hdu 2448 Mining Station on the Sea (floyd+KM)
题意：n个船n个港口，一个港口只能承接一个船，m个油田，给出n个船各自在哪个油田，然后给出m个油田之间的无相图，然后给出油田和港口之间的有向图。求n个船到达港口的最小距离之和。
思路：想到了用floyd先更新一下距离，然后KM.不过思维不够严谨，只更新了港口通过油田到达油田的距离，而没有更新油田通过油田到达油田的距离QAQ.
所以应该先更新油田通过油田到达油田的距离，然后再更新港口通过油田到达油田的距离。。。
哦，还有。。。不要把n个船所对应的港口作为下标。。而是转化成1..n，这样写KM里会比较好写。。。不然总得带着那个id[i].
1/* ***********************************************
2Author :111qqz
3Created Time :2016年06月03日 星期五 03时07分16秒
4File Name :code/hdu/2448.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=205;
34int n,m,k,p;
35int id[N];
36int a[N][N],b[N][N];
37int lx[N],ly[N];
38int link[N];
39bool visx[N],visy[N];
40int slk[N];
41int w[N][N];
42
43bool find( int u)
44{
45 // cout<<"u:"<<u<<endl;
46 visx[u] = true;
47
48 for ( int v = 1 ; v <= n ; v++)
49 {
50 if (visy[v]) continue;
51 int tmp = lx[u] + ly[v] - w[u][v];
52 if (tmp==0)
53 {
54 visy[v] = true;
55 if (link[v]==-1||find(link[v]))
56 {
57 link[v] = u;
58 return true;
59 }
60 }else if(tmp<slk[v]) slk[v] = tmp;
61 }
62 return false;
63}
64LL KM()
65{
66 ms(link,-1);
67 ms(lx,0xc0);
68 ms(ly,0);
69
70 for ( int i = 1 ; i <= n ; i++)
71 for ( int j = 1 ; j <= n ; j++)
72 lx[i] = max(lx[i],w[i][j]);
73
74
75 for ( int i = 1 ; i <= n ; i++)
76 {
77 ms(slk,0x3f);
78
79 while (1)
80 {
81 ms(visx,false);
82 ms(visy,false);
83
84 if (find(i)) break;
85
86 int d = inf;
87
88 for ( int j = 1 ; j <= n ; j++)
89 if (!visy[j]&&slk[j]<d) d = slk[j];
90
91 for ( int j = 1 ; j <= n ; j++)
92 if (visx[j]) lx[j]-=d;
93 for ( int j = 1 ; j <= n ; j++)
94 if (visy[j]) ly[j]+=d ; else slk[j]-=d;
95 }
96 }
97 LL res = 0LL ;
98
99 for ( int i= 1 ; i <= n ; i++)
100 if (link[i]>-1) res += LL(w[link[i]][i]);
101
102 return -res;
103
104}
105int main()
106{
107 #ifndef ONLINE_JUDGE
108 freopen("code/in.txt","r",stdin);
109 #endif
110
111 while (scanf("%d%d%d%d",&n,&m,&k,&p)!=EOF)
112 {
113 ms(a,0x3f);
114 ms(b,0x3f);
115 ms(id,0);
116 ms(w,0x3f);
117 for ( int i = 1 ;i <= m ; i++) a[i][i] = 0 ;
118 for ( int i = 1 ; i <= n ; i++) scanf("%d",&id[i]);
119
120 for ( int i = 1 ; i <= k ; i++)
121 {
122 int u,v,cost;
123 scanf("%d%d%d",&u,&v,&cost);
124 //a[u][v] = max(a[u][v],-w);
125 a[u][v] = min(a[u][v],cost);
126 a[v][u] = min(a[v][u],cost);
127 }
128
129 for ( int kk = 1 ; kk <= m ; kk++)
130 for ( int i = 1 ; i <= m ; i++)
131 for ( int j = 1 ; j <= m ; j++)
132 a[i][j] = min(a[i][j],a[i][kk]+a[kk][j]);
133 for ( int i = 1 ; i <= p ; i++)
134 {
135 int u,v,cost;
136 scanf("%d%d%d",&u,&v,&cost);
137 for ( int j = 1 ; j <= n ; j++)
138 w[j][u] = min(w[j][u],a[id[j]][v]+cost);
139 }
140
141
142// for ( int kk = 1 ; kk <= m ; kk++) //这只是港口经过中间station到达station的距离近了，还有station经过中间station到达station距离近的情况没考虑,必须放在一起更新。
143// for ( int i = 1 ; i <= n ; i++)
144// for ( int j = 1 ; j <= m ; j++)
145// if (b[i][kk]+a[kk][j]<b[i][j]) b[i][j] = b[i][kk] + a[kk][j];
146
147 for ( int i = 1 ; i <= n ; i++)
148 for ( int j = 1 ; j <= n ; j++)
149 w[i][j]=-w[i][j];
150
151 LL ans = KM();
152 printf("%lld\n",ans);
153
154
155 }
156
157 #ifndef ONLINE_JUDGE
158 fclose(stdin);
159 #endif
160 return 0;
161}