上下界 最小流

做法：
1、和最大流有点相反，先ss tt连边（和最大流一样）但是这个和最大流不同的是先不用建立 t-s 的边
2、Dinic(ss,tt)
3、现在建立 t-s （t, s, 0, oo)
4、maxflow=Dinic(s,t)  if（sum==maxflow）成立有解不成立无解 sum和最大流一样表示ss流出的所有边流量和

5、输出边的流量

#include<iostream>
#include<algorithm>
#include<stdlib.h>
#include<string.h>
#include<stdio.h>
#include<math.h>
#include<string>
#include<vector>
#include<queue>
#include<list>
using namespace std;
typedef long long lld;
typedef unsigned int ud;
#define INF_MAX 0x3f3f3f3f
#define eatline() char chch;while((chch=getchar())!='\n')continue;
#define MemsetMax(a) memset(a,0x3f,sizeof a)
#define MemsetZero(a) memset(a,0,sizeof a)
#define MemsetMin(a) memset(a,-1,sizeof a)
#define MemsetFalse(a) MemsetZero(a)
#define PQ priority_queue
#define Q queue
#define maxn 308
#define maxm 40800
struct Edge
{
int v, f;
int id;
int next;
}E[maxm];
int k;
int head[maxn], h[maxn];
int level[maxn], stack[maxn];
int inout[maxn], low[maxn];
int ans[maxn];


bool BFS(int s, int t)
{
memset(level, 0, sizeof level);
level[s] = 1;
Q<int>q;
q.push(s);
while (!q.empty())
{
int u = q.front(); q.pop();
if (u == t)
return true;
for (int i = head[u]; i != -1; i = E[i].next)
{
int v = E[i].v;
if (!level[v] && E[i].f > 0)
{
level[v] = level[u] + 1;
q.push(v);
}
}
}
return false;
}


int Dinic(int s, int t)
{
int maxflow = 0;
while (BFS(s, t))
{
memcpy(h, head, sizeof h);
int top = 0;
int u = s;
while (true)
{


if (u == t)
{
int minflow = INF_MAX, flag = 0;
for (int i = 0; i < top; i++)
{
if (minflow>E[stack[i]].f)
{
minflow = E[stack[i]].f;
flag = i;
}
}
for (int i = 0; i < top; i++)
{
E[stack[i]].f -= minflow;
E[stack[i] ^ 1].f += minflow;
}
top = flag;
maxflow += minflow;
u = E[stack[top] ^ 1].v;
}
for (int i = h[u]; i != -1; i = h[u] = E[i].next)
{
int v = E[i].v;
if (level[v] == level[u] + 1 && E[i].f)
break;
}
if (h[u] != -1)
{
stack[top++] = h[u];
u = E[h[u]].v;
}
else
{
if (top == 0)
break;
level[u] = 0;
u = E[stack[--top] ^ 1].v;
}
}
}
return maxflow;
}


void add_edge(int u, int v, int low = 0, int up = INF_MAX)
{
E[k].v = v;
E[k].f = up - low;
E[k].next = head[u];
head[u] = k++;


E[k].v = u;
E[k].f = 0;
E[k].next = head[v];
head[v] = k++;
}


int main()
{
int n, m;
int s, t, ss, tt;
int u, v, c, z;
while (scanf("%d%d", &n, &m) != EOF)
{
s = 1;
t = n;
ss = t + 1;
tt = t + 2;
k = 0;
memset(head, -1, sizeof head);
for (int i = 1; i <= m; i++)
{
scanf("%d%d%d%d", &u, &v, &z, &c);
if (c == 1)
{
inout[u] -= z;
inout[v] += z;
low[i] = z;
}
else
{
add_edge(u, v, 0, z);
low[i] = 0;
}
}
int sum = 0;
for (int i = 1; i <= n; i++)
{
if (inout[i] > 0)
add_edge(ss, i, 0, inout[i]), sum += inout[i];
else
add_edge(i, tt, 0, -inout[i]);
}
Dinic(ss, tt); //printf("Dinic\n");
add_edge(t, s, 0, INF_MAX);
if (sum != Dinic(ss, tt))
printf("Impossible\n");
else
{
for (int i = head[t]; i != -1; i = E[i].next)
{
if (E[i].v == s)
{
printf("%d\n", E[i ^ 1].f);
break;
}
}
for (int i = 0; i < m; i++)
{
if (i!=m-1)
printf("%d ", E[i*2+1].f);
else
printf("%d\n", E[i*2+1].f);
}
}

}
return 0;
}
/*
4 4
1 2 2 0
2 4 1 1
1 3 2 1
3 4 3 0
4 4
1 2 1 0
2 4 2 1
1 3 3 1
3 4 2 0
*/