Less Time, More profit
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Problem Description
The city planners plan to build N plants in the city which has M shops. Each shop needs products from some plants to make profit of proi units.Building ith plant needs investment of payi units and it takes ti days.Two or more plants can be built simultaneously, so that the time for building multiple plants is maximum of their periods(ti).You should make a plan to make profit of at least L units in the shortest period.InputFirst line contains T, a number of test cases.For each test case, there are three integers N, M, L described above.And there are N lines and each line contains two integers payiti(1<= i <= N).Last there are M lines and for each line, first integer is proi, and there is an integer k and next k integers are index of plants which can produce material to make profit for the shop. 1 <= T <= 30 1 <= N, M <= 200
Output
For each test case, first line contains a line “Case #x: t p”, x is the number of the case, t is the shortest period and p is maximum profit in t hours. You should minimize t first and then maximize p. If this plan is impossible, you should print “Case #x: impossible”
Sample Input
2
1 1 2
1 5
3 1 1
1 1 3
1 5
3 1 1
Sample Output
Case #1: 5 2
Case #2: impossible
Author
金策工业综合大学（DPRK）
Source
2016 Multi-University Training Contest 9
M个商店，N个工厂，每个商店获利的条件是建设了指定的k个工厂。求总获利不小于L，工厂建设的时间最大值最小是多少。
工厂到汇点建一条边pay[i]，源点到商店建一条边pro[i]，商店到需要的工厂建一条边INF，商店的总收益-最小割就是答案。
可以看看国家集训队论文：Amber《最小割模型在信息学竞赛中的应用》
代码语言：javascript复制#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <queue>
#define N 205<<1
#define sf(a) scanf("%d",&a);
using namespace std;
const int INF=0x3f3f3f3f;
struct plant{
int pay,t,id;
}pt[N];
struct shop{
int pro,k,pt[N],t;
}s[N];
int t,n,m,l,st,ed,tot;
int arc[N][N], d[N];
int ans,tans;
bool bfs()
{
memset(d, -1, sizeof d);
queue<int>q;
q.push(st);
d[st] = 0;
while(!q.empty())
{
int i,k=q.front();
q.pop();
for(i = 1; i <= ed; i )
if(arc[k][i] > 0 && d[i] == -1)
{
d[i] = d[k] 1;
q.push(i);
}
}
return d[ed]>0;
}
int dinic (int k, int low)
{
if(k == ed)return low;
int a,i;
for(i = 1; i <= ed; i )
if(d[i] == d[k] 1&& arc[k][i] > 0 &&(a = dinic(i, min(low, arc[k][i]))))
{
arc[k][i] -= a;
arc[i][k] = a;
return a;
}
return 0;
}
int cmp(plant a,plant b){
return a.t<b.t;
}
int main() {
sf(t);
for(int cas=1;cas<=t;cas ){
printf("Case #%d: ",cas);
scanf("%d%d%d",&n,&m,&l);
for(int i=1;i<=n;i ){
scanf("%d%d",&pt[i].pay,&pt[i].t);
pt[i].id=i;
}
for(int i=1;i<=m;i ){
sf(s[i].pro);
sf(s[i].k);
s[i].t=0;
for(int j=1;j<=s[i].k;j ){
int x;
sf(x);
s[i].pt[j]=x;
s[i].t=max(s[i].t,pt[x].t);
}
}
sort(pt 1,pt 1 n,cmp);
int ok=0;
st=n m 1,ed=st 1;
for(int i=1;i<=n;i ){
memset(arc,0,sizeof arc);
for(int j=1;j<=i;j )
arc[pt[j].id][ed]=pt[j].pay;
tot=0;
for(int j=1;j<=m;j )if(s[j].t<=pt[i].t){
tot =s[j].pro;
arc[st][j n]=s[j].pro;
for(int k=1;k<=s[j].k;k )
arc[j n][s[j].pt[k]]=INF;
}
ans = 0;
while(bfs())
while(tans=dinic(st, INF)) ans = tans;
ans=tot-ans;
if(ans>=l){
printf("%d %dn",pt[i].t,ans);
ok=1;
break;
}
}
if(!ok)puts("impossible");
}
}工厂按时间排序再依次增大最大时间，这样是140ms，改成二分以为会更快，结果960ms，去了注释们再交一次居然超时了。可能是dinic用的是邻接矩阵的原因。
