UVaLive 3487 Duopoly (最小割)

题意：有两个公司A和B在申请一些资源，现在给出两个公司所申请的内容，内容包括价钱和申请的资源 ，现在你做为官方，你只能拒绝一个申请或者接受一个申请，同一个资源不能两个公司都拥有，且申请的资源不能只给部分，问：作为官方，你能得到的最大利益是多少

析：就是一个最小割，因为AB两个公司，资源不能共用，只能给一个，也就是官方要舍弃一些利益让他们不共用资源，要这个舍弃的最小。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#include <list>
#include <assert.h>
#include <bitset>
#include <numeric>
#define debug() puts("++++")
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a, b, sizeof a)
#define sz size()
#define pu push_up
#define pd push_down
#define cl clear()
#define all 1,n,1
#define FOR(i,x,n)  for(int i = (x); i < (n); ++i)
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e17;
const double inf = 1e20;
const double PI = acos(-1.0);
const double eps = 1e-3;
const int maxn = 6000 + 10;
const int maxm = 3e5 + 10;
const int mod = 1000000007;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, -1, 0, 1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c) {
return r >= 0 && r < n && c >= 0 && c < m;
}

struct Edge{
int from, to, cap, flow;
};

struct Dinic{
int n, m, s, t;
vector<Edge> edges;
vector<int> G[maxn];
int d[maxn];
bool vis[maxn];
int cur[maxn];

void init(int n){
this-> n = n;
FOR(i, 0, n)  G[i].cl;
edges.cl;
}

void addEdge(int from, int to, int cap){
edges.pb((Edge){from, to, cap, 0});
edges.pb((Edge){to, from, 0, 0});
m = edges.sz;
G[from].pb(m - 2);
G[to].pb(m - 1);
}

bool bfs(){
ms(vis, 0);  vis[s] = 1;
d[s] = 0;
queue<int> q;  q.push(s);

while(!q.empty()){
int u = q.front();  q.pop();
for(int i = 0; i < G[u].sz; ++i){
Edge &e = edges[G[u][i]];
if(!vis[e.to] && e.cap > e.flow){
d[e.to] = d[u] + 1;
vis[e.to] = 1;
q.push(e.to);
}
}
}
return vis[t];
}

int dfs(int u, int a){
if(u == t || a == 0)  return a;
int flow = 0, f;
for(int &i = cur[u]; i < G[u].sz; ++i){
Edge &e = edges[G[u][i]];
if(d[e.to] == d[u] + 1 && (f = dfs(e.to, min(a, e.cap - e.flow))) > 0){
e.flow += f;
edges[G[u][i]^1].flow -= f;
flow += f;
a -= f;
if(a == 0)  break;
}
}
return flow;
}

int maxflow(int s, int t){
this-> s = s;
this-> t = t;
int flow = 0;
while(bfs()){ ms(cur, 0);   flow += dfs(s, INF); }
return flow;
}
};

Dinic dinic;

bool vis[maxn>>1][maxn>>1];
int a[maxm], b[maxm];

int main(){
ios::sync_with_stdio(false);
int T;  cin >> T;
for(int kase = 1; kase <= T; ++kase){
cin >> n;  cin.get();
int s = 0, t = n + 3000 + 1;
int mmax = -INF, ans = 0;
dinic.init(t + 2);
string line;
ms(vis, 0);  ms(a, 0);  ms(b, 0);
for(int i = 1; i <= n; ++i){
getline(cin, line);
stringstream ss(line);
int x;  ss >> x;
dinic.addEdge(s, i, x);
ans += x;
while(ss >> x){
mmax = max(mmax, x);
a[x] = i;
}
}
cin >> m;  cin.get();
for(int i = 1; i <= m; ++i){
getline(cin, line);
stringstream ss(line);
int x;  ss >> x;
dinic.addEdge(i + n, t, x);
ans += x;
while(ss >> x){
mmax = max(mmax, x);
b[x] = i;
}
}
for(int i = 1; i <= mmax; ++i){
if(!a[i] || !b[i] || vis[a[i]][b[i]])  continue;
vis[a[i]][b[i]] = 1;
dinic.addEdge(a[i], b[i] + n, INF);
}
if(kase > 1)  cout << endl;
cout << "Case " << kase << ":\n";
cout << ans - dinic.maxflow(s, t) << endl;
}
return 0;
}