Flow construction
有N個點,M條邊,每條邊有{u, v, w, c},如果c==1表示這條邊一定要滿流流過去,否則,這條邊的流量可以是0~w的任意。求最小可行流。
於是,我們很簡單的構造了一個可行流網絡,這些都是簡單的,關鍵是求出可行流之後的步驟,有很多的細節上的東西。
首先,題目中給的源匯點是1和N,但是如果有直接從1到N的邊,會形成環流這樣的形式,作爲DAG圖的網絡流顯然是不允許這樣的情況出現的,所以,我們開始的時候獨立創建一個s、t的源匯點。s鏈接1號點,N號點鏈接t,流量爲INF,無上下界。然後,我們就避免了源匯點之間構成了直接環的情況,這時候再去跑就滿足網絡流的條件了。
現在,既然是要求最小流,那麼我們將起點設置爲t,將終點設置爲s,相當於反過來了,同時不要忘記刪除我們加入的t到s的“使有源匯圖看作無源匯圖”的這條特殊邊。
#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <limits>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <bitset>
#include <unordered_map>
#include <unordered_set>
#define lowbit(x) ( x&(-x) )
#define pi 3.141592653589793
#define e 2.718281828459045
#define INF 0x3f3f3f3f3f3f3f3f
#define HalF (l + r)>>1
#define lsn rt<<1
#define rsn rt<<1|1
#define Lson lsn, l, mid
#define Rson rsn, mid+1, r
#define QL Lson, ql, qr
#define QR Rson, ql, qr
#define myself rt, l, r
using namespace std;
typedef unsigned long long ull;
typedef unsigned int uit;
typedef long long ll;
const int maxN = 107, maxM = 2e4 + 7;
int N, M, head[maxN], cnt;
ll du[maxN], fl[maxM], sum;
struct Eddge
{
int nex, to; ll flow;
Eddge(int a=-1, int b=0, ll c=0):nex(a), to(b), flow(c) {}
}edge[maxM];
inline void addEddge(int u, int v, ll f)
{
edge[cnt] = Eddge(head[u], v, f);
head[u] = cnt++;
}
inline void _add(int u, int v, ll f) { addEddge(u, v, f); addEddge(v, u, 0); }
struct ISAP
{
int S, T, gap[maxN], cur[maxN], deep[maxN], que[maxN], ql, qr, node;
inline void init()
{
for(int i=0; i<=node + 1; i++)
{
gap[i] = deep[i] = 0;
cur[i] = head[i];
}
++gap[deep[T] = 1];
que[ql = qr = 1] = T;
while(ql <= qr)
{
int u = que[ql ++];
for(int i=head[u], v; ~i; i=edge[i].nex)
{
v = edge[i].to;
if(!deep[v]) { ++gap[deep[v] = deep[u] + 1]; que[++qr] = v; }
}
}
}
inline ll aug(int u, ll Flow)
{
if(u == T) return Flow;
ll flow = 0;
for(int &i = cur[u], v; ~i; i=edge[i].nex)
{
v = edge[i].to;
if(deep[u] == deep[v] + 1)
{
ll tmp = aug(v, min(edge[i].flow, Flow));
flow += tmp; Flow -= tmp; edge[i].flow -= tmp; edge[i ^ 1].flow += tmp;
if(!Flow) return flow;
}
}
if(!(--gap[deep[u]])) deep[S] = node + 1;
++gap[++deep[u]]; cur[u] = head[u];
return flow;
}
inline ll Max_Flow()
{
init();
ll ret = aug(S, INF);
while(deep[S] <= node) ret += aug(S, INF);
return ret;
}
} mf;
int s, t;
inline void init()
{
cnt = 0; s = 0; t = N + 1; mf.S = N + 2; mf.T = N + 3;
for(int i=0; i<=N + 3; i++) { head[i] = -1; du[i] = 0; }
}
int main()
{
scanf("%d%d", &N, &M);
init();
for(int i=1, u, v, w, op; i<=M; i++)
{
scanf("%d%d%d%d", &u, &v, &w, &op);
if(op)
{
_add(u, v, 0); fl[i] = w;
du[u] -= w; du[v] += w;
}
else
{
_add(u, v, w); fl[i] = 0;
}
}
sum = 0;
for(int i=1; i<=N; i++)
{
if(du[i] > 0)
{
_add(mf.S, i, du[i]);
sum += du[i];
}
else if(du[i] < 0)
{
_add(i, mf.T, -du[i]);
sum -= du[i];
}
}
_add(s, 1, INF); _add(N, t, INF);
_add(t, s, INF); mf.node = N + 4;
if(mf.Max_Flow() * 2LL != sum) { printf("Impossible\n"); return 0; }
ll tmp = edge[cnt - 1].flow; edge[cnt - 2].flow = edge[cnt - 1].flow = 0;
head[mf.S] = head[mf.T] = -1;
mf.S = t; mf.T = s; mf.node = N + 2;
ll ans = tmp - mf.Max_Flow();
printf("%lld\n", ans);
for(int i=1; i<=M; i++)
{
printf("%lld%c", edge[i * 2 - 1].flow + fl[i], i == M ? '\n' : ' ');
}
return 0;
}