UVa 1212 - Duopoly

链接

传送门

题意

两家公司有多个申请,每个申请包含一些资源的使用权并且会给政府一些资金,同一个公司的申请中不含相同的资源。资源的使用权不能同时批给两个公司,求政府的最大收益。

思路

从原点向T公司的申请连容量为申请金额的边,从M公司的申请向汇点连容量为申请金额的边,两公司冲突的申请之间连容量为无穷大的边。政府解决冲突的最小代价就是源点到汇点的最小割。

代码

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#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>
using namespace std;
typedef long long LL;
const int inf = 0x3f3f3f3f;
const int maxn = 6010;
const int maxm = 300010;
int p[maxn], vis[maxm];
int g[maxn >> 1][maxn >> 1];
struct Edge {
int from, to, cap, flow;
Edge() {}
Edge(int u, int v, int c, int f): from(u), to(v), cap(c), flow(f) {}
};
struct ISAP {
int n, m, s, t;
vector<Edge> edges;
vector<int> G[maxn];
bool vis[maxn];
int p[maxn];
int d[maxn];
int num[maxn];
int cur[maxn];
void AddEdge(int from, int to, int cap) {
edges.push_back(Edge(from, to, cap, 0));
edges.push_back(Edge(to, from, 0, 0));
m = int(edges.size());
G[from].push_back(m - 2);
G[to].push_back(m - 1);
}
void ClearAll(int n) {
this -> n = n;
for (int i = 0; i < n; ++i) {
G[i].clear();
}
edges.clear();
}
bool BFS() {
memset(vis, 0, sizeof vis);
vis[t] = true;
d[t] = 0;
queue<int> Q;
Q.push(t);
while (!Q.empty()) {
int x = Q.front();
Q.pop();
for (int i = 0; i < G[x].size(); ++i) {
Edge& e = edges[G[x][i] ^ 1];
if (!vis[e.from] && e.cap > e.flow) {
vis[e.from] = true;
d[e.from] = d[x] + 1;
Q.push(e.from);
}
}
}
return vis[s];
}
int Augment() {
int x = t, a = inf;
while (x != s) {
Edge& e = edges[p[x]];
a = min(a, e.cap - e.flow);
x = e.from;
}
x = t;
while (x != s) {
edges[p[x]].flow += a;
edges[p[x] ^ 1].flow -= a;
x = edges[p[x]].from;
}
return a;
}
int Maxflow(int s, int t) {
this -> s = s, this -> t = t;
int flow = 0;
BFS();
memset(num, 0, sizeof num);
for (int i = 0; i < n; ++i) {
++num[d[i]];
}
memset(cur, 0, sizeof cur);
int x = s;
while (d[s] < n) {
if (x == t) {
flow += Augment();
x = s;
}
bool ok = false;
for (int i = cur[x]; i < G[x].size(); ++i) {
Edge& e = edges[G[x][i]];
if (e.cap > e.flow && d[x] == d[e.to] + 1) {
ok = true;
p[e.to] = G[x][i];
cur[x] = i;
x = e.to;
break;
}
}
if (!ok) {
int m = n - 1;
for (int i = 0; i < G[x].size(); ++i) {
Edge& e = edges[G[x][i]];
if (e.cap > e.flow) {
m = min(m, d[e.to]);
}
}
if (--num[d[x]] == 0) {
break;
}
++num[d[x] = m + 1];
cur[x] = 0;
if (x != s) {
x = edges[p[x]].from;
}
}
}
return flow;
}
void ClearFlow() {
for (int i = 0; i < m; ++i) {
edges[i].flow = 0;
}
}
} isap;
int main() {
int t, tt = 1;
scanf("%d", &t);
while (t--) {
int n, m, ans = 0;
scanf("%d", &n);
memset(vis, -1, sizeof vis);
for (int i = 0; i < n; ++i) {
scanf("%d", &p[i]);
ans += p[i];
while (getchar() != '\n') {
int x;
scanf("%d", &x);
vis[x] = i;
}
}
scanf("%d", &m);
int k = n + m;
for (int i = n; i < k; ++i) {
scanf("%d", &p[i]);
ans += p[i];
while (getchar() != '\n') {
int x;
scanf("%d", &x);
if (vis[x] != -1) {
if (g[vis[x]][i - n] != tt) {
isap.AddEdge(vis[x], i, inf);
g[vis[x]][i - n] = tt;
}
}
}
}
isap.n = k + 5;
for (int i = 0; i < n; ++i) {
isap.AddEdge(k, i, p[i]);
}
for (int i = n; i < k; ++i) {
isap.AddEdge(i, k + 1, p[i]);
}
ans -= isap.Maxflow(k, k + 1);
printf("Case %d:\n%d\n", tt++, ans);
if (t) {
puts("");
}
isap.ClearAll(k + 5);
}
return 0;
}