hdu 3722 Card Game (有向环覆盖,拆点,二分图最佳匹配,KM算法)
题意:n个串,a串放在b串前面的val值是“The score of sticking two cards is the longest common prefix of the second card and the reverse of the first card”.问如何放使得总的val最大。
思路:先暴力处理出每两个的权值。。2002001000的复杂度。。还是可以接受的。。
然后把每个串看成了一个点,由于一个串最多可以被放在前面一次,被放在后面一次,所以可以类比图论中的环的入度和出度为1.
然后跑一遍KM. 1A,开心。
/* ***********************************************
Author :111qqz
Created Time :2016年06月02日 星期四 23时37分55秒
File Name :code/hdu/3722.cpp
************************************************ */
1#include <cstdio>
2#include <cstring>
3#include <iostream>
4#include <algorithm>
5#include <vector>
6#include <queue>
7#include <set>
8#include <map>
9#include <string>
10#include <cmath>
11#include <cstdlib>
12#include <ctime>
13#define fst first
14#define sec second
15#define lson l,m,rt<<1
16#define rson m+1,r,rt<<1|1
17#define ms(a,x) memset(a,x,sizeof(a))
18typedef long long LL;
19#define pi pair < int ,int >
20#define MP make_pair
1using namespace std;
2const double eps = 1E-8;
3const int dx4[4]={1,0,0,-1};
4const int dy4[4]={0,-1,1,0};
5const int inf = 0x3f3f3f3f;
6const int N=205;
7int n;
8char st[N][1005];
9int w[N][N];
10int link[N];
11int lx[N],ly[N];
12bool visx[N],visy[N];
13int slk[N];
14int solve(string a,string b)
15{
16 int la = a.length();
17 int lb = b.length();
18 int len = min(la,lb);
19 int res = 0 ;
20 for ( int i = 0 ; i < len ; i++)
21 {
22 if (b[i]==a[la-1-i]) res++;
23 else break;
24 }
25 return res;
26}
1bool find( int u)
2{
3 // cout<<"u:"<<u<<endl;
4 visx[u] = true;
5 for ( int v = 1 ; v <= n ; v++)
6 {
7 if (visy[v]) continue;
1 int tmp = lx[u] + ly[v] - w[u][v];
2 if (tmp==0)
3 {
4 visy[v] = true;
5 if (link[v]==-1||find(link[v]))
6 {
7 link[v] = u ;
8 return true;
9 }
10 }else if (tmp<slk[v]) slk[v] = tmp;
11 }
12 return false;
13}
14int KM()
15{
16 ms(lx,0);
17 ms(ly,0);
18 ms(link,-1);
1 for ( int i = 1 ; i <= n ; i++)
2 for ( int j = 1 ; j <= n ; j++ )
3 lx[i] = max(lx[i],w[i][j]);
1 for ( int i = 1; i <= n ; i++)
2 {
3 ms(slk,0x3f);
1 while (1)
2 {
3 ms(visx,false);
4 ms(visy,false);
if (find(i)) break;
int d = inf;
for ( int j = 1 ; j <= n ; j++)
if (!visy[j]&&slk[j]<d) d= slk[j];
for ( int j = 1 ; j <= n ; j++)
if (visx[j]) lx[j]-=d;
1 for ( int j = 1 ; j <= n ; j++)
2 if (visy[j]) ly[j]+=d ; else slk[j]-=d;
3 }
4 }
5 int res = 0 ;
6 for ( int i = 1 ; i <= n ; i++)
7 if (link[i]>-1) res += w[link[i]][i];
1 return res;
2}
3int main()
4{
5 #ifndef ONLINE_JUDGE
6 freopen("code/in.txt","r",stdin);
7 #endif
1 while (~scanf("%d",&n))
2 {
3 for ( int i = 1 ; i <= n ; i++) scanf("%s",st[i]);
ms(w,0);
1 for ( int i = 1 ; i <= n ; i++)
2 for ( int j = 1 ; j <= n ; j++)
3 if (i!=j) w[i][j] = solve(string(st[i]),string(st[j]));
// for ( int i = 1 ; i <= n ; i++)
// for ( int j = 1 ; j <= n ; j++) if (i!=j) cout<<"w[i][j]:"<<w[i][j]<<endl;
1 int ans = KM();
2 printf("%d\n",ans);
3 }
1 #ifndef ONLINE_JUDGE
2 fclose(stdin);
3 #endif
4 return 0;
5}