The Exchange of Items
Time Limit: 2 Seconds
Memory Limit: 65536 KB
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 one Yith 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
题意：现在有N个物品，进行物物交换，告诉每个物品 i 的初始时的个数A[i]和最终想得到的个数B[i]，M种交换方式，问为了达到目的最少的交换次数是多少。
思路：最近在做最小割，看到什么都想往最小割上套
发现不行，恩，好像费用流可以搞。添加源点ｓ和汇点t，如果A[i]>B[i]，也就是物品i的初始个数大于最终个数，那么我们从源点向i连边，容量为A[i]-B[i]，费用为0，同样如果B[i]>A[i]，i向汇点连边，容量为B[i]-A[i]，费用为0，另外M种交换方式建双向边，容量为INF（因为理论上可以进行无数次交换嘛），费用为1，跑一遍费用流。
代码：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#pragma comment (linker,"/STACK:102400000,102400000")
#define pi acos(-1.0)
#define eps 1e-6
#define lson rt<<1,l,mid
#define rson rt<<1|1,mid+1,r
#define FRE(i,a,b) for(i = a; i <= b; i++)
#define FREE(i,a,b) for(i = a; i >= b; i--)
#define FRL(i,a,b) for(i = a; i < b; i++)
#define FRLL(i,a,b) for(i = a; i > b; i--)
#define mem(t, v) memset ((t) , v, sizeof(t))
#define sf(n) scanf("%d", &n)
#define sff(a,b) scanf("%d %d", &a, &b)
#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
#define pf printf
#define DBG pf("Hi\n")
typedef long long ll;
using namespace std;
#define INF 0x3f3f3f3f
#define mod 1000000009
const int maxn = 10005;
const int MAXN = 2005;
const int MAXM = 200010;
int n,m;
int A[MAXN],B[MAXN],X[MAXN],Y[MAXN];
struct Edge
{
int to,next,cap,flow,cost;
}edge[MAXM];
int head[MAXN],tol;
int pre[MAXN],dis[MAXN];
bool vis[MAXN];
int N;
void init(int n)
{
N=n;
tol=0;
memset(head,-1,sizeof(head));
}
void addedge(int u,int v,int cap,int cost)
{
edge[tol].to=v;
edge[tol].cap=cap;
edge[tol].cost=cost;
edge[tol].flow=0;
edge[tol].next=head[u];
head[u]=tol++;
edge[tol].to=u;
edge[tol].cap=0;
edge[tol].cost=-cost;
edge[tol].flow=0;
edge[tol].next=head[v];
head[v]=tol++;
}
bool spfa(int s,int t)
{
queue<int>q;
for (int i=0;i<N;i++)
{
dis[i]=INF;
vis[i]=false;
pre[i]=-1;
}
dis[s]=0;
vis[s]=true;
q.push(s);
while (!q.empty())
{
int u=q.front();
q.pop();
vis[u]=false;
for (int i=head[u];i!=-1;i=edge[i].next)
{
int v=edge[i].to;
if (edge[i].cap > edge[i].flow && dis[v] > dis[u] + edge[i].cost)
{
dis[v]=dis[u] + edge[i].cost;
pre[v]=i;
if (!vis[v])
{
vis[v]=true;
q.push(v);
}
}
}
}
if (pre[t]==-1) return false;
else return true;
}
//返回的是最大流，cost存的是最小费用
int minCostMaxflow(int s,int t,int &cost)
{
int flow=0;
cost=0;
while (spfa(s,t))
{
int Min=INF;
for (int i=pre[t];i!=-1;i=pre[edge[i^1].to])
{
if (Min > edge[i].cap-edge[i].flow)
Min=edge[i].cap-edge[i].flow;
}
for (int i=pre[t];i!=-1;i=pre[edge[i^1].to])
{
edge[i].flow+=Min;
edge[i^1].flow-=Min;
cost+=edge[i].cost*Min;
}
flow+=Min;
}
return flow;
}
int main()
{
#ifndef ONLINE_JUDGE
freopen("C:/Users/lyf/Desktop/IN.txt","r",stdin);
#endif
int i,j,u,v;
while (~sff(n,m))
{
int all=0;
init(n+2);
int s=0,t=n+1;
for (i=1;i<=n;i++)
{
sff(A[i],B[i]);
if (A[i]>B[i])
addedge(s,i,A[i]-B[i],0);
else if (A[i]<B[i])
{
addedge(i,t,B[i]-A[i],0);
all+=(B[i]-A[i]);
}
}
for (i=0;i<m;i++)
{
sff(u,v);
addedge(u,v,INF,1);
addedge(v,u,INF,1);
}
int cost,ans;
ans=minCostMaxflow(s,t,cost);
if (ans<all)
pf("-1\n");
else
pf("%d\n",cost);
}
return 0;
}
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