UVA 12534 - Binary Matrix 2 (网络流‘最小费用最大流’ZKW)

题目：

http://acm.hust.edu.cn/vjudge/contest/view.action?cid=93745#problem/A

题意：

给出r*c的01矩阵，可以翻转格子使得0表成1,1变成0，求出最小的步数使得每一行中1的个数相等，每一列中1的个数相等。

思路：

网络流。容量可以保证每一行和每一列的1的个数相等，费用可以算出最小步数。

行向列建边，如果该格子是1，则建一条费用为-1，容量为1的边，若为0，则建一条费用为1，容量为1的边。

源点向行建边，从1~c 枚举每一行中1 的个数作为边的容量cap，费用为0，而每一列到汇点的容量为cap*r/c（整数才可用）。

最后答案为原图中1的个数sum+ 最小费用。

题目需要跑zkw算法才不会超时。

AC.

#include <iostream>
#include <algorithm>
#include <cstring>
#include <string>
#include <cstdio>
#include <cmath>
#include <queue>
#include <map>
#include <set>
#define eps 1e-5

using namespace std;
const int maxn = 1000;
const int maxm = 50000;
const int INF = 0x3f3f3f3f;

typedef long long ll;

struct MaxFlow
{
int size, n;
int st, en, maxflow, mincost;
bool vis[maxn];
int net[maxn], pre[maxn], cur[maxn], dis[maxn];
std::queue <int> Q;

struct EDGE
{
int v, cap, cost, next;
EDGE() {}
EDGE(int a, int b, int c, int d)
{
v = a, cap = b, cost = c, next = d;
}
} E[maxm << 1];

void init(int _n)
{
n = _n, size = 0;
memset(net, -1, sizeof(net));
}

void add(int u, int v, int cap, int cost)
{
E[size] = EDGE(v, cap, cost, net[u]);
net[u] = size++;
E[size] = EDGE(u, 0, -cost, net[v]);
net[v] = size++;
}

bool adjust()
{
int v, min = INF;
for (int i = 0; i <= n; i++)
{
if (!vis[i])
continue;
for (int j = net[i]; v = E[j].v, j != -1; j = E[j].next)
if (E[j].cap)
if (!vis[v] && dis[v] - dis[i] + E[j].cost < min)
min = dis[v] - dis[i] + E[j].cost;
}
if (min == INF)
return false;
for (int i = 0; i <= n; i++)
if (vis[i])
cur[i] = net[i], vis[i] = false, dis[i] += min;
return true;
}

int augment(int i, int flow)
{
if (i == en)
{
mincost += dis[st] * flow;
maxflow += flow;
return flow;
}
vis[i] = true;
for (int j = cur[i], v; v = E[j].v, j != -1; j = E[j].next)
{
if (!E[j].cap)
continue;
if (vis[v] || dis[v] + E[j].cost != dis[i])
continue;
int delta = augment(v, std::min(flow, E[j].cap));
if (delta)
{
E[j].cap -= delta;
E[j ^ 1].cap += delta;
cur[i] = j;
return delta;
}
}
return 0;
}

void spfa()
{
int u, v;
for (int i = 0; i <= n; i++)
vis[i] = false, dis[i] = INF;
dis[st] = 0;
Q.push(st);
vis[st] = true;
while (!Q.empty())
{
u = Q.front(), Q.pop();
vis[u] = false;
for (int i = net[u]; v = E[i].v, i != -1; i = E[i].next)
{
if (!E[i].cap || dis[v] <= dis[u] + E[i].cost)
continue;
dis[v] = dis[u] + E[i].cost;
if (!vis[v])
{
vis[v] = true;
Q.push(v);
}
}
}
for (int i = 0; i <= n; i++)
dis[i] = dis[en] - dis[i];
}

int zkw(int s, int t, int need)
{
st = s, en = t;
spfa();
mincost = maxflow = 0;
for (int i = 0; i <= n; i++)
vis[i] = false, cur[i] = net[i];

do
{
while (augment(st, INF))
memset(vis, false, sizeof(vis));
} while (adjust());
if (maxflow < need)
return -1;
return mincost;
}
} zkw;

int r, c;
char ss[maxn][maxn];

int main()
{
//freopen("in", "r", stdin);
int T, ca = 1;
scanf("%d", &T);
while(T--) {
scanf("%d%d", &r, &c);
int sum = 0;
// zkw.init(r+c+2);
for(int i = 1; i <= r; ++i) {
scanf("%s", ss[i]+1);
for(int j = 1; j <= c; ++j) {
if(ss[i][j] == '1') sum++;
}
}
int src = r+c+1, des = src+1;
int n = des;
int ans = INF;

for(int p = 0; p <= c; ++p) {
if((p*r)%c != 0) continue;

zkw.init(n+1);
for(int i = 1; i <= r; ++i) {

for(int j = 1; j <= c; ++j) {
if(ss[i][j] == '1') {
zkw.add(i, r+j, 1, -1);
}
else zkw.add(i, r+j, 1, 1);
}
}
for(int j = 1; j <= r; ++j) {
zkw.add(src, j, p, 0);
}
for(int j = 1; j <= c; ++j) {
zkw.add(r+j, des, p*r/c, 0);
}

int cost = zkw.zkw(src, des, -1);
ans = min(ans, sum+cost);
}
printf("Case %d: %d\n", ca++, ans);
}
return 0;
}