直接按求可行流的方法建图，然后求一遍最大流，判断一下就可以了。。
#include<algorithm>
#include<iostream>
#include<cstring>
#include<vector>
#include<cstdio>
#include<cmath>
#include<queue>
#include<stack>
#include<map>
#include<set>
#define LL long long
#define CLR(a, b) memset(a, b, sizeof(a))
using namespace std;
const int maxn = 330;
const int INF = 0x3f3f3f3f;
struct Edge
{
int from, to, cap, flow;
Edge() {}
Edge(int from, int to, int cap, int flow)
:from(from), to(to), cap(cap), flow(flow) {}
};
bool operator < (const Edge& a, const Edge& b)
{
return a.from < b.from || (a.from == b.from && a.to < b.to);
}
struct ISAP
{
int n, m, s, t;
vector<Edge> edges;
vector<int> G[maxn]; // 邻接表，G[i][j]表示结点i的第j条边在e数组中的序号
bool vis[maxn]; // BFS使用
int d[maxn]; // 从起点到i的距离
int cur[maxn]; // 当前弧指针
int p[maxn]; // 可增广路上的上一条弧
int num[maxn]; // 距离标号计数
void AddEdge(int from, int to, int cap)
{
edges.push_back(Edge(from, to, cap, 0));
edges.push_back(Edge(to, from, 0, 0));
m = edges.size();
G[from].push_back(m-2);
G[to].push_back(m-1);
}
bool BFS()
{
memset(vis, 0, sizeof(vis));
queue<int> Q;
Q.push(t);
vis[t] = 1;
d[t] = 0;
while(!Q.empty())
{
int x = Q.front();
Q.pop();
for(int i = 0; i < G[x].size(); i++)
{
Edge& e = edges[G[x][i]^1];
if(!vis[e.from] && e.cap > e.flow)
{
vis[e.from] = 1;
d[e.from] = d[x] + 1;
Q.push(e.from);
}
}
}
return vis[s];
}
void init(int n)
{
this->n = n;
for(int i = 0; i < n; i++) G[i].clear();
edges.clear();
}
int Augment()
{
int x = t, a = INF;
while(x != s)
{
Edge& e = edges[p[x]];
a = min(a, e.cap-e.flow);
x = edges[p[x]].from;
}
x = t;
while(x != s)
{
edges[p[x]].flow += a;
edges[p[x]^1].flow -= a;
x = edges[p[x]].from;
}
return a;
}
int Maxflow(int s, int t, int need)
{
this->s = s;
this->t = t;
int flow = 0;
BFS();
memset(num, 0, sizeof(num));
for(int i = 0; i < n; i++) num[d[i]]++;
int x = s;
memset(cur, 0, sizeof(cur));
while(d[s] < n)
{
if(x == t)
{
flow += Augment();
if(flow >= need) return flow;
x = s;
}
int ok = 0;
for(int i = cur[x]; i < G[x].size(); i++)
{
Edge& e = edges[G[x][i]];
if(e.cap > e.flow && d[x] == d[e.to] + 1) // Advance
{
ok = 1;
p[e.to] = G[x][i];
cur[x] = i; // 注意
x = e.to;
break;
}
}
if(!ok) // Retreat
{
int m = n-1; // 初值注意
for(int i = 0; i < G[x].size(); i++)
{
Edge& e = edges[G[x][i]];
if(e.cap > e.flow) m = min(m, d[e.to]);
}
if(--num[d[x]] == 0) break;
num[d[x] = m+1]++;
cur[x] = 0; // 注意
if(x != s) x = edges[p[x]].from;
}
}
return flow;
}
} sol;
int row[maxn], col[maxn];
int mx[maxn][maxn], mn[maxn][maxn];
int m, n, c;
void gao(int r, int q, int v, int flag)
{
if(r == 0 && q == 0)
{
for(int i = 1; i <= m; i ++)
for(int j = 1; j <= n; j ++)
{
if(flag) mx[i][j] = min(mx[i][j], v);
else mn[i][j] = max(mn[i][j], v);
}
}
else if(r == 0)
{
for(int i = 1; i <= m; i ++)
{
if(flag) mx[i][q] = min(mx[i][q], v);
else mn[i][q] = max(mn[i][q], v);
}
}
else if(c == 0)
{
for(int i = 1; i <= n; i ++)
{
if(flag) mx[r][i] = min(mx[r][i], v);
else mn[r][i] = max(mn[r][i], v);
}
}
else
{
if(flag) mx[r][q] = min(mx[r][q], v);
else mn[r][q] = max(mn[r][q], v);
}
}
bool check(int l, int r)
{
for(int i = l; i < r; i += 2)
{
Edge e = sol.edges[i];
if(e.flow < e.cap) return false;
}
for(int i = 1; i <= m; i ++)
for(int j = 1; j <= n; j ++)
if(mn[i][j] > mx[i][j]) return false;
return true;
}
void work()
{
scanf("%d%d", &m, &n);
int sumc = 0, sumr = 0;
for(int i = 1; i <= m; i ++) scanf("%d", &row[i]), sumr += row[i];
for(int i = 1; i <= n; i ++) scanf("%d", &col[i]), sumc += col[i];
CLR(mn, 0); CLR(mx, INF);
scanf("%d", &c);
for(int i = 0; i < c; i ++)
{
int r, q, v; char op[4];
scanf("%d%d%s%d", &r, &q, op, &v);
if(op[0] == ‘=‘)
{
gao(r, q, v, 1);
gao(r, q, v, 0);
}
else if(op[0] == ‘<‘)
gao(r, q, v - 1, 1);
else
gao(r, q, v + 1, 0);
}
sol.init(n + m + 5);
int s = 0, t = n + m + 1, ss = m + n + 2, st = n + m + 3;
for(int i = 1; i <= m; i ++)
for(int j = 1; j <= n; j ++)
{
sol.AddEdge(i, m + j, mx[i][j] - mn[i][j]);
}
for(int i = 1; i <= m; i ++)
{
sol.AddEdge(ss, i, row[i]);
int sum = 0;
for(int j = 1; j <= n; j ++)
sum += mn[i][j];
sol.AddEdge(i, st, sum);
}
for(int i = 1; i <= n; i ++)
{
sol.AddEdge(i + m, st, col[i]);
int sum = 0;
for(int j = 1; j <= m; j ++)
sum += mn[j][i];
sol.AddEdge(ss, i + m, sum);
}
sol.AddEdge(s, st, sumr);
sol.AddEdge(ss, t, sumc);
for(int i = 1; i <= m; i ++)
sol.AddEdge(s, i, 0);
for(int i = 1; i <= n; i ++)
sol.AddEdge(i + m, t, 0);
sol.AddEdge(t, s, INF);
sol.Maxflow(ss, st, INF);
if(check(n*m*2, n*m*2+4*n+4*m+4))
{
for(int i = 0; i < m*n*2; i += 2)
{
int r = i/(n*2) + 1, c = i%(n*2)/2 + 1;
printf("%d", sol.edges[i].flow + mn[r][c]);
if(i % (n * 2) == n*2 - 2) puts("");
else printf(" ");
}
}
else puts("IMPOSSIBLE");
}
int main()
{
int T, flag = false;
scanf("%d", &T);
while(T --)
{
if(flag) puts("");
flag = true;
work();
}
}
poj 2396 Budget(可行流),布布扣,bubuko.com
原文：http://blog.csdn.net/ok_again/article/details/23843891