POJ ~ 2396 ~ Budget （有源汇上下界可行流）

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题意

T组测试数据，给你一个n*m的矩阵，输入每一行的和，每一列的和，在输入一些形如x y > z限制条件，如果x为0表示所有行，y为0表示所有列，x==0，y==0表示整个矩阵。求是否村子满足要求的矩阵，存在就输出矩阵。

思路

转化为有源汇上下界可行流来做。

行列看作点，行为Xi，列为Yi，建立源点s，汇点t。

①源点向行Xi建边，上下界均为行和

②列Yi向汇点Yi建边，上下界均为列和

③根据限制条件建边，由于可能对某一行列或者点可能有多种限制，大于1，大于2，小于4，小于3...所以我们先处理出来每个点的上下界（过程中需要判断给的数据是否合法，有可能某个点要求大于4且小于3，这种就是IMPOSSIBLE），然后根据每个点（X，Y）的上下界建边。

接下来就是有源汇的上下界可行流的部分了，相比

多加一条 t ~ s 边权为INF的边，这样就转化为无源汇上下界可行流了。

//#include<bits/stdc++.h>
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <vector>
#include <queue>
using namespace std;

const int MAXN = 1e5 + 5;
const int INF = 0x3f3f3f3f;

struct Edge
{
    int from, to, cap, flow;       //起点,终点,容量,流量
    Edge(int u, int v, int c, int f) : from(u), to(v), cap(c), flow(f) {}
};
struct Dinic
{
    int n, m, s, t;                //结点数,边数(包括反向弧),源点s,汇点t
    vector<Edge> edges;            //边表。edges[e]和edges[e^1]互为反向弧
    vector<int> G[MAXN];           //邻接表，G[i][j]表示结点i的第j条边在edges数组中的序号
    int d[MAXN];                   //从起点到i的距离（层数差）
    int cur[MAXN];                 //当前弧下标
    bool vis[MAXN];                //BFS分层使用

    void init(int n)
    {
        this->n = n;
        edges.clear();
        for (int i = 0; i <= n; i++) G[i].clear();
    }

    void add_edge(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()//构造分层网络
    {
        memset(vis, 0, sizeof(vis));
        queue<int> Q;
        d[s] = 0;
        vis[s] = true;
        Q.push(s);
        while (!Q.empty())
        {
            int x = Q.front(); Q.pop();
            for (int i = 0; i < G[x].size(); i++)
            {
                Edge& e = edges[G[x][i]];
                if (!vis[e.to] && e.cap > e.flow)
                {
                    vis[e.to] = true;
                    d[e.to] = d[x] + 1;
                    Q.push(e.to);
                }
            }
        }
        return vis[t];
    }
    int DFS(int x, int a)//沿阻塞流增广
    {
        if (x == t || a == 0) return a;
        int flow = 0, f;
        for (int& i = cur[x]; i < G[x].size(); i++)//从上次考虑的弧
        {
            Edge& e = edges[G[x][i]];
            if (d[x] + 1 == d[e.to] && (f = DFS(e.to, min(a, e.cap - e.flow))) > 0)//多路增广
            {
                e.flow += f;
                edges[G[x][i]^1].flow -= f;
                flow += f;
                a -= f;
                if (a == 0) break;
            }
        }
        return flow;
    }
    int max_flow(int s, int t)
    {
        this->s = s; this->t = t;
        int flow = 0;
        while (BFS())
        {
            memset(cur, 0, sizeof(cur));
            flow += DFS(s, INF);
        }
        return flow;
    }

}solve;

const int maxn = 205;
int n, m, down[maxn][maxn], up[maxn][maxn], sum[maxn*2];

void work(char op, int x, int y, int v, bool& ok)
{
    if (op == '>') down[x][y] = max(down[x][y], v+1);
    else if (op == '<') up[x][y] = min(up[x][y], v-1);
    else
    {
        if (v < down[x][y] || v > up[x][y]) ok = false;
        down[x][y] = up[x][y] = v;
    }
    if (down[x][y] > up[x][y]) ok = false;
}

int main()
{
    int T; scanf("%d", &T);
    while (T--)
    {
        int tot = 0;
        memset(sum, 0, sizeof(sum));
        scanf("%d%d", &n, &m);
        int s = 0, t = n+m+1, vs = n+m+2, vt = n+m+3;
        solve.init(vt);
        for (int i = 1; i <= n; i++)
        {
            int x; scanf("%d", &x);
            solve.add_edge(s, i, 0);
            sum[s] -= x;
            sum[i] += x;
        }
        for (int i = 1; i <= m; i++)
        {
            int x; scanf("%d", &x);
            solve.add_edge(i+n, t, 0);
            sum[i+n] -= x;
            sum[t] += x;
        }
        memset(down, 0, sizeof(down));
        memset(up, INF, sizeof(up));
        bool ok = true;
        int k; scanf("%d", &k);
        while (k--)
        {
            int x, y, v; char op;
            scanf("%d %d %c %d", &x, &y, &op, &v);
            if (x == 0 && y)
            {
                for (int i = 1; i <= n; i++) work(op, i, y, v, ok);
            }
            else if (x && y == 0)
            {
                for (int i = 1; i <= m; i++) work(op, x, i, v, ok);
            }
            else if (x == 0 && y == 0)
            {
                for (int i = 1; i <= n; i++)
                {
                    for (int j = 1; j <= m; j++) work(op, i, j, v, ok);
                }
            }
            else work(op, x, y, v, ok);
        }
        if (!ok) { printf("IMPOSSIBLE\n"); continue; }

        int cur = solve.edges.size();//记录矩阵点对应的边的起始编号
        for (int i = 1; i <= n; i++)
        {
            for (int j = 1; j <= m; j++)
            {
                solve.add_edge(i, j+n, up[i][j]-down[i][j]);
                sum[i] -= down[i][j];
                sum[j+n] += down[i][j];
            }
        }
        solve.add_edge(t, s, INF);
        for (int i = s; i <= t; i++)
        {
            if (sum[i] < 0) solve.add_edge(i, vt, -sum[i]);
            else solve.add_edge(vs, i, sum[i]), tot += sum[i];
        }
        int ans = solve.max_flow(vs, vt);
        if (ans != tot) printf("IMPOSSIBLE\n");
        else
        {
            int ss = n+m, x = 1, y = 1;
            for (int i = cur; i < cur+(n*m)*2; i += 2)
            {
                printf("%d", solve.edges[i].flow+down[x][y]);
                if (y == m) printf("\n"), x++, y = 1;
                else printf(" "), y++;
            }
        }
        printf("\n");
    }
    return 0;
}

/*
2

2 3
8 10
5 6 7
4
0 2 > 2
2 1 = 3
2 3 > 2
2 3 < 5

2 2
4 5
6 7
1
1 1 > 10
*/