SGU 176 Flow construction 有源汇有上下界的最小流

题目：http://acm.hust.edu.cn/vjudge/problem/11025

题意：有n个点用m个导管连接，物质可以在导管中流动，从起点1流到终点n。每次输入u v z c描述一个导管，u v代表导管连接u v两点且从u流向v，c有0 1两种值，为0时导管的流量不大于z，为1时导管内的流量必须为z，即上下界都为z。问满足要求的最小流量

思路：有源汇有上下界的最小流问题。

/*du[i]表示i节点的入流之和与出流之和的差。 
*增设附加源点sups和附加汇点supt，连边（sups，du[i]（为正）），（-du[i]（为负），supt）。
*对附加源点和附加汇点做一次最大流
*从汇点tt到源点ss连一条容量为无穷大的边
*再次对附加源点和附加汇点做一次最大流
*当且仅当所有附加弧满载时有可行流，最后答案为经过边tt -> ss的流量
*/
#include 
#include 
#include 
#include 
using namespace std;

const int N = 110;
const int INF = 0x3f3f3f3f;
struct edge
{
    int to, cap, next, id;
}g[N*N*2];
int cnt, nv, head[N], level[N], gap[N], cur[N], pre[N];
int du[N], dn[N*N];
void add_edge(int v, int u, int cap, int id = 0)
{
    g[cnt].to = u, g[cnt].cap = cap, g[cnt].next = head[v], g[cnt].id = 0, head[v] = cnt++;
    g[cnt].to = v, g[cnt].cap = 0, g[cnt].next = head[u], g[cnt].id = id, head[u] = cnt++;
}
int sap(int s, int t)
{
    memset(level, 0, sizeof level);
    memset(gap, 0, sizeof gap);
    memcpy(cur, head, sizeof head);
    gap[0] = nv;
    int v = pre[s] = s, flow = 0, aug = INF;
    while(level[s] < nv)
    {
        bool flag = false;
        for(int &i = cur[v]; i != -1; i = g[i].next)
        {
            int u = g[i].to;
            if(g[i].cap > 0 && level[v] == level[u] + 1)
            {
                flag = true;
                pre[u] = v;
                v = u;
                aug = min(aug, g[i].cap);
                if(v == t)
                {
                    flow += aug;
                    while(v != s)
                    {
                        v = pre[v];
                        g[cur[v]].cap -= aug;
                        g[cur[v]^1].cap += aug;
                    }
                    aug = INF;
                }
                break;
            }
        }
        if(flag) continue;
        int minlevel = nv;
        for(int i = head[v]; i != -1; i = g[i].next)
        {
            int u = g[i].to;
            if(g[i].cap > 0 && minlevel > level[u])
                minlevel = level[u], cur[v] = i;
        }
        if(--gap[level[v]] == 0) break;
        level[v] = minlevel + 1;
        gap[level[v]]++;
        v = pre[v];
    }
    return flow;
}

int main()
{
    int n, m, u, v, z, c;
    while(~ scanf("%d%d", &n, &m))
    {
        cnt = 0;
        memset(head, -1, sizeof head);
        memset(du, 0, sizeof du);
        memset(dn, 0, sizeof dn);
        for(int i = 1; i <= m; i++)
        {
            scanf("%d%d%d%d", &u, &v, &z, &c);
            if(c == 1) du[v] += z, du[u] -= z, dn[i] = z;
            else add_edge(u, v, z, i);
        }
        int ss = 1, tt = n, num = cnt;
        int sups = 0, supt = n + 1;
        for(int i = 1; i <= n; i++)
        {
            if(du[i] > 0) add_edge(sups, i, du[i]);
            if(du[i] < 0) add_edge(i, supt, -du[i]);
        }
        nv = supt + 1;
        sap(sups, supt);
        add_edge(tt, ss, INF);
        int tmp = cnt - 1;
        sap(sups, supt);
        bool flag = true;
        for(int i = head[sups]; i != -1; i = g[i].next)
            if(g[i].cap > 0)
            {
                flag = false; break;
            }
        if(!flag) puts("Impossible");
        else
        {
            printf("%d\n", g[tmp].cap);
            for(int i = 0; i < num; i++) dn[g[i].id] = g[i].cap;
            for(int i = 1; i <= m; i++) printf("%d%c", dn[i], i == m ? '\n' : ' ');
        }
    }
    return 0;
}