首页 > 代码库 > ZOJ 2314 带上下界的可行流
ZOJ 2314 带上下界的可行流
对于无源汇问题,方法有两种.
1 从边的角度来处理. 新建超级源汇, 对于每一条有下界的边,x->y, 建立有向边 超级源->y ,容量为x->y下界,建立有向边 x-> 超级汇,容量为x->y下界.建立有向边 x->y,容量为x->y的上界减下界.
2 从点的角度来处理. 新建超级源汇,对于每个点流进的下界和为 in, 流出此点的下界和为out.如果in > out. 建立有向边 超级源->i,容量为in-out.反之,建立有向边 i->超级汇,容量为out-in.
如果超级源的每一条出边都满流,则存在一个可行流,可行流的流量就是每一条逆向边的流量+ 此边下界.
从点的角度的处理方法是边的处理方法的拓展.基于边的处理比较容易理解,但是复杂度较高.
下面是基于点的处理的程序:
1: /*
2: 带上下界的可行流
3: */
4: #include<iostream>
5: #include<cmath>
6: #include<memory>
7: #include <string.h>
8: #include <cstdio>
9: using namespace std;
10:
11: #define V 300 // vertex
12: #define E 40800 // edge
13: #define INF 0x3F3F3F3F // 1061109567
14:
15: int i,j,k;
16: #define REP(i,n) for((i)=0;(i)<(int)(n);(i)++)
17: #define snuke(c,itr) for(__typeof((c).begin()) itr=(c).begin();itr!=(c).end();itr++)
18:
19: struct MaxFlow
20: {
21: struct Edge
22: {
23: int v, w, next; //w for capicity
24: int lb,up;
25: } edge[E];
26:
27: int head[V]; // head[u]表示顶点u第一条邻接边的序号, 若head[u] = -1, u没有邻接边
28: int e; // the index of the edge
29: int src, sink;
30: int net[V]; // 流入此节点的流的下界和 - 流出此节点的流的下界和,对于带上下界的来进行使用
31:
32:
33: void addedge(int u, int v, int w, int lb = 0, int up = INF, int rw = 0)
34: {
35: edge[e].v = v;
36: edge[e].w= w;
37: edge[e].next = head[u];
38: edge[e].lb = lb, edge[e].up = up;
39: head[u] = e++;
40: // reverse edge v -> u
41: edge[e].v = u;
42: edge[e].w = rw;
43: edge[e].lb = lb, edge[e].up = up;
44: edge[e].next = head[v];
45: head[v] = e++;
46: }
47:
48: int ISAP(int VertexNum )
49: {
50: int u, v, max_flow, aug, min_lev;
51: int curedge[V], parent[V], level[V];
52: int count[V], augment[V];
53:
54: memset(level, 0, sizeof(level));
55: memset(count, 0, sizeof(count));
56: REP(i,VertexNum+1) curedge[i] = head[i];
57: max_flow = 0;
58: augment[src] = INF;
59: parent[src] = -1;
60: u = src;
61:
62: while (level[src] < VertexNum)
63: {
64: if (u == sink)
65: {
66: max_flow += augment[sink];
67: aug = augment[sink];
68: for (v = parent[sink]; v != -1; v = parent[v])
69: {
70: i = curedge[v];
71: edge[i].w -= aug;
72: edge[i^1].w += aug;
73: augment[edge[i].v] -= aug;
74: if (edge[i].w == 0) u = v;
75: }
76: }
77: for (i = curedge[u]; i != -1; i = edge[i].next)
78: {
79: v = edge[i].v;
80: if (edge[i].w > 0 && level[u] == (level[v]+1))
81: {
82: augment[v] = min(augment[u], edge[i].w);
83: curedge[u] = i;
84: parent[v] = u;
85: u = v;
86: break;
87: }
88: }
89: if (i == -1)
90: {
91: if (--count[level[u]] == 0) break;
92: curedge[u] = head[u];
93: min_lev = VertexNum;
94: for (i = head[u]; i != -1; i = edge[i].next)
95: if (edge[i].w > 0)
96: min_lev = min(level[edge[i].v], min_lev);
97: level[u] = min_lev + 1;
98: count[level[u]]++;
99: if (u != src ) u = parent[u];
100: }
101: }
102: return max_flow;
103: }
104: void solve()
105: {
106: int T; cin>>T;
107: for(int cas =1; cas <=T; cas++)
108: {
109: int N,M; cin>>N>>M;
110: e = 0;
111: memset(head, -1, sizeof(head));
112: memset(net, 0, sizeof(net));
113: int a,b,c,d;
114: for(int i=1; i<=M;i++)
115: {
116: scanf("%d%d%d%d", &a, &b, &c, &d);
117: net[b] += c; net[a] -= c;
118: addedge(a,b,d-c, c,d);
119: }
120: // 无源汇,这两个是super src, super sink.
121: src =http://www.mamicode.com/N+1; sink = N+2;
122: for(int i=1; i<=N; i++)
123: {
124: if(net[i] > 0)
125: addedge(src, i, net[i]);
126: else
127: addedge(i, sink, -net[i]);
128: }
129: ISAP(N+2);
130: bool flag = true;
131: for (i = head[src]; i != -1; i = edge[i].next)
132: {
133: if(edge[i].w !=0)
134: {
135: flag = false;
136: break;
137: }
138: }
139: if(flag)
140: {
141: cout<<"YES"<<endl;
142: for(int i=0; i<M;i++)
143: {
144: cout<<edge[i*2+1].w + edge[i*2+1].lb<<endl;
145: }
146: cout<<endl;
147: }else
148: {
149: cout<<"NO"<<endl;
150: cout<<endl;
151: }
152: }
153:
154: }
155: }sap;
156:
157: int main()
158: {
159: // freopen("1.txt","r",stdin);
160: sap.solve();
161: return 0;
162: }
声明:以上内容来自用户投稿及互联网公开渠道收集整理发布,本网站不拥有所有权,未作人工编辑处理,也不承担相关法律责任,若内容有误或涉及侵权可进行投诉: 投诉/举报 工作人员会在5个工作日内联系你,一经查实,本站将立刻删除涉嫌侵权内容。