poj1456Supermarket 贪心解法

传送门

题意:

给出n个产品的利润和贩卖截止时间,单位时间只能贩卖一个产品.求最终获得的最大利润.

分析:

要获得最大利润,在同一时间肯定优先考虑利润较大的产品;

所以可以根据利润进行排序,优先处理利润大的产品,最终获得的肯定是最大利润.

#include <cstdio>#include <cstdlib>#include <iostream>#include <stack>#include <queue>#include <algorithm>#include <cstring>#include <string>#include <cmath>#include <vector>#include <bitset>#include <list>#include <sstream>#include <set>#include <functional>using namespace std;#define INF 0x3f3f3f3f#define MAX 100typedef long long ll;struct node{int p,d;bool operator < (const node&a)const{return p < a.p;}}v[10005];int n;int place[10005];int use[10005];int main(int argc, char const* argv[]){#ifndef ONLINE_JUDGE    freopen("in.txt", "r", stdin);    freopen("out.txt", "w", stdout);#endifwhile (scanf("%d",&n)!=EOF){ll ans = 0;for (int i = 0; i < n; i += 1) {scanf("%d%d",&v[i].p,&v[i].d);}sort(v,v+n);memset(place,0,sizeof(place));memset(use,0,sizeof(use));for (int i = n-1; i >= 0; i -= 1){int t = v[i].d;while (use[t]&&t>0){t = t-1;}if (t>0){ans += v[i].p;use[t] = 1;}}printf("%lld\n",ans);}return 0;}

可以对上述代码进行优化,每次处理产品后记录相同截止时间的下个可用时间点.这样后面截止时间相同的就不用从后往前查找了.

#include <cstdio>#include <cstdlib>#include <iostream>#include <stack>#include <queue>#include <algorithm>#include <cstring>#include <string>#include <cmath>#include <vector>#include <bitset>#include <list>#include <sstream>#include <set>#include <functional>using namespace std;#define INF 0x3f3f3f3f#define MAX 100typedef long long ll;struct node{int p,d;bool operator < (const node&a)const{return p < a.p;}}v[10005];int n;int nex[10005];int use[10005];int main(int argc, char const* argv[]){#ifndef ONLINE_JUDGE    freopen("in.txt", "r", stdin);    freopen("out.txt", "w", stdout);#endifwhile (scanf("%d",&n)!=EOF){ll ans = 0;memset(nex,0,sizeof(nex));memset(use,0,sizeof(use));for (int i = 0; i < n; i += 1) {scanf("%d%d",&v[i].p,&v[i].d);nex[v[i].d] = v[i].d;}sort(v,v+n);for (int i = n-1; i >= 0; i -= 1){int t = nex[v[i].d];while (use[t]&&t>0) t = t-1;//找到可用时间点if (t>0){ans += v[i].p;use[t] = 1;nex[v[i].d] = t-1;}}printf("%lld\n",ans);}return 0;}