[CQOI2014]危桥

嘟嘟嘟

看到这个数据范围。就很容易想到网络流。
首先对于正常的桥，这条边的容量相当于(INF)，对于危桥，容量为(2)。然后按这个方式建无向图就行。
对于在(a1)和(a2)之间往返(an)次，相当于(a1)向(a2)流了至少(2 * an)的流量。那么就从源点向(a1)连一条容量为(2 * an)的边，从(a2)向汇点连一条容量为(2 * an)的边；对于(b)也同理。最后判断总流量是否为(2 * (an + bn))即可。
但这个算法不一定对，因为可能从(a1)流到(b2)，所以我们再重建图，把(a1)和(b2)都连到源点，(a2)和(b1)都连向汇点。这样就保证不会从(a1)流到(b2)了。

#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<vector>
#include<stack>
#include<queue>
using namespace std;
#define enter puts("") 
#define space putchar(' ')
#define Mem(a, x) memset(a, x, sizeof(a))
#define rg register
typedef long long ll;
typedef double db;
const int INF = 0x3f3f3f3f;
const db eps = 1e-8;
const int maxn = 55;
inline ll read()
{
  ll ans = 0;
  char ch = getchar(), last = ' ';
  while(!isdigit(ch)) last = ch, ch = getchar();
  while(isdigit(ch)) ans = (ans << 1) + (ans << 3) + ch - '0', ch = getchar();
  if(last == '-') ans = -ans;
  return ans;
}
inline void write(ll x)
{
  if(x < 0) x = -x, putchar('-');
  if(x >= 10) write(x / 10);
  putchar(x % 10 + '0');
}

int n, a1, a2, an, b1, b2, bn;
char a[maxn][maxn];

int s, t;
struct Edge
{
  int nxt, from, to, cap, flow;
}e[(maxn * maxn) << 1];
int head[maxn], ecnt = -1;
void init()
{
  Mem(head, -1); ecnt = -1;
}
void addEdge(int x, int y, int w)
{
  e[++ecnt] = (Edge){head[x], x, y, w, 0};
  head[x] = ecnt;
  e[++ecnt] = (Edge){head[y], y, x, 0, 0};
  head[y] = ecnt;
}

int dis[maxn];
bool bfs()
{
  Mem(dis, 0); dis[s] = 1;
  queue<int> q; q.push(s);
  while(!q.empty())
    {
      int now = q.front(); q.pop();
      for(int i = head[now], v; i != -1; i = e[i].nxt)
	{
	  v = e[i].to;
	  if(!dis[v] && e[i].cap > e[i].flow)
	    {
	      dis[v] = dis[now] + 1;
	      q.push(v);
	    }
	}
    }
  return dis[t];
}
int cur[maxn];
int dfs(int now, int res)
{
  if(now == t || res == 0) return res;
  int flow = 0, f;
  for(int& i = cur[now], v; i != -1; i = e[i].nxt)
    {
      v = e[i].to;
      if(dis[v] == dis[now] + 1 && (f = dfs(v, min(res, e[i].cap - e[i].flow))) > 0)
	{
	  e[i].flow += f; e[i ^ 1].flow -= f;
	  flow += f; res -= f;
	  if(res == 0) break;
	}
    }
  return flow;
}

int maxflow()
{
  int flow = 0;
  while(bfs())
    {
      memcpy(cur, head, sizeof(head));
      flow += dfs(s, INF);
    }
  return flow;
}

int main()
{
  while(scanf("%d%d%d%d%d%d%d", &n, &a1, &a2, &an, &b1, &b2, &bn) != EOF)
    {
      init();
      bool flg = 0;
      s = n; t = n + 1;
      for(int i = 0; i < n; ++i) scanf("%s", a[i]);
      for(int i = 0; i < n; ++i)
	for(int j = 0; j < n; ++j)
	  if(a[i][j] == 'O') addEdge(i, j, 2);
	  else if(a[i][j] == 'N') addEdge(i, j, INF);
      addEdge(s, a1, an << 1); addEdge(s, b1, bn << 1);
      addEdge(a2, t, an << 1); addEdge(b2, t, bn << 1);
      if(maxflow() >= ((an + bn) << 1)) flg = 1;
      if(flg)
	{
	  init();
	  for(int i = 0; i < n; ++i)
	    for(int j = 0; j < n; ++j)
	      if(a[i][j] == 'O') addEdge(i, j, 2);
	      else if(a[i][j] == 'N') addEdge(i, j, INF);
	  addEdge(s, a1, an << 1); addEdge(s, b2, bn << 1);
	  addEdge(a2, t, an << 1); addEdge(b1, t, bn << 1);	 
	  if(maxflow() < ((an + bn) << 1)) flg = 0;
	}
      puts(flg ? "Yes" : "No");
    }
  return 0;
}