POJ 2396 Budget

POJ_2396

    第一次接触有上下界的网络流问题，感觉只要按套路来就可以了，具体的算法可以参考：http://blog.csdn.net/water_glass/article/details/6823741。

#include<stdio.h>
#include<string.h>
#include<algorithm>
#define MAXN 210
#define MAXM 30
#define MAXV 230
#define MAXE 24900
#define INF 0x3f3f3f3f
int N, M, low[MAXN][MAXM], high[MAXN][MAXM], R[MAXN], C[MAXM];
int S, T, SS, TT, first[MAXV], e, next[MAXE], v[MAXE], flow[MAXE];
int d[MAXV], q[MAXV], work[MAXV];
int id[MAXN][MAXM];
void update(int x1, int x2, int y1, int y2, char op, int z)
{
    int i, j;
    for(i = x1; i <= x2; i ++)
        for(j = y1; j <= y2; j ++)
        {
            if(op == '=')
                low[i][j] = std::max(low[i][j], z), high[i][j] = std::min(high[i][j], z);
            else if(op == '<')
                high[i][j] = std::min(high[i][j], z - 1);
            else
                low[i][j] = std::max(low[i][j], z + 1);
        }
}
void init()
{
    int i, j, k, x, y, z, n;
    char op[5];
    scanf("%d%d", &N, &M);
    memset(low, 0, sizeof(low)), memset(high, 0x3f, sizeof(high));
    for(i = 1; i <= N; i ++)
        scanf("%d", &R[i]);    
    for(i = 1; i <= M; i ++)
        scanf("%d", &C[i]);
    scanf("%d", &n);
    for(k = 0; k < n; k ++)
    {
        scanf("%d%d%s%d", &x, &y, op, &z);
        if(x == 0 && y == 0)
            update(1, N, 1, M, op[0], z);
        else if(x == 0)
            update(1, N, y, y, op[0], z);
        else if(y == 0)
            update(x, x, 1, M, op[0], z);
        else
            update(x, x, y, y, op[0], z);
    }
}
int check()
{
    for(int i = 1; i <= N; i ++) for(int j = 1; j <= M; j ++) if(low[i][j] > high[i][j]) return 0;
    return 1;
}
void add(int x, int y, int z)
{
    v[e] = y, flow[e] = z;
    next[e] = first[x], first[x] = e ++;    
}
int build()
{
    int i, j, sum = 0;
    S = 0, T = N + M + 1, SS = T + 1, TT = SS + 1;
    memset(first, -1, sizeof(first[0]) * (TT + 1)), e = 0;
    add(T, S, INF), add(S, T, 0);
    for(i = 1; i <= N; i ++)
    {
        add(SS, i, R[i]), add(i, SS, 0), add(S, TT, R[i]), add(TT, S, 0);
        sum += R[i];
    }
    for(i = 1; i <= M; i ++)
    {
        add(SS, T, C[i]), add(T, SS, 0), add(N + i, TT, C[i]), add(TT, N + i, 0);
        sum += C[i];
    }
    for(i = 1; i <= N; i ++)
        for(j = 1; j <= M; j ++)
        {
            add(i, N + j, high[i][j] - low[i][j]), id[i][j] = e, add(N + j, i, 0);
            add(SS, N + j, low[i][j]), add(N + j, SS, 0), add(i, TT, low[i][j]), add(TT, i, 0);
            sum += low[i][j];
        }
    return sum;    
}
int bfs(int S, int T)
{
    int i, j, rear = 0;
    memset(d, -1, sizeof(d[0]) * (T + 1));
    d[S] = 0, q[rear ++] = S;
    for(i = 0; i < rear; i ++)
        for(j = first[q[i]]; j != -1; j = next[j])
            if(flow[j] && d[v[j]] == -1)
            {
                d[v[j]] = d[q[i]] + 1, q[rear ++] = v[j];
                if(v[j] == T)
                    return 1;    
            }    
    return 0;
}
int dfs(int cur, int a, int T)
{
    if(cur == T) return a;
    for(int &i = work[cur]; i != -1; i = next[i])
        if(flow[i] && d[v[i]] == d[cur] + 1)
            if(int t = dfs(v[i], std::min(a, flow[i]), T))
            {
                flow[i] -= t, flow[i ^ 1] += t;
                return t;
            }
    return 0;
}
int dinic(int S, int T)
{
    int ans = 0, t;
    while(bfs(S, T))
    {
        memcpy(work, first, sizeof(first[0]) * (T + 1));
        while(t = dfs(S, INF, T))
            ans += t;    
    }
    return ans;
}
void print()
{
    int i, j;
    for(i = 1; i <= N; i ++)
    {
        printf("%d", flow[id[i][1]] + low[i][1]);
        for(j = 2; j <= M; j ++)
            printf(" %d", flow[id[i][j]] + low[i][j]);
        printf("\n");
    }
}
void solve()
{
    if(!check())
    {
        printf("IMPOSSIBLE\n");
        return ;    
    }
    int sum = build();
    if(sum != dinic(SS, TT))
        printf("IMPOSSIBLE\n");
    else
        print();
}
int main()
{
    int t;
    scanf("%d", &t);
    while(t --)
    {
        init();
        solve();
        if(t) printf("\n");
    }
    return 0;
}