题意:N天钱序列,最大报销额为M,对于其中的一天可选择报不不报,问最多能报销多少钱。(每天的钱不可拆分)(1 <= N <= 30, 0 <= M <= 10000000)
题目链接:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3631
——>>背包呀,不过要优化,寒假时优化到160ms,今天重刷,优化到了80ms,痛快!
优化思想:
1、从大到小排序;
2、到了最大报销额,更新MAX;
3、将后面的加起来,看是否比现在的MAX小;
4、第3步中将后面的加起来,可事先打个前N项和的表。
#include <cstdio>
#include <algorithm>
using namespace std;
const int maxn = 30 + 10;
int N, M, MAX, a[maxn], sum[maxn];
bool cmp(const int& e1, const int& e2)
{
return e1 > e2;
}
void dp(int cur, int cur_sum)
{
if(cur_sum == M)
{
MAX = cur_sum;
return;
}
if(cur == N+1)
{
MAX = max(MAX, cur_sum);
return;
}
if(cur_sum + sum[N] - sum[cur-1] < MAX) return;
if(cur_sum + a[cur] <= M) dp(cur + 1, cur_sum + a[cur]);
dp(cur + 1, cur_sum);
}
int main()
{
int i;
while(scanf("%d%d", &N, &M) == 2)
{
for(i = 1; i <= N; i++) scanf("%d", &a[i]);
sort(a + 1, a + 1 + N, cmp);
sum[0] = 0;
sum[1] = a[1];
for(i = 2; i <= N; i++) sum[i] = sum[i-1] + a[i];
MAX = 0;
dp(1, 0);
printf("%d\n", MAX);
}
return 0;
}
再次优化,0ms过!!!痛快!!!
优化思想5:
增加恰好到达M的标记。
#include <cstdio>
#include <algorithm>
using namespace std;
const int maxn = 30 + 10;
int N, M, MAX, a[maxn], sum[maxn];
bool ok;
bool cmp(const int& e1, const int& e2)
{
return e1 > e2;
}
void dp(int cur, int cur_sum)
{
if(ok) return;
if(cur_sum == M)
{
MAX = cur_sum;
ok = 1;
return;
}
if(cur == N+1)
{
MAX = max(MAX, cur_sum);
return;
}
if(cur_sum + sum[N] - sum[cur-1] < MAX) return;
if(cur_sum + a[cur] <= M) dp(cur + 1, cur_sum + a[cur]);
dp(cur + 1, cur_sum);
}
int main()
{
int i;
while(scanf("%d%d", &N, &M) == 2)
{
for(i = 1; i <= N; i++) scanf("%d", &a[i]);
sort(a + 1, a + 1 + N, cmp);
sum[0] = 0;
sum[1] = a[1];
for(i = 2; i <= N; i++) sum[i] = sum[i-1] + a[i];
MAX = 0;
ok = 0;
dp(1, 0);
printf("%d\n", MAX);
}
return 0;
}