sgu 176 Flow construction
题意：有一个加工厂，里面有n台机器，起点为1终点为n。中间的生产环节有货物限制，输入m个环节。每个环节有u,v,z,c四个数字。u表示起始机器，v表示终止机器，如果c为1，那么这条边的流量必须为z。如果c为0，那么流量在[0, z]之间。保证没有自环，没有重边，只有1号机器能够生产货物，n号机器消耗货物，其他机器不能积累货物。问满足条件的情况下，1号机器最少要生产多少货物。
思路：
有源汇有上下界的最小流
1. 增设一个超级源点，一个超级汇点，以同样的方式将原网络转化为没有流量下界限制的新网络。
2. 不连接原先网络的源汇点，并以超级源汇点为起点，终点做一次最大流maxflow1。
3. 此时从汇点向源点连接一条流量为INF的边，再以超级源汇点做一次网络流maxflow2。
4. 如果从源点出发的流均为满流，则有最小流，最小流的值为从汇点流向源点的流量。
* 在判断是否为满流时，条件为 maxflow1 + maxflow2 = 超级源点流出的流量
* 可以理解为第一遍做网络流时并无T->S的边，所以S->T的流量在第一遍时都已尽力往其他边流了. 于是加上T->S这条边后，都是些剩余的流不到其他边的流量. 从而达到尽可能减少T->S这边上的流量的效果，即减小了最终答案.
代码：
/*
ID: wuqi9395@126.com
PROG:
LANG: C++
*/
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<cmath>
#include<cstdio>
#include<vector>
#include<string>
#include<fstream>
#include<cstring>
#include<ctype.h>
#include<iostream>
#include<algorithm>
using namespace std;
#define INF (1 << 30)
#define LINF (1LL << 60)
#define PI acos(-1.0)
#define mem(a, b) memset(a, b, sizeof(a))
#define rep(i, a, n) for (int i = a; i < n; i++)
#define per(i, a, n) for (int i = n - 1; i >= a; i--)
#define eps 1e-6
#define debug puts("===============")
#define pb push_back
#define mkp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
#define POSIN(x,y) (0 <= (x) && (x) < n && 0 <= (y) && (y) < m)
typedef long long ll;
typedef unsigned long long ULL;
const int maxn = 1000;
const int maxm = 100000;
struct node {
int v; // vertex
int cap; // capacity
int flow; // current flow in this arc
int nxt;
} e[maxm * 2];
int g[maxn], cnt;
int st, ed, n;
int S, D;
int N, M;
int tot[maxn], low[maxm], id[maxm];
void add(int u, int v, int c, int k) {
e[++cnt].v = v;
e[cnt].cap = c;
e[cnt].flow = 0;
e[cnt].nxt = g[u];
g[u] = cnt;
id[k] = cnt;
e[++cnt].v = u;
e[cnt].cap = 0;
e[cnt].flow = 0;
e[cnt].nxt = g[v];
g[v] = cnt;
id[0] = 0;
}
void init() {
memset(g, 0, sizeof(g));
cnt = 1;
S = 1, D = N, st = N + 1, ed = N + 2;
int u, v, z, c;
for (int i = 1; i <= M; i++) {
scanf("%d%d%d%d", &u, &v, &z, &c);
if (c == 0) {
add(u, v, z, i);
low[i] = 0;
} else {
low[i] = z;
tot[u] -= z;
tot[v] += z;
}
}
n = N + 4;
}
int dist[maxn], numbs[maxn], q[maxn];
void rev_bfs() {
int font = 0, rear = 1;
for (int i = 0; i <= n; i++) { //n为总点数
dist[i] = maxn;
numbs[i] = 0;
}
q[font] = ed;
dist[ed] = 0;
numbs[0] = 1;
while(font != rear) {
int u = q[font++];
for (int i = g[u]; i; i = e[i].nxt) {
if (e[i ^ 1].cap == 0 || dist[e[i].v] < maxn) continue;
dist[e[i].v] = dist[u] + 1;
++numbs[dist[e[i].v]];
q[rear++] = e[i].v;
}
}
}
int maxflow() {
rev_bfs();
int u, totalflow = 0;
int curg[maxn], revpath[maxn];
for(int i = 0; i <= n; ++i) curg[i] = g[i];
u = st;
while(dist[st] < n) {
if(u == ed) { // find an augmenting path
int augflow = INF;
for(int i = st; i != ed; i = e[curg[i]].v)
augflow = min(augflow, e[curg[i]].cap);
for(int i = st; i != ed; i = e[curg[i]].v) {
e[curg[i]].cap -= augflow;
e[curg[i] ^ 1].cap += augflow;
e[curg[i]].flow += augflow;
e[curg[i] ^ 1].flow -= augflow;
}
totalflow += augflow;
u = st;
}
int i;
for(i = curg[u]; i; i = e[i].nxt)
if(e[i].cap > 0 && dist[u] == dist[e[i].v] + 1) break;
if(i) { // find an admissible arc, then Advance
curg[u] = i;
revpath[e[i].v] = i ^ 1;
u = e[i].v;
} else { // no admissible arc, then relabel this vertex
if(0 == (--numbs[dist[u]])) break; // GAP cut, Important!
curg[u] = g[u];
int mindist = n;
for(int j = g[u]; j; j = e[j].nxt)
if(e[j].cap > 0) mindist = min(mindist, dist[e[j].v]);
dist[u] = mindist + 1;
++numbs[dist[u]];
if(u != st)
u = e[revpath[u]].v; // Backtrack
}
}
return totalflow;
}
int lowbound_minflow() {
long long sum = 0;
for (int i = 1; i <= N; i++) {
if (tot[i] > 0) add(st, i, tot[i], 0), sum += tot[i];
if (tot[i] < 0) add(i, ed, -tot[i], 0);
}
int dd = maxflow();
add(D, S, INF, M + 1);
dd += maxflow();
if (dd != sum) return -1;
return e[id[M + 1]].flow;
}
int main () {
scanf("%d%d", &N, &M);
init();
int mn = lowbound_minflow();
if (mn == -1) puts("Impossible");
else {
printf("%d\n", mn);
for (int i = 1; i < M; i++) {
if (low[i] == 0) printf("%d ", e[id[i]].flow);
else printf("%d ", low[i]);
}
if (low[M] == 0) printf("%d\n", e[id[M]].flow);
else printf("%d\n", low[M]);
}
return 0;
}