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题目链接:C. Booking System
题意:n个旅游团,每个团有一定人数,和开销,现在一个餐馆有k个桌子,每个桌子能坐一定人数,要把这些桌子分配给旅游团,一定要能坐的人数大于旅游团人数才能坐下,问最多能赚的钱,并输出旅游团桌子匹配方案。
思路:贪心,从钱最多的旅游团开始放,每次从最小的桌子开始找,然后注意最后输出的id号,所以排序前要多存一个id
代码:
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
const int N = 1005;int n, vis[N], k;
struct Visit {int num, value, id;
} v[N];struct Table {int num, id;
} t[N];bool cmp(Visit a, Visit b) {if (a.value != b.value)return a.value > b.value;return a.num > b.num;
}bool cmp2(Table a, Table b) {return a.num < b.num;
}int main() {scanf("%d", &n);for (int i = 0; i < n; i++) {scanf("%d%d", &v[i].num, &v[i].value);v[i].id = i;}sort(v, v + n, cmp);scanf("%d", &k);for (int i = 0; i < k; i++) {scanf("%d", &t[i].num);t[i].id = i;}sort(t, t + k, cmp2);int ans1 = 0, ans2 = 0, ans[N];memset(ans, -1, sizeof(ans));for (int i = 0; i < n; i++) {for (int j = 0; j < k; j++) {if (vis[j]) continue;if (t[j].num >= v[i].num) {vis[j] = 1;ans[i] = t[j].id;ans1++;ans2 += v[i].value;break;}}}printf("%d %d\n", ans1, ans2);for (int i = 0; i < n; i++) {if (ans[i] == -1) continue;printf("%d %d\n", v[i].id + 1, ans[i] + 1);}return 0;
}
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