SGU 176 Flow construction 有源汇有上下界的最小流

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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 <iostream>#include <cstdio>#include <cstring>#include <algorithm>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;}

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