题目就是类似背包问题,给容量为10的书架选情感值较大的书,并且这些书相邻的是不能在相同位置是相同的字母。
TLE超时了,今天是个失败的题解,求帮忙呀!
#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<vector>
using namespace std;
struct node {
int val;
char pre[101];
char cur[101];
}a[2501];
bool cmp(node a, node b)
{
if (strcmp(a.cur, b.cur) < 0)
return 1;
else
return 0;
}
bool del(char* a, char* b)
{
int i=0;
while (a[i] != '\0' && b[i] != '\0')
{
if (a[i] == b[i])
return true;
i++;
}
return false;
}
int main()
{
int n;
scanf("%d", &n);
for (int i = 0; i < n; i++)
{
scanf("%d",&a[i].val); int sub = 0; int sub1 = 0;
char ch;
ch = getchar();
cin.getline(a[i].pre,100);
while((ch=a[i].pre[sub++])!='\0')
{
if(ch!='\''&&ch!=' ')
{
if(ch>='A'&&ch<='Z')
a[i].cur[sub1++]=ch+32;
else
a[i].cur[sub1++]=ch;
}
}
}
sort(a, a + n, cmp);
vector<node>ans[2501][11];
int data[2501][11] = {
0 };
for (int i = 1; i <= n; i++)
{
for (int j = 1; j <= 10; j++)
{
int k = i-1;
while (ans[k][j - 1].size() != 0 && del(a[i - 1].cur, ans[k][j - 1][ans[k][j - 1].size() - 1].cur))
{
k -= 1;
if (k == 0)
break;
}
data[i][j] = data[k ][j - 1] + a[i - 1].val;
ans[i][j] = ans[k][j - 1];
ans[i][j].push_back(a[i - 1]);
if (data[i][j] < data[i - 1][j])
{
data[i][j] = data[i - 1][j];
ans[i][j] = ans[i - 1][j];
}
}
}
printf("%d\n%d\n", ans[n][10].size(), data[n][10]);
for (int i = 0; i < ans[n][10].size(); i++)
{
printf("%s\n", ans[n][10][i].pre);
}
return 0;
}