POJ2396 Budget

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上下界网络流之可行流。

对于这种矩阵的题，做过就应该知道怎么建图。
像二分图一样，左边(n)个点代表行，右边(m)个点代表列。对于点((i, j))的限制，就从左边第(i)个点向右边第(j)个点连边。
然后这题基本也就完事了。
建图虽然不难，但写起来比较麻烦，因为数据较小，推荐邻接矩阵存每一条边的最小最大容量，这样方便更新取最紧的限制。
刚开始我WA了是为了减小复杂度，建图的时候如果当前条件不符合就break，导致后面的数据没有读入，从而影响了后面的几组。结果RE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 = 205;
const int maxe = 1e6 + 5;
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, m, s, t, ss, tt;
char ch[2];
int id[maxn][maxn];

struct Edge
{
  int nxt, from, to, cap, flow;
}e[maxe];
int head[maxn << 1], ecnt = -1;
void addEdge(int x, int y, int w, int i, int j)
{
  e[++ecnt] = (Edge){head[x], x, y, w, 0};
  head[x] = id[i][j] = ecnt;
  e[++ecnt] = (Edge){head[y], y, x, 0, 0};
  head[y] = ecnt;
}

int dis[maxn << 1];
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)
	if(!dis[v = e[i].to] && e[i].cap > e[i].flow)
	  {
	    dis[v] = dis[now] + 1;
	    q.push(v);
	  }
    }
  return dis[t];
}
int cur[maxn << 1];
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)
    {
      if(dis[v = e[i].to] == dis[now] + 1 && (f = dfs(v, min(e[i].cap - e[i].flow, res))) > 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));
      int x = dfs(s, INF);
      flow += x;
    }
  return flow;
}

bool flg = 1;
int sumi[maxn], sumo[maxn], tot = 0;
int b[maxn][maxn], c[maxn][maxn];
void Set(int x, int y, char ch, int w)
{
  if(ch == '=')
    {
      if(b[x][y] > w || c[x][y] < w) flg = 0;
      else b[x][y] = c[x][y] = w;
    }
  else if(ch == '<')
    {
      if(b[x][y] >= w) flg = 0;
      else c[x][y] = min(c[x][y], w - 1);
    }
  else if(ch == '>')
    {
      if(c[x][y] <= w) flg = 0;
      else b[x][y] = max(b[x][y], w + 1);
    }
}
void build()
{
  for(int i = 1; i <= n; ++i)
    for(int j = 1; j <= m; ++j)
      addEdge(i, j + n, c[i][j] - b[i][j], i, j);
  for(int i = 1; i <= n; ++i)
    {
      int sum = sumi[i];
      for(int j = 1; j <= m; ++j) sum -= b[i][j];
      if(sum > 0) addEdge(s, i, sum, 0, 0), tot += sum;
      else addEdge(i, t, -sum, 0, 0);
    }
  for(int j = 1; j <= m; ++j)
    {
      int sum = -sumo[j];
      for(int i = 1; i <= n; ++i) sum += b[i][j];
      if(sum > 0) addEdge(s, n + j, sum, 0, 0), tot += sum;
      else addEdge(n + j, t, -sum, 0, 0);
    }
}

void init()
{
  Mem(head, -1); ecnt = -1;
  Mem(b, 0); Mem(c, 0x3f); tot = 0; flg = 1;
  s = 0; t = n + m + 1;
}

int main()
{
  int T = read();
  while(T--)
    {
      n = read(); m = read();
      init();
      for(int i = 1; i <= n; ++i) sumi[i] = read();
      for(int i = 1; i <= m; ++i) sumo[i] = read();
      int q = read();
      for(int i = 1, x, y, w; i <= q; ++i)
	{
	  x = read(), y = read(); scanf("%s", ch); w = read();
	  if(!x && !y)
	    for(int j = 1; j <= n; ++j)
	      for(int k = 1; k <= m; ++k) Set(j, k, ch[0], w);
	  else if(!x && y)
	    for(int j = 1; j <= n; ++j) Set(j, y, ch[0], w);
	  else if(x && !y)
	    for(int j = 1; j <= m; ++j) Set(x, j, ch[0], w);
	  else Set(x, y, ch[0], w);
	}
      if(!flg) {puts("IMPOSSIBLE"); continue;}
      build();
      if(maxflow() < tot) puts("IMPOSSIBLE");
      else
	{
	  for(int i = 1; i <= n; ++i)
	    {
	      for(int j = 1; j <= m; ++j) write(e[id[i][j]].flow + b[i][j]), space;
	      enter;
	    }
	}
    }
  return 0;
}