Road constructions
Time Limit: 6000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1475 Accepted Submission(s): 489
Problem Description
N cities are required to connect with each other by a new transportation system. After several rounds of bidding, we have selected M constructions companies and
decided which section is assigned to which company, the associated cost and the direction of each road.
Due to the insufficiency of national fiscal revenue and special taxation system （the tax paid by each company pays is a fixed amount and tax payment occurs at the
beginning of the construction of the project） The government wishes to complete the project in several years and collects as much tax as possible to support the public
expense
For the restrictions of construction and engineering techniques, if a company is required to start the construction, then itself and its associated companies have to
complete all the tasks they commit (if company A constructs a road
from city 1 to city 2, company B constructs a road from city 2 to city 3, company C constructs a road from city 1 to city 3, we call
companies A and B are associated and other company pairs have no such relationship, pay attention, in this example and a are not associated, in other words,’
associated‘ is a directed relationship).
Now the question is what the maximum income the government can obtain in the first year is？
Input
There are multiple cases (no more than 50).
Each test case starts with a line, which contains 2 positive integers, n and m (1<=n<=1000, 1<=m<=5000).
The next line contains m integer which means the tax of each company.
The Third line has an integer k (1<=k<=3000)which indicates the number of the roads.
Then k lines fellow, each contains 4 integers, the start of the roads, the end of the road, the company is responsible for this road and the cost of the road.
The end of the input with two zero
Output
For each test case output the maximum income in a separate line, and if you can not get any income, please output 0.
Sample Input
4 2 500 10 4 1 2 1 10 2 3 1 20 4 3 1 30 1 4 2 60 4 2 500 100 5 1 2 1 10 2 3 1 20 4 3 1 30 4 3 2 10 1 4 2 60 3 1 10 3 1 2 1 100 2 3 1 100 3 1 1 100 0 0
Sample Output
440 470 0 Hint for second test case, if you choose company 2 responsible ways, then you must choose the path of responsible company 1, but if you choose company 1, then you do not have to choose company 2.
Source
2011 Multi-University Training Contest 8 - Host
by HUST
题意：国家现有K条有向路可能被建设（可建可不建），每个有向路由某个公司担任修建，且国家需要支付建设此路的花费，从M个公司中选出一些公司去完成建设 这些被选出的公司所担任修建的所有相关路，如果A公司修u->2，而B公司修2->v，那么选了A公司也必须选B公（即A公司与B公司有关联）。并且这些被选出的公司需要向国家交不同的税，问国家能得到的利益最多是多少。
解题：建图：vs＝0为源点，vt = m+1 为汇点。i , j 为公司，tex[ i ]为交的税，cost[ i ]为 i 公司建设道路国家需支付的花费。
1：< vs , i , tex[ i ] >
2：< i , vt , cost [ i ] >
3：< i , j , INF > （如果 i 公司与 j 公司有关联）
ans = sum{ tex[i]: 1<= i <= m} - maxflow（即最小割）。
#include<stdio.h>
#include<string.h>
#include<queue>
#include<vector>
#include<algorithm>
using namespace std;
#define captype int
const int MAXN = 5010; //点的总数
const int MAXM = 400010; //边的总数
const int INF = 1<<30;
struct EDG{
int to,next;
captype cap;
} edg[MAXM];
int eid,head[MAXN];
int gap[MAXN]; //每种距离(或可认为是高度)点的个数
int dis[MAXN]; //每个点到终点eNode 的最短距离
int cur[MAXN]; //cur[u] 表示从u点出发可流经 cur[u] 号边
int pre[MAXN];
void init(){
eid=0;
memset(head,-1,sizeof(head));
}
//有向边 三个参数，无向边4个参数
void addEdg(int u,int v,captype c,captype rc=0){
edg[eid].to=v; edg[eid].next=head[u];
edg[eid].cap=c; head[u]=eid++;
edg[eid].to=u; edg[eid].next=head[v];
edg[eid].cap=rc; head[v]=eid++;
}
captype maxFlow_sap(int sNode,int eNode, int n){//n是包括源点和汇点的总点个数，这个一定要注意
memset(gap,0,sizeof(gap));
memset(dis,0,sizeof(dis));
memcpy(cur,head,sizeof(head));
pre[sNode] = -1;
gap[0]=n;
captype ans=0; //最大流
int u=sNode;
while(dis[sNode]<n){ //判断从sNode点有没有流向下一个相邻的点
if(u==eNode){ //找到一条可增流的路
captype Min=INF ;
int inser;
for(int i=pre[u]; i!=-1; i=pre[edg[i^1].to]) //从这条可增流的路找到最多可增的流量Min
if(Min>=edg[i].cap){
Min=edg[i].cap;
inser=i;
}
for(int i=pre[u]; i!=-1; i=pre[edg[i^1].to]){
edg[i].cap-=Min;
edg[i^1].cap+=Min; //可回流的边的流量
}
ans+=Min;
u=edg[inser^1].to;
continue;
}
bool flag = false; //判断能否从u点出发可往相邻点流
int v;
for(int i=cur[u]; i!=-1; i=edg[i].next){
v=edg[i].to;
if(edg[i].cap>0 && dis[u]==dis[v]+1){
flag=true;
cur[u]=pre[v]=i;
break;
}
}
if(flag){
u=v;
continue;
}
//如果上面没有找到一个可流的相邻点，则改变出发点u的距离（也可认为是高度）为相邻可流点的最小距离+1
int Mind= n;
for(int i=head[u]; i!=-1; i=edg[i].next)
if(edg[i].cap>0 && Mind>dis[edg[i].to]){
Mind=dis[edg[i].to];
cur[u]=i;
}
gap[dis[u]]--;
if(gap[dis[u]]==0) return ans; //当dis[u]这种距离的点没有了，也就不可能从源点出发找到一条增广流路径
//因为汇点到当前点的距离只有一种，那么从源点到汇点必然经过当前点，然而当前点又没能找到可流向的点，那么必然断流
dis[u]=Mind+1;//如果找到一个可流的相邻点，则距离为相邻点距离+1，如果找不到，则为n+1
gap[dis[u]]++;
if(u!=sNode) u=edg[pre[u]^1].to; //退一条边
}
return ans;
}
struct NODE{
int to,comp;
};
vector<NODE>mp[1005];
int main()
{
int n,m,k,u,v,company,cost[MAXN],c;
NODE now ;
while(scanf("%d%d",&n,&m)>0&&n+m!=0)
{
init();
int vs = 0 , vt = m+1 , ans = 0;
for(int i=1; i<=m; i++){
scanf("%d",&v);
addEdg( vs , i , v );
cost[i] = 0;
ans += v;
}
for(int i=1; i<=n; i++)
mp[i].clear();
scanf("%d",&k);
while(k--){
scanf("%d%d%d%d",&u,&v,&company ,&c);
cost[company] += c;
now.to = v ; now.comp = company ;
mp[u].push_back(now);
}
for(int i=1; i<=m; i++)
addEdg( i , vt , cost[i]);
for(int i=1; i<=n; i++)
for(int l=mp[i].size()-1; l>=0; l-- )
{
u = mp[i][l].to;
int comp1 = mp[i][l].comp;
for(int r=mp[u].size()-1; r>=0; r--)
{
int comp2 = mp[u][r].comp;
if(comp1==comp2)continue;
addEdg( comp1 , comp2 , INF );
}
}
ans -= maxFlow_sap( vs , vt , vt+1 );
printf("%d\n",ans );
}
}
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