Light OJ 1304 The Best Contest Site Ever 行列匹配变形

题目来源：Light OJ 1304 The Best Contest Site Ever

题意：。的位置可以放人 W是障碍物 R除了不能放人之外和。一样 求最多放几个人 

思路：行列匹配变形 要重新编行号和列号每一行 因为被W隔断了

..RR.

.W.RW

.RRR. 

对应行号是

1 1 0 0 2

3 0 4 0 0

5 0 0 0 5

模拟一下预处理

#include <cstdio>
#include <cstring>
#include <vector>
using namespace std;
const int maxn = 110;
int vis[maxn*maxn];
int y[maxn*maxn];
vector <int> G[maxn*maxn];
int n, m;
char a[maxn][maxn];
int b[maxn][maxn];
int c[maxn][maxn];
struct node
{
	int x, y;
	node(){}
	node(int x, int y): x(x), y(y) {}
};
vector <node> X[10010];
bool dfs(int u)
{
    for(int i = 0; i < G[u].size(); i++)
    {
        int v = G[u][i];
        if(vis[v])
            continue;
        vis[v] = true;
        if(y[v] == -1 || dfs(y[v]))
        {
            y[v] = u;
            return true;
        }
    }
    return false;
}
int match()
{
    int ans = 0;
    memset(y, -1, sizeof(y));
    for(int i = 1; i <= n; i++)
    {
		memset(vis, 0, sizeof(vis));
		if(dfs(i))
		    ans++;
    }
    return ans;
}

int main()
{
	int cas = 1;
	int T;
	scanf("%d", &T);
	while(T--)
	{
		int r, C;
		scanf("%d %d", &r, &C);
		for(int i = 0; i < r; i++)
			scanf("%s", a[i]);
		
		
		n = 0, m = 0;
		memset(b, 0, sizeof(b));
		for(int i = 0; i < r; i++)
		{
			int j = 0, flag = 1, flag2 = 0;
			while(j < C)
			{
				while(a[i][j] == 'W' && j < C)
				{
					flag2 = 1;
					flag++;
					j++;
				}	
				if(j < C && a[i][j] == '.')
				{
					if(!flag2)
					{
						flag2 = 1;
						flag = 1;
					}
					if(flag)
					{
						++n;
						flag = 0;
					}
					b[i][j] = n;
				}
				j++;
			}
		}
		memset(c, 0, sizeof(c));
		for(int i = 0; i < C; i++)
		{
			int j = 0, flag = 1, flag2 = 0;
			while(j < r)
			{
				while(a[j][i] == 'W' && j < r)
				{
					flag++;
					j++;
				}
				if(j < r && a[j][i] == '.')
				{
					if(!flag2)
					{
						flag2 = 1;
						flag = 1;
					}
					if(flag)
					{
						++m;
						flag = 0;
					}
					c[j][i] = m;
				}
				j++;
			}
		}
		for(int i = 0; i <= n; i++)
		{
			G[i].clear();
			X[i].clear();
		}
		for(int i = 0; i < r; i++)
		{
			for(int j = 0; j < C; j++)
			{
				//printf("%d...", b[i][j]);
				if(a[i][j] == '.')
				{
					G[b[i][j]].push_back(c[i][j]);
					X[b[i][j]].push_back(node(i, j));
				}
			}
			//puts("");
		}
		printf("Case %d: %d\n", cas++, match());
		for(int i = 1; i <= m; i++)
		{
			if(y[i] == -1)
				continue;
			int u = y[i];
			for(int j = 0; j < X[u].size(); j++)
			{
				node no = X[u][j];
				if(c[no.x][no.y] == i)
				{
					a[no.x][no.y] = 'T';
					break;
				}
			}
			
		}
		for(int i = 0; i < r; i++)
			puts(a[i]);
	}
    return 0;
}