There will be several test cases in the input. Each test case will begin with a line with three integers:
N A B
Where N is the number of teams (1
N1, 000), and A and B are the number of balloons in rooms A and B, respectively (0A, B10, 000). On each of the next N lines there will be three integers, representing information for each team:
K DA DB
Where K is the total number of balloons that this team will need, DA is the distance of this team from room A, and DB is this team's distance from room B (0DA, DB1, 000). You may assume that there are enough balloons - that is,
(K's)A + B. The input will end with a line with three 0s.
Output
For each test case, output a single integer, representing the minimum total distance that must be traveled to deliver all of the balloons. Count only the outbound trip, from room A or room B to the team. Don't count the distance that a runner must travel to return to room A or room B. Print each integer on its own line with no spaces. Do not print any blank lines between answers.
Sample Input
3 15 35 10 20 10 10 10 30 10 40 10 0 0 0
Sample Output
300
费用流代码：
#include <cstdio> #include <cstring> using namespace std; #define MAXN 2000 #define MAXM 4000 #define INF 0x3f3f3f3f #define MIN(a,b) (a<b?a:b) #define V(p) edge[(p)].v #define F(p) edge[(p)].f #define C(p) edge[(p)].c #define Nx(p) edge[(p)].next int n,A,B,s,t,ans,ecnt; int dis[MAXN],adj[MAXN]; bool vis[MAXN]; struct node { int v,f,c,next; }edge[MAXM*2]; void Addedge(int u,int v,int f,int c) { ++ecnt; V(ecnt)=v; F(ecnt)=f; C(ecnt)=c; Nx(ecnt)=adj[u]; adj[u]=ecnt; ++ecnt; V(ecnt)=u; F(ecnt)=0; C(ecnt)=-c; Nx(ecnt)=adj[v]; adj[v]=ecnt; } int Aug(int u,int lim) { if(u==t){ ans+=lim*dis[s]; return lim; } vis[u]=true; int p,v,f,c,delta,sum=0; for(p=adj[u];p;p=Nx(p)){ v=V(p); f=F(p); c=C(p); if(vis[v] || !f || dis[v]+c!=dis[u]) continue; delta=Aug(v,MIN(f,(lim-sum))); F(p)-=delta; F(p^1)+=delta; sum+=delta; if(sum==lim) break; } return sum; } bool Update() { int i,p,Min=INF; for(i=s;i<=t;++i){ if(!vis[i]) continue; for(p=adj[i];p;p=Nx(p)){ if(!F(p) || vis[V(p)]) continue; Min=MIN(Min,(dis[V(p)]+C(p)-dis[i])); } } if(Min==INF) return false; for(i=s;i<=t;++i){ if(!vis[i]) continue; dis[i]+=Min; } return true; } void ZKW() { do{ for(memset(vis,0,sizeof(vis));Aug(s,INF);memset(vis,0,sizeof(vis))); }while(Update()); printf("%d\n",ans); } void Init() { memset(adj,0,sizeof(adj)); memset(dis,0,sizeof(dis)); ans=0; ecnt=1; s=0; t=n+3; int i,k,da,db; Addedge(s,n+1,A,0); Addedge(s,n+2,B,0); for(i=1;i<=n;++i){ scanf("%d%d%d",&k,&da,&db); Addedge(n+1,i,INF,da); Addedge(n+2,i,INF,db); Addedge(i,t,k,0); } } int main() { while(true){ scanf("%d%d%d",&n,&A,&B); if(!n && !A && !B) break; Init(); ZKW(); } return 0; }
贪心代码：
#include <cstdio> #include <cmath> #include <cstring> #include <ctime> #include <iostream> #include <algorithm> #include <set> #include <vector> #include <sstream> #include <queue> #include <typeinfo> #include <fstream> typedef long long ll; using namespace std; //freopen("D.in","r",stdin); //freopen("D.out","w",stdout); #define sspeed ios_base::sync_with_stdio(0);cin.tie(0) #define maxn 10011 const int inf=0x7fffffff; //无限大 struct node { int total; int da; int db; int dis; }; bool cmp(node a,node b) { return a.dis>b.dis; } node team[maxn]; int main() { sspeed; int n,a,b; while(cin>>n>>a>>b) { if(n==0&&a==0&&b==0) break; int sum=0; for(int i=0;i<n;i++) { cin>>team[i].total>>team[i].da>>team[i].db; team[i].dis=fabs(team[i].da-team[i].db); } sort(team,team+n,cmp); for(int i=0;i<n;i++) { if(team[i].da<team[i].db) { if(a>=team[i].total) { sum+=team[i].da*team[i].total; a-=team[i].total; } else { sum+=team[i].da*a; sum+=(team[i].total-a)*team[i].db; b-=(team[i].total-a); a=0; } } else { if(b>=team[i].total) { sum+=team[i].db*team[i].total; b-=team[i].total; } else { sum+=team[i].db*b; sum+=(team[i].total-b)*team[i].da; a-=(team[i].total-b); b=0; } } } cout<<sum<<endl; } }
