题目:http://acm.hdu.edu.cn/showproblem.php?pid=3722
大意就是给n个字符串,然后对于任意两个字符串进行匹配,第一个倒序,第二个正序,找它们的最长公共前缀长度,就是它们的分数,自己和自己匹配分数为0,然后把这些字符串组成一些圈,求能得到的最大分数
思路:预先处理出任意的两个字符串的分数,然后KM。。。
#include <iostream>
#include <string>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cctype>
#include <vector>
using namespace std;
const int N = 210;
const int INF = 0x3f3f3f3f;
int nx, ny;
int lx[N], ly[N], match[N], slack[N];
bool visx[N], visy[N];
int s[N][N];
char str[N][1010];
bool hungary(int v)
{
visx[v] = true;
for(int i = 0; i < ny; i++)
{
if(visy[i]) continue;
if(lx[v] + ly[i] == s[v][i])
{
visy[i] = true;
if(match[i] == -1 || hungary(match[i]))
{
match[i] = v;
return true;
}
}
else slack[i] = min(slack[i], lx[v] + ly[i] - s[v][i]);
}
return false;
}
void km()
{
memset(match, -1, sizeof match);
memset(ly, 0, sizeof ly);
memset(lx, 0, sizeof lx);
for(int i = 0; i < nx; i++)
for(int j = 0; j < ny; j++)
lx[i] = max(lx[i], s[i][j]);
for(int i = 0; i < nx; i++)
{
memset(slack, 0x3f, sizeof slack);
while(true)
{
memset(visx, 0, sizeof visx);
memset(visy, 0, sizeof visy);
if(hungary(i)) break;
else
{
int tmp = INF;
for(int j = 0; j < ny; j++)
if(!visy[j]) tmp = min(tmp, slack[j]);
for(int j = 0; j < nx; j++)
if(visx[j]) lx[j] -= tmp;
for(int j = 0; j < ny; j++)
if(visy[j]) ly[j] += tmp;
else slack[j] -= tmp;
}
}
}
}
int judge(char s1[], char s2[])
{
int res = 0;
for(int i = strlen(s1) - 1, j = 0; s1[i] && s2[j]; i--, j++)
if(s1[i] != s2[j]) return res;
else res++;
return res;
}
int main()
{
int n;
while(~ scanf("%d", &n))
{
for(int i = 0; i < n; i++)
scanf(" %s", str[i]);
for(int i = 0; i < n; i++)
for(int j = 0; j < n; j++)
if(i == j) s[i][j] = 0;
else s[i][j] = judge(str[i], str[j]);
nx = ny = n;
km();
int res = 0;
for(int i = 0; i < n; i++)
res += s[match[i]][i];
printf("%d\n", res);
}
return 0;
}