sgu194 Reactor Cooling【无源汇有上下界可行流】

  这是模板题了吧,先建立附加源汇,然后保留每个点的in-out,如果这个值是正的,那么就从附加源先这个点连一个边权为in-out的边,否则从这个点向附加汇连一条相反数的边,剩下题目中的边就用上界-下界连就好了。

 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 = 205, maxm = 40205;
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() {
34     static int que[maxn]; int qh(0), qt(0);
35     clr(level);
36     level[que[++qt] = s] = 1;
37     while (qh != qt) {
38         int x = que[++qh]; if (qh == maxn - 1) qh = 0;
39         for (int i = G[x]; i; i = E[i].nx) if (E[i].c && !level[E[i].v]) {
40             level[que[++qt] = E[i].v] = level[x] + 1;
41             if (qt == maxn - 1) qt = 0;
42         }
43     }
44     return !!level[t];
45 }
46 int dfs(int u, int rm) {
47     if (u == t) return rm;
48     int rm1 = rm;
49     for (int i = G[u]; i; i = E[i].nx) {
50         if (E[i].c && level[E[i].v] == level[u] + 1) {
51             int flow = dfs(E[i].v, min(E[i].c, rm));
52             E[i].c -= flow, E[i ^ 1].c += flow;
53             if ((rm -= flow) == 0) break;
54         }
55     }
56     if (rm1 == rm) level[u] = 0;
57     return rm1 - rm;
58 }
59 
60 int l[maxm], in[maxn], out[maxn];
61 int cnt;
62 int main() {
63     clr(G), edtot = 1, clr(in), clr(out);
64     int n, m;
65     scanf("%d%d", &n, &m);
66     s = n + 1, t = n + 2;
67     rep(i, 1, m) {
68         int x, y, a, b; scanf("%d%d%d%d", &x, &y, &a, &b); l[i] = a;
69         addedge(x, y, b - a);
70         in[y] += a, out[x] += a;
71     }
72     int sum(0);
73     rep(i, 1, n) {
74         if (in[i] > out[i]) addedge(s, i, in[i] - out[i]), sum += in[i] - out[i];
75         else addedge(i, t, out[i] - in[i]);
76     }
77     int ans(0);
78     while (bfs()) ans += dfs(s, 0x3f3f3f3f);
79     if (ans != sum) { puts("NO"); return 0; }
80     puts("YES");
81     for (int i = 1; i <= m; i++) printf("%d\n", E[(i << 1) ^ 1].c + l[i]);
82     return 0;
83 }
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posted @ 2015-12-17 02:01  y7070  阅读(498)  评论(0编辑  收藏  举报