sgu176 Flow Construction【有源汇有上下界最小流】

 1 #include <bits/stdc++.h>
 2 #define rep(i, a, b) for (int i = a; i <= b; i++)
 3 #define drep(i, a, b) for (int i = a; i >= b; i--)
 4 #define REP(i, a, b) for (int i = a; i < b; i++)
 5 #define mp make_pair
 6 #define pb push_back
 7 #define clr(x) memset(x, 0, sizeof(x))
 8 #define xx first
 9 #define yy second
10 using namespace std;
11 typedef long long i64;
12 typedef pair<int, int> pii;
13 const int inf = ~0U >> 1;
14 const i64 INF = ~0ULL >> 1;
15 //******************************
16 
17 const int maxn = 105, maxm = 10105;
18 
19 struct Ed {
20     int u, v, nx, c; Ed() {}
21     Ed(int _u, int _v, int _nx, int _c) :
22         u(_u), v(_v), nx(_nx), c(_c) {}
23 } E[maxm << 1];
24 int G[maxn], edtot = 1;
25 void addedge(int u, int v, int c) {
26     E[++edtot] = Ed(u, v, G[u], c);
27     G[u] = edtot;
28     E[++edtot] = Ed(v, u, G[v], 0);
29     G[v] = edtot;
30 }
31 
32 int level[maxn], S, T;
33 bool bfs(int s, int t) {
34     static int que[maxn]; int qh(0), qt(0);
35     clr(level); level[que[++qt] = s] = 1;
36     while (qh != qt) {
37         int x = que[++qh]; if (qh == maxn - 1) qh = 0;
38         for (int i = G[x]; i; i = E[i].nx) if (E[i].c && !level[E[i].v]) {
39             level[que[++qt] = E[i].v] = level[x] + 1;
40             if (qt == maxn - 1) qt = 0;
41         }
42     }
43     return !!level[t];
44 }
45 int dfs(int u, int rm, int t) {
46     if (u == t) return rm;
47     int rm1 = rm;
48     for (int i = G[u]; i; i = E[i].nx) {
49         if (E[i].c && level[E[i].v] == level[u] + 1) {
50             int flow = dfs(E[i].v, min(E[i].c, rm), t);
51             E[i].c -= flow, E[i ^ 1].c += flow;
52             if ((rm -= flow) == 0) break;
53         }
54     }
55     if (rm1 == rm) level[u] = 0;
56     return rm1 - rm;
57 }
58 
59 int l[maxm], in[maxn];
60 int main() {
61     int n, m;
62     scanf("%d%d", &n, &m);
63     rep(i, 1, m) {
64         int x, y, a, b; scanf("%d%d%d%d", &x, &y, &a, &b);
65         l[i] = a * b;
66         if (b) in[y] += a, in[x] -= a;
67         addedge(x, y, a - a * b);
68     }
69     S = n + 1, T = n + 2;
70     int sum(0);
71     rep(i, 1, n) {
72         if (in[i] > 0) sum += in[i];
73         if (in[i] > 0) addedge(S, i, in[i]);
74         else addedge(i, T, -in[i]);
75     }
76     addedge(n, 1, 0x3f3f3f3f);
77     while (bfs(S, T)) sum -= dfs(S, 0x3f3f3f3f, T);
78     if (sum) { puts("Impossible"); return 0; }
79     int ans = E[edtot].c;
80     E[edtot].c = E[edtot ^ 1].c = 0;
81     int tmp(0);
82     while (bfs(n, 1)) tmp += dfs(n, 0x3f3f3f3f, 1);
83     ans -= tmp;
84     if (ans < 0) {
85         addedge(S, 1, -ans);
86         while (bfs(S, n)) dfs(S, 0x3f3f3f3f, n);
87         ans = 0;
88     }
89     printf("%d\n", ans);
90     rep(i, 1, m) {
91         printf(i == 1 ? "%d" : " %d", E[(i << 1) ^ 1].c + l[i]);
92     }
93     puts("");
94     return 0;
95 }