POJ 3189 Steady Cow Assignment(最大流）

POJ 3189 Steady Cow Assignment

题目链接

题意：一些牛，每个牛心目中都有一个牛棚排名，然后给定每个牛棚容量，要求分配这些牛给牛棚，使得所有牛对牛棚的排名差距尽量小

思路：这种题的标准解法都是二分一个差值，枚举下界确定上界，然后建图判断，这题就利用最大流进行判断，值得一提的是dinic的效率加了减枝还是是卡着时间过的，这题理论上用sap或者二分图多重匹配会更好

代码：

#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
using namespace std;

const int MAXNODE = 1025;
const int MAXEDGE = 200005;

typedef int Type;
const Type INF = 0x3f3f3f3f;

struct Edge {
int u, v;
Type cap, flow;
Edge() {}
Edge(int u, int v, Type cap, Type flow) {
this->u = u;
this->v = v;
this->cap = cap;
this->flow = flow;
}
};

struct Dinic {
int n, m, s, t;
Edge edges[MAXEDGE];
int first[MAXNODE];
int next[MAXEDGE];
bool vis[MAXNODE];
Type d[MAXNODE];
int cur[MAXNODE];
vector<int> cut;

void init(int n) {
this->n = n;
memset(first, -1, sizeof(first));
m = 0;
}
void add_Edge(int u, int v, Type cap) {
edges[m] = Edge(u, v, cap, 0);
next[m] = first[u];
first[u] = m++;
edges[m] = Edge(v, u, 0, 0);
next[m] = first[v];
first[v] = m++;
}

bool bfs() {
memset(vis, false, sizeof(vis));
queue<int> Q;
Q.push(s);
d[s] = 0;
vis[s] = true;
while (!Q.empty()) {
int u = Q.front(); Q.pop();
for (int i = first[u]; i != -1; i = next[i]) {
Edge& e = edges[i];
if (!vis[e.v] && e.cap > e.flow) {
vis[e.v] = true;
d[e.v] = d[u] + 1;
Q.push(e.v);
}
}
}
return vis[t];
}

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

Type Maxflow(int s, int t) {
this->s = s; this->t = t;
Type flow = 0;
while (bfs()) {
for (int i = 0; i < n; i++)
cur[i] = first[i];
flow += dfs(s, INF);
}
return flow;
}

void MinCut() {
cut.clear();
for (int i = 0; i < m; i += 2) {
if (vis[edges[i].u] && !vis[edges[i].v])
cut.push_back(i);
}
}
} gao;

const int N = 1005;
const int M = 25;

int n, b, g
[M], s[M];

bool build(int x, int y) {
gao.init(n + b + 2);
for (int i = 1; i <= n; i++) {
gao.add_Edge(0, i, 1);
for (int j = x; j <= y; j++)
gao.add_Edge(i, g[i][j] + n, 1);
}
for (int i = 1; i <= b; i++)
gao.add_Edge(i + n, n + b + 1, s[i]);
return gao.Maxflow(0, n + b + 1) == n;
}

int main() {
while (~scanf("%d%d", &n, &b)) {
for (int i = 1; i <= n; i++)
for (int j = 1; j <= b; j++)
scanf("%d", &g[i][j]);
for (int i = 1; i <= b; i++)
scanf("%d", &s[i]);
int ans = 100;
for (int i = 1; i <= b; i++) {
for (int j = b; j >= i; j--) {
if (ans <= j - i + 1) continue;
if (build(i, j)) ans = min(ans, j - i + 1);
}
}
printf("%d\n", ans);
}
return 0;
}