
return 0;
}
