Codeforces 416C Booking System(贪心)
2014-04-15 07:31
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题目链接:Codeforces 416C Booking System
题目大意:餐馆有n批客人,每批客人给出人数和消费金额。然后餐馆有m个桌子,给出每个桌子可以坐的人数。问说在这一时刻,餐馆可以最大盈利多少,并且接待尽量多的客人。
解题思路:将客人按照消费金额、人数排序,然后桌子按照从小到大排序。消费高的客人尽量找最小满足的桌子给他们。注意输出的时候客人有序号,桌子同样有序号,不能打乱。
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
const int N = 1005;
struct state {
int id, val, sum;
state () {
}
}g
, t
;
int n, m, v
;
bool cmpS (const state& a, const state& b) {
if (a.val != b.val) return a.val > b.val;
return a.sum > b.sum;
}
bool cmpT (const state& a, const state& b) {
return a.sum < b.sum;
}
void init () {
memset(v, 0, sizeof(v));
scanf("%d", &n);
for (int i = 0; i < n; i++) {
scanf("%d%d", &g[i].sum, &g[i].val);
g[i].id = i;
}
scanf("%d", &m);
for (int i = 0; i < m; i++) {
scanf("%d", &t[i].sum);
t[i].id = i;
}
sort(g, g + n, cmpS);
sort(t, t + m, cmpT);
}
void solve () {
int ans = 0, cnt = 0;
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
if (g[i].sum <= t[j].sum) {
cnt++;
ans += g[i].val;
v[g[i].id] = t[j].id + 1;
t[j].sum = 0;;
break;
}
}
}
printf("%d %d\n", cnt, ans);
for (int i = 0; i < n; i++) {
if (v[i]) printf("%d %d\n", i + 1, v[i]);
}
}
int main () {
init ();
solve ();
return 0;
}
题目大意:餐馆有n批客人,每批客人给出人数和消费金额。然后餐馆有m个桌子,给出每个桌子可以坐的人数。问说在这一时刻,餐馆可以最大盈利多少,并且接待尽量多的客人。
解题思路:将客人按照消费金额、人数排序,然后桌子按照从小到大排序。消费高的客人尽量找最小满足的桌子给他们。注意输出的时候客人有序号,桌子同样有序号,不能打乱。
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
const int N = 1005;
struct state {
int id, val, sum;
state () {
}
}g
, t
;
int n, m, v
;
bool cmpS (const state& a, const state& b) {
if (a.val != b.val) return a.val > b.val;
return a.sum > b.sum;
}
bool cmpT (const state& a, const state& b) {
return a.sum < b.sum;
}
void init () {
memset(v, 0, sizeof(v));
scanf("%d", &n);
for (int i = 0; i < n; i++) {
scanf("%d%d", &g[i].sum, &g[i].val);
g[i].id = i;
}
scanf("%d", &m);
for (int i = 0; i < m; i++) {
scanf("%d", &t[i].sum);
t[i].id = i;
}
sort(g, g + n, cmpS);
sort(t, t + m, cmpT);
}
void solve () {
int ans = 0, cnt = 0;
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
if (g[i].sum <= t[j].sum) {
cnt++;
ans += g[i].val;
v[g[i].id] = t[j].id + 1;
t[j].sum = 0;;
break;
}
}
}
printf("%d %d\n", cnt, ans);
for (int i = 0; i < n; i++) {
if (v[i]) printf("%d %d\n", i + 1, v[i]);
}
}
int main () {
init ();
solve ();
return 0;
}
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