The Exchange of Items
64-bit integer IO format: %lld Java class name: Main
Bob lives in an ancient village, where transactions are done by one item exchange with another. Bob is very clever and he knows what items will become more valuable later on. So, Bob has decided to do some business with villagers.
At first, Bob has N kinds of items indexed from 1 to N, and each item has Ai. There are M ways to exchanges items. For the ith way (Xi, Yi), Bob can exchange one Xith item to oneYith item, vice versa. Now Bob wants that his ith item has exactly Bi, and he wonders what the minimal times of transactions is.
Input
There are multiple test cases.
For each test case: the first line contains two integers: N and M (1 <= N, M <= 100).
The next N lines contains two integers: Ai and Bi (1 <= Ai, Bi <= 10,000).
Following M lines contains two integers: Xi and Yi (1 <= Xi, Yi <= N).
There is one empty line between test cases.
Output
For each test case output the minimal times of transactions. If Bob could not reach his goal, output -1 instead.
Sample Input
2 1 1 2 2 1 1 2 4 2 1 3 2 1 3 2 2 3 1 2 3 4
Sample Output
1 -1
Source
Author
1 #include <bits/stdc++.h> 2 using namespace std; 3 const int INF = 0x3f3f3f3f; 4 const int maxn = 500; 5 struct arc{ 6 int to,flow,cost,next; 7 arc(int x = 0,int y = 0,int z = 0,int nxt = -1){ 8 to = x; 9 flow = y; 10 cost = z; 11 next = nxt; 12 } 13 }e[maxn*maxn]; 14 int head[maxn],p[maxn],tot; 15 void add(int u,int v,int flow,int cost){ 16 e[tot] = arc(v,flow,cost,head[u]); 17 head[u] = tot++; 18 e[tot] = arc(u,0,-cost,head[v]); 19 head[v] = tot++; 20 } 21 bool in[maxn]; 22 int d[maxn],S,T; 23 bool spfa(){ 24 queue<int>q; 25 memset(d,0x3f,sizeof d); 26 memset(in,false,sizeof in); 27 memset(p,-1,sizeof p); 28 d[S] = 0; 29 q.push(S); 30 while(!q.empty()){ 31 int u = q.front(); 32 q.pop(); 33 in[u] = false; 34 for(int i = head[u]; ~i; i = e[i].next){ 35 if(e[i].flow && d[e[i].to] > d[u] + e[i].cost){ 36 d[e[i].to] = d[u] + e[i].cost; 37 p[e[i].to] = i; 38 if(!in[e[i].to]){ 39 in[e[i].to] = true; 40 q.push(e[i].to); 41 } 42 } 43 } 44 } 45 return p[T] > -1; 46 } 47 void solve(int sum){ 48 int cost = 0,flow = 0; 49 while(spfa()){ 50 int minF = INF; 51 for(int i = p[T]; ~i; i = p[e[i^1].to]) 52 minF = min(minF,e[i].flow); 53 for(int i = p[T]; ~i; i = p[e[i^1].to]){ 54 e[i].flow -= minF; 55 e[i^1].flow += minF; 56 } 57 flow += minF; 58 cost += d[T]*minF; 59 } 60 if(sum == flow) printf("%d ",cost); 61 else puts("-1"); 62 } 63 int main(){ 64 int n,m,u,v,sum; 65 bool flag = false; 66 while(~scanf("%d%d",&n,&m)){ 67 //if(flag) puts(""); 68 memset(head,-1,sizeof head); 69 sum = S = tot = 0; 70 T = n + 1; 71 for(int i = 1; i <= n; ++i){ 72 scanf("%d%d",&u,&v); 73 add(S,i,u,0); 74 add(i,T,v,0); 75 sum += v; 76 } 77 for(int i = 0; i < m; ++i){ 78 scanf("%d%d",&u,&v); 79 add(u,v,INF,1); 80 add(v,u,INF,1); 81 } 82 solve(sum); 83 } 84 return 0; 85 }