题目：
2 4 4 *ooo o### **#* ooo* 4 4 #*** *#** **#* ooo#
3 5
题意：
给你一个图，"*"代表海，"#"代表冰山，"o"代表浮冰。问你最多能停多少艘军舰在海上。但是有条件：任意两艘军舰不能在同一行或同一列，在同一列的唯一条件是中间有冰山隔开，浮冰不起隔作用，浮冰上面不停船。
思路：
二分匹配中的行列匹配，但是要变形，所谓的变形在于拆点，先是给行编号，遇到冰山就拆点。接着是行编号，也是遇到冰山就拆点。
然后再进行二分匹配，就可以达到目标。
代码：
#include<cstdio>
#include<cstring>
#include<iostream>
#include<sstream>
#include<algorithm>
#include<vector>
#include<bitset>
#include<set>
#include<queue>
#include<stack>
#include<map>
#include<cstdlib>
#include<cmath>
#define PI 2*asin(1.0)
#define LL __int64
const int MOD = 1e9 + 7;
const int N = 3e3 + 15;
const int INF = (1 << 30) - 1;
const int letter = 130;
using namespace std;
int tc;
int n, m;
int s , t;
short vis1[N][N], vis2[N][N];
char dis[N][N];
int vis[N];
int x[N], y[N];
int cur[N];
struct node {
int from, to, cap, flow;
};
vector<node>edges;
vector<int>G[N];
void init() {
for(int i = 0; i < N - 10; i++)
G[i].clear();
edges.clear();
}
void addedge(int from, int to, int cap) {
edges.push_back((node) {
from, to, cap, 0
});
edges.push_back((node) {
to, from, 0, 0
});
int v = edges.size();
G[from].push_back(v - 2);
G[to].push_back(v - 1);
}
bool bfs() {
memset(vis, -1, sizeof(vis));
queue<int>q;
while(!q.empty()) q.pop();
vis[s] = 0;
q.push(s);
while(!q.empty()) {
int x = q.front();
q.pop();
for(int i = 0; i < G[x].size(); i++) {
node &e = edges[G[x][i]];
if(vis[e.to] < 0 && e.cap > e.flow) {
vis[e.to] = vis[x] + 1;
if(e.to == t) return 1;
q.push(e.to);
}
}
}
return 0;
}
int dfs(int x, int a) {
if(x == t || a == 0) return a;
int flow = 0, f;
for(int &i = cur[x]; i < G[x].size(); i++) {
node &e = edges[G[x][i]];
if(vis[x] + 1 == vis[e.to] && (f = dfs(e.to, min(a, e.cap - e.flow))) > 0) {
e.flow += f;
flow += f;
edges[G[x][i] ^ 1].flow -= f;
a -= f;
if(a == 0) break;
}
}
return flow;
}
int maxflow(int s, int t) {
int flow = 0;
while(bfs()) {
memset(cur, 0, sizeof(cur));
flow += dfs(s, INF);
}
return flow;
}
int main() {
cin >> tc;
while(tc--) {
init();
s = 0;
int num = 1;
scanf("%d%d", &n, &m);
memset(x, 0, sizeof(x));
memset(y, 0, sizeof(y));
memset(vis1, 0, sizeof(vis1));
memset(vis2, 0, sizeof(vis2));
getchar();
for(int i = 0; i < n; i++) {
for(int j = 0; j < m; j++) {
scanf("%c", &dis[i][j]);
}
getchar();
}
/// hang
for(int i = 0; i < n; i++) {
int flag = 1, q = 0;
for(int j = 0; j < m; j++) {
if(dis[i][j] == '*') vis1[i][j] = num, flag = 0, q = 1, x[num] = 1;
if(dis[i][j] == '#' && flag == 0) num++, flag = 1;
}
if(q == 1 && i != n - 1)num++;
}
num++;
for(int j = 0; j < m; j++) {
int flag = 1, q = 0;
for(int i = 0; i < n; i++) {
if(dis[i][j] == '*') vis2[i][j] = num, flag = 0, q = 1, y[num] = 1;
if(dis[i][j] == '#' && flag == 0) num++, flag = 1;
}
if(q == 1) num++;
}
t = n + m + 15;
for(int i = 0; i < N; i++)
if(x[i]) addedge(s, i, 1);
for(int i = 0; i < N; i++)
if(y[i]) addedge(i, t, 1);
for(int i = 0; i < n; i++) {
for(int j = 0; j < m; j++) {
if(dis[i][j] == '*') addedge(vis1[i][j],vis2[i][j],1);
}
}
printf("%d\n",maxflow(s,t));
}
return 0;
}HDU 5093Battle ships(2014上海邀请赛)
原文地址：http://blog.csdn.net/u014325920/article/details/42975523
