Road constructions
Time Limit: 6000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1436 Accepted Submission(s): 467
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
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题意：N个城市，m条道路需要建设，每条道路由某个公司负责并花费一定的费用，每个公司会向国家纳税，若公司A造1到2的路，B造2到3的路，那么就说A和B有联系。求国家能得到的钱的最大值。ans=纳税之和-建路花费之和。
思路：做最大权闭合图的题就是要找到有相互依赖关系的两个集合。题目中告诉了公司之间有依赖关系，那么就以公司为节点，源点和每个公司连边权为纳税，公司和会点连边权为花费，公司与公司之间有联系的连边权为INF。
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 maxn 1005
#define MAXN 5005
#define mod 1000000009
#define INF 0x3f3f3f3f
#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 FRL(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")
const int MAXM = 200010;
typedef long long ll;
using namespace std;
struct Edge
{
int to,next,cap,flow;
}edge[MAXM];
int tol;
int head[MAXN];
int gap[MAXN],dep[MAXN],pre[MAXN],cur[MAXN];
void init()
{
tol=0;
memset(head,-1,sizeof(head));
}
//加边，单向图三个参数，双向图四个参数
void addedge(int u,int v,int w,int rw=0)
{
edge[tol].to=v; edge[tol].cap=w; edge[tol].next=head[u];
edge[tol].flow=0; head[u]=tol++;
edge[tol].to=u; edge[tol].cap=rw; edge[tol].next=head[v];
edge[tol].flow=0; head[v]=tol++;
}
//输入参数：起点，终点，点的总数
//点的编号没有影响，只要输入点的总数
int sap(int start,int end,int N)
{
memset(gap,0,sizeof(gap));
memset(dep,0,sizeof(dep));
memcpy(cur,head,sizeof(head));
int u=start;
pre[u]=-1;
gap[0]=N;
int ans=0;
while (dep[start]<N)
{
if (u==end)
{
int Min=INF;
for (int i=pre[u];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[u];i!=-1;i=pre[edge[i^1].to])
{
edge[i].flow+=Min;
edge[i^1].flow-=Min;
}
u=start;
ans+=Min;
continue;
}
bool flag=false;
int v;
for (int i=cur[u];i!=-1;i=edge[i].next)
{
v=edge[i].to;
if (edge[i].cap-edge[i].flow && dep[v]+1==dep[u])
{
flag=true;
cur[u]=pre[v]=i;
break;
}
}
if (flag)
{
u=v;
continue;
}
int Min=N;
for (int i=head[u];i!=-1;i=edge[i].next)
if (edge[i].cap-edge[i].flow && dep[edge[i].to]<Min)
{
Min=dep[edge[i].to];
cur[u]=i;
}
gap[dep[u]]--;
if (!gap[dep[u]]) return ans;
dep[u]=Min+1;
gap[dep[u]]++;
if (u!=start) u=edge[pre[u]^1].to;
}
return ans;
}
struct Node
{
int u,v,c,cost;
}node[MAXN];
int n,m,num;
int cost[MAXN];
int main()
{
// freopen("C:/Users/asus1/Desktop/IN.txt","r",stdin);
int i,j,k,u,v,c,co;
while (scanf("%d%d",&n,&m),(n+m))
{
init();
int all=0;
for (i=1;i<=m;i++)
{
cost[i]=0;
scanf("%d",&c);
addedge(0,i,c);
all+=c;
}
scanf("%d",&k);
for (i=1;i<=k;i++)
{
scanf("%d%d%d%d",&node[i].u,&node[i].v,&node[i].c,&node[i].cost);
cost[node[i].c]+=node[i].cost;
}
for (i=1;i<=m;i++)
addedge(i,m+1,cost[i]);
for (i=1;i<=k;i++)
{
for (j=1;j<=k;j++)
{
if (i!=j)
{
if (node[i].c!=node[j].c&&node[i].v==node[j].u)
addedge(node[i].c,node[j].c,INF);
}
}
}
printf("%d\n",all-sap(0,m+1,m+2));
}
return 0;
}