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Apple Transportation
Time Limit: 1 Second Memory Limit: 32768 KB
There‘s a big apple tree in the forest. In the tree there are N nodes (numbered from 0 to N - 1), and the nodes are connected by branches. On each node of the tree, there is a squirrel. In the autumn, some apples will grow on the nodes. After all apples are ripe, each squirrel will collect all the apples of their own node and store them. For the demand to be harmonic, they decide to redistribute the apples to minimize the variance (please refer to the hint) of apples in all nodes. Obviously, an apple cannot be divided into several parts. To reach this goal, some transportation should be taken. The cost of transporting x apples from node u to node v is x * distance (node u, node v). Now given the current amount of apples of each node and the structure of the apple tree, you should help the squirrels to find the minimal cost to redistribute the apples.
Input
Input consists of multiple test cases (less than 80 cases)!
For each test case, the first line contains an integer N (1 <= N <= 100), which is the number of the nodes in the tree.
The following line contains N integers a0,a1,...,aN-1 (0 <= ai <= 10000), representing the amount of the i-th node‘s apples.
The following N - 1 lines each contain three integers u, v, c (0 <= u,v <= N - 1, 0 <= c <= 1000), which means node u and node v are connected by a branch, the length of the branch is c.
There is a blank line between consecutive cases.
Output
For each case output the minimal total transportation cost. The minimal cost is guaranteed to be less than 231.
Sample Input
3 1 2 3 0 1 1 0 2 1 3 1 3 3 0 1 3 0 2 4 2 1 2 0 1 1
Sample Output
1 3 0
Hint
The formula to calculate the variance 
of x1, x2, ..., xn:
Author: ZHOU, Yilun
Source: ZOJ Monthly, August 2009
上下界费用流的水题。。。懒的写题解了。。。sad
1 //#####################
2 //Author:fraud
3 //Blog: http://www.cnblogs.com/fraud/
4 //#####################
5 #include <iostream>
6 #include <sstream>
7 #include <ios>
8 #include <iomanip>
9 #include <functional>
10 #include <algorithm>
11 #include <vector>
12 #include <string>
13 #include <list>
14 #include <queue>
15 #include <deque>
16 #include <stack>
17 #include <set>
18 #include <map>
19 #include <cstdio>
20 #include <cstdlib>
21 #include <cmath>
22 #include <cstring>
23 #include <climits>
24 #include <cctype>
25 using namespace std;
26 #define XINF INT_MAX
27 #define INF 0x3FFFFFFF
28 #define MP(X,Y) make_pair(X,Y)
29 #define PB(X) push_back(X)
30 #define REP(X,N) for(int X=0;X<N;X++)
31 #define REP2(X,L,R) for(int X=L;X<=R;X++)
32 #define DEP(X,R,L) for(int X=R;X>=L;X--)
33 #define CLR(A,X) memset(A,X,sizeof(A))
34 #define IT iterator
35 typedef long long ll;
36 typedef pair<int,int> PII;
37 typedef vector<PII> VII;
38 typedef vector<int> VI;
39 struct edge
40 {
41 int to,cap,cost,rev;
42 edge(int _to,int _cap,int _cost,int _rev)
43 {
44 to=_to;cap=_cap;cost=_cost;rev=_rev;
45 }
46 };
47 int V;
48 const int MAX_V=410;
49 vector<edge> G[MAX_V];
50 int dis[MAX_V];
51 int prevv[MAX_V],preve[MAX_V];//最短路中的前驱结点和对应的边
52 void add_edge(int from,int to,int cap,int cost)
53 {
54 G[from].push_back(edge(to,cap,cost,G[to].size()));
55 G[to].push_back(edge(from,0,-cost,G[from].size()-1));
56 }
57 int vis[MAX_V];
58 int min_cost_flow(int s,int t,int f)//如果不能在增广则返回-1
59 {
60 int res=0;
61 while(f>0)
62 {
63 fill(dis,dis+V,INF);
64 dis[s]=0;
65 queue<int>q;
66 CLR(vis,0);
67 q.push(s);
68 while(!q.empty())
69 {
70 int v=q.front();
71 q.pop();
72 vis[v]=0;
73 for(int i=0;i<G[v].size();i++)
74 {
75 edge &e=G[v][i];
76 if(e.cap>0&&dis[e.to]>dis[v]+e.cost)
77 {
78 dis[e.to]=dis[v]+e.cost;
79 prevv[e.to]=v;
80 preve[e.to]=i;
81 if(!vis[e.to])
82 {
83 q.push(e.to);
84 vis[e.to]=1;
85 }
86 }
87 }
88 }
89 /* bool update=1;
90 while(update)
91 {
92 update=false;
93 for(int v=0;v<V;v++)
94 {
95 if(dis[v]==INF) continue;
96 for(int i=0;i<G[v].size();i++)
97 {
98 edge &e=G[v][i];
99 if(e.cap>0&&dis[e.to]>dis[v]+e.cost)
100 {
101 dis[e.to]=dis[v]+e.cost;
102 prevv[e.to]=v;
103 preve[e.to]=i;
104 update=1;
105 }
106 }
107 }
108 }*/
109 if(dis[t]==INF)
110 {
111 return -1;
112 }
113 int d=f;
114 for(int v=t;v!=s;v=prevv[v])
115 {
116 d=min(d,G[prevv[v]][preve[v]].cap);
117 }
118 f-=d;
119 res+=d*dis[t];
120 for(int v=t;v!=s;v=prevv[v])
121 {
122 edge &e=G[prevv[v]][preve[v]];
123 e.cap-=d;
124 G[v][e.rev].cap+=d;
125 }
126 //cout<<f<<endl;
127 // cout<<"ok"<<endl;
128 }
129 return res;
130 }
131 int a[MAX_V];
132 int main()
133 {
134 int n;
135 while(scanf("%d",&n)!=EOF){
136 int sum=0;
137 for(int i=1;i<=n;i++){
138 scanf("%d",a+i);
139 sum+=a[i];
140 }
141 CLR(prevv,-1);
142 CLR(preve,-1);
143 for(int i=0;i<n+3;i++)G[i].clear();
144 int s=0,tt=n+1,t=n+2;
145 int temp=sum/n;
146 V=t+1;
147 for(int i=1;i<=n;i++){
148 add_edge(s,i,a[i],0);
149 add_edge(i,tt,1,0);
150 add_edge(i,t,temp,0);
151 }
152 add_edge(tt,t,sum-n*temp,0);
153 int u,v,d;
154
155 for(int i=0;i<n-1;i++){
156 scanf("%d%d%d",&u,&v,&d);
157 u++;
158 v++;
159 add_edge(u,v,INF,d);
160 add_edge(v,u,INF,d);
161 }
162 printf("%d\n",min_cost_flow(s,t,sum));
163
164
165 }
166 return 0;
167 }
代码君