hdu 3435 A new Graph Game (有向环覆盖,拆点,二分图最优匹配,KM算法)
题意:给你一张图,图上可能有多个哈密顿回路。叫你求出形成多个哈密顿回路的总距离最小值
思路:题意杀啊。。。什么鬼了。。。然后时间。。1000的数据。。n3复杂度。。。还多组数据。。。。不是很懂这个时间是怎么算的。。为毛才2600MS啊。。。。
做法同hdu 1853…
1/* ***********************************************
2Author :111qqz
3Created Time :2016年06月02日 星期四 20时13分25秒
4File Name :code/hdu/1853.cpp
5************************************************ */
6
7#include <cstdio>
8#include <cstring>
9#include <iostream>
10#include <algorithm>
11#include <vector>
12#include <queue>
13#include <set>
14#include <map>
15#include <string>
16#include <cmath>
17#include <cstdlib>
18#include <ctime>
19#define fst first
20#define sec second
21#define lson l,m,rt<<1
22#define rson m+1,r,rt<<1|1
23#define ms(a,x) memset(a,x,sizeof(a))
24typedef long long LL;
25#define pi pair < int ,int >
26#define MP make_pair
27
28using namespace std;
29const double eps = 1E-8;
30const int dx4[4]={1,0,0,-1};
31const int dy4[4]={0,-1,1,0};
32const int inf = 0x3f3f3f3f;
33const int N=1E3+4;
34int n,m;
35bool visx[N],visy[N];
36int link[N];
37int lx[N],ly[N];
38int slk[N];
39int w[N][N];
40
41
42bool find( int u)
43{
44 visx[u] = true;
45 for ( int v = 1 ; v <= n ; v++)
46 {
47 if (visy[v]) continue;
48 int tmp = lx[u] + ly[v] - w[u][v];
49 if (tmp==0)
50 {
51 visy[v] = true;
52
53 if (link[v]==-1||find(link[v]))
54 {
55 link[v] = u;
56 return true;
57 }
58 }else if (tmp<slk[v]) slk[v] = tmp;
59 }
60 return false;
61}
62int KM()
63{
64 ms(lx,0xc0);
65 ms(ly,0);
66 ms(link,-1);
67
68 for ( int i = 1 ; i <= n ; i++)
69 for ( int j = 1 ; j <= n ; j++)
70 lx[i] = max(lx[i],w[i][j]);
71 for ( int i = 1 ; i <= n ; i++) if (lx[i]==-inf-1) return -1; //?
72
73 for ( int i = 1 ; i <= n ; i++)
74 {
75 ms(slk,0x3f);
76
77 while (1)
78 {
79 ms(visx,false);
80 ms(visy,false);
81
82 if (find(i)) break;
83
84 int d = inf;
85
86 for ( int j = 1 ; j <= n ; j++)
87 if (!visy[j]&&slk[j]<d) d = slk[j];
88
89 for ( int j = 1 ; j <= n ; j++)
90 if (visx[j]) lx[j]-=d;
91 for ( int j = 1 ; j <= n ; j++)
92 if (visy[j]) ly[j]+=d ; else slk[j]-=d;
93 }
94 }
95 int res = 0 ;
96
97 for ( int i = 1 ; i <= n ; i++)
98 if (link[i]>-1) res += w[link[i]][i];
99
100 return -res;
101
102}
103int main()
104{
105 #ifndef ONLINE_JUDGE
106 freopen("code/in.txt","r",stdin);
107 #endif
108 int T;
109 int cas = 0 ;
110 cin>>T;
111 while (T--)
112 {
113 scanf("%d %d",&n,&m);
114 ms(w,0xc0);
115 for ( int i = 1 ; i <= m ; i++)
116 {
117 int u,v,cost;
118 scanf("%d%d%d",&u,&v,&cost);
119 w[u][v] =max(w[u][v],-cost); //会不会有重边。..果然有。
120 w[v][u] =max(w[v][u],-cost);
121 }
122
123 int ans = KM();
124 printf("Case %d: ",++cas);
125 if (ans==-1) puts("NO");
126 else printf("%d\n",ans);
127
128
129 }
130
131
132
133
134
135 #ifndef ONLINE_JUDGE
136 fclose(stdin);
137 #endif
138 return 0;
139}
