poj1171

pairs的组合只可能是3+3或者3+4

所以其余长度的单词无需记录

把记录后的单词n^2枚举pair

View Code
#include <iostream>
#include
<cstdlib>
#include
<cstring>
#include
<cstdio>
using namespace std;

char pre1[30] = "qwertyuiopasdfghjklzxcvbnm";
int pre2[30] =
{
7, 6, 1, 2, 2, 5, 4, 1, 3, 5, 2, 1, 4, 6, 5, 5, 7, 6, 3, 7, 7, 4, 6, 5, 2, 5 };
int value[300];
char collected[10];
char word[40005][10];
int n, ans;

int cal(char * st, int len)
{
int ret = 0;
for (int i = 0; i < len; i++)
ret
+= value[st[i]];
return ret;
}

bool ok(char *a, char *b)
{
int len1 = strlen(a);
int len2 = strlen(b);
int len = strlen(collected);
if (len1 + len2 > len)
return false;
bool vis[10];
memset(vis,
0, sizeof(vis));
for (int i = 0; i < len1; i++)
{
bool did = false;
for (int j = 0; j < len; j++)
if (collected[j] == a[i] && !vis[j])
{
vis[j]
= true;
did
= true;
break;
}
if (!did)
return false;
}
for (int i = 0; i < len2; i++)
{
bool did = false;
for (int j = 0; j < len; j++)
if (collected[j] == b[i] && !vis[j])
{
vis[j]
= true;
did
= true;
break;
}
if (!did)
return false;
}
return true;
}

int main()
{
//freopen("t.txt", "r", stdin);
for (int i = 0; i < 26; i++)
value[pre1[i]]
= pre2[i];
scanf(
"%s", collected);
int lenc = strlen(collected);
char now[10];
n
= 0;
ans
= 0;
while (scanf("%s", now), strcmp(".", now))
{
int len = strlen(now);
if (len > lenc || !ok(now, ""))
continue;
ans
= max(ans, cal(now, len));
if (len <= lenc / 2 + (lenc & 1))
strcpy(word[n
++], now);
}
for (int i = 0; i < n - 1; i++)
for (int j = i; j < n; j++)
if (ok(word[i], word[j]))
ans
= max(ans, cal(word[i], strlen(word[i])) + cal(word[j], strlen(word[j])));
printf(
"%d\n", ans);
return 0;
}
原文地址:https://www.cnblogs.com/rainydays/p/2158826.html