HDU 3917 Road constructions 最大权闭合图

题目：http://acm.hdu.edu.cn/showproblem.php?pid=3917

题意：给出n个工程，m个公司，给出每个公司施工需要上交的税务tax，给出k个工程，以及相应的政府补助，工程可有谁完成！并且要求如果选取i公司担任这个工程，那么i公司负责的项目也相应必须有它负责！同时，与其” 相连 “的公司j 也必须担任” 相连“的任务，问政府可以获得的最高利润

思路：最大权闭合子图。从源点向每个公司连边，容量为税款，对于有相连关系的公司连一条边，容量为无穷大，然后把公司和汇点连边，容量为该公司所有工程的补助，最后税款和减去最大流

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <vector>
using namespace std;

typedef long long ll;
const int N = 5010;
const int INF = 0x3f3f3f3f;
struct edge
{
int to, cap, next;
}g[N*200];
int head
, pre
, cur
, gap
, level
;
int cnt, nv, cas = 0;
void add_edge(int v, int u, int cap)
{
g[cnt].to = u, g[cnt].cap = cap, g[cnt].next = head[v], head[v] = cnt++;
g[cnt].to = v, g[cnt].cap = 0, g[cnt].next = head[u], head[u] = cnt++;
}
int sap(int s, int t)
{
memset(level, 0, sizeof level);
memset(gap, 0, sizeof gap);
memcpy(cur, head, sizeof head);
gap[0] = nv;
int v = pre[s] = s, flow = 0, aug = INF;
while(level[s] < nv)
{
bool flag = false;
for(int &i = cur[v]; i != -1; i = g[i].next)
{
int u = g[i].to;
if(g[i].cap > 0 && level[v] == level[u] + 1)
{
flag = true;
pre[u] = v;
v = u;
aug = min(aug, g[i].cap);
if(v == t)
{
flow += aug;
while(v != s)
{
v = pre[v];
g[cur[v]].cap -= aug;
g[cur[v]^1].cap += aug;
}
aug = INF;
}
break;
}
}
if(flag) continue;
int minlevel = nv;
for(int i = head[v]; i != -1; i = g[i].next)
{
int u = g[i].to;
if(g[i].cap > 0 && minlevel > level[u])
minlevel = level[u], cur[v] = i;
}
if(--gap[level[v]] == 0) break;
level[v] = minlevel + 1;
gap[level[v]]++;
v = pre[v];
}
return flow;
}

int main()
{
int n, m, k, tmp;
int a
, b
, c
, d
, ans
;
while(scanf("%d%d", &n, &m), n || m)
{
int sum = 0;
cnt = 0;
memset(head, -1, sizeof head);
memset(ans, 0, sizeof ans);
for(int i = 1; i <= m; i++)
{
scanf("%d", &tmp);
add_edge(0, i, tmp);
sum += tmp;
}
scanf("%d", &k);
vector<int> st
, en
;
for(int i = 1; i <= k; i++)
{
scanf("%d%d%d%d", &a[i], &b[i], &c[i], &d[i]);
ans[c[i]] += d[i];
st[a[i]].push_back(c[i]);
en[b[i]].push_back(c[i]);
}
for(int i = 1; i <= n; i++)
for(int j = 0; j < st[i].size(); j++)
for(int k = 0; k < en[i].size(); k++)
add_edge(en[i][k], st[i][j], INF);
for(int i = 1; i <= m; i++)
add_edge(i, m + 1, ans[i]);
nv = m + 2;
printf("%d\n", sum - sap(0, m + 1));
}
return 0;
}