[ACM-ICPC 2018 沈阳网络赛] F. Fantastic Graph （有源汇上下界可行流+建图)
"Oh, There is a bipartite graph.""Make it Fantastic."
X wants to check whether a bipartite graph is a fantastic graph. He has two fantastic numbers, and he wants to let all the degrees to between the two boundaries. You can pick up several edges from the current graph and try to make the degrees of every point to between the two boundaries. If you pick one edge, the degrees of two end points will both increase by one. Can you help X to check whether it is possible to fix the graph?
Input
There are at most 3030 test cases.
For each test case,The first line contains three integers NN the number of left part graph vertices, MM the number of right part graph vertices, and KK the number of edges ( 1 \le N \le 20001≤N≤2000,0 \le M \le 20000≤M≤2000,0 \le K \le 60000≤K≤6000 ). Vertices are numbered from 11 to NN.
The second line contains two numbers L, RL,R (0 \le L \le R \le 300)(0≤L≤R≤300). The two fantastic numbers.
Then KK lines follows, each line containing two numbers UU, VV (1 \le U \le N,1 \le V \le M)(1≤U≤N,1≤V≤M). It shows that there is a directed edge from UU-th spot to VV-th spot.
Note. There may be multiple edges between two vertices.
Output
One line containing a sentence. Begin with the case number. If it is possible to pick some edges to make the graph fantastic, output "Yes" (without quote), else output "No" (without quote).
样例输入
3 3 7
2 3
1 2
2 3
1 3
3 2
3 3
2 1
2 1
3 3 7
3 4
1 2
2 3
1 3
3 2
3 3
2 1
2 1
样例输出
Case 1: Yes
Case 2: No
SOLUTION:
开一个源点s和一个汇点t，对于每个N点，从s连一条[L,R]的边，
对于每一个M点，连向t一条[L,R]的边，
对于图中原本的NM之间的边，连一条[0,1]的边，然后跑有源汇点有上下界的网络流就可以。
CODE:
1 #include <cstdio>
2 #include <cstring>
3 #include <algorithm>
4 #include <iostream>
5 #include <queue>
6
7 using namespace std;
8 const int INF=2147000000;
9 const int maxn=4000+100;
10 const int maxm=100000+100;
11 struct Dinic
12 {
13 int n,m,s,t,sz;
14 int head[maxn],Next[maxm*2],to[maxm*2],cap[2*maxm],flow[2*maxm];
15 int vis[maxn],d[maxn],cur[maxn];
16 void init(int n)
17 {
18 this->n=n;
19 sz=-1;
20 memset(head,-1,sizeof(head));
21 }
22 void add_edge(int a,int b,int c)
23 {
24 ++sz;
25 to[sz]=b;
26 cap[sz]=c;
27 flow[sz]=c;
28 Next[sz]=head[a];
29 head[a]=sz;
30 ++sz;
31 to[sz]=a;
32 cap[sz]=c;
33 flow[sz]=0;
34 Next[sz]=head[b];
35 head[b]=sz;
36 }
37 bool BFS()
38 {
39 memset(vis,0,sizeof(vis));
40 queue<int>q;
41 vis[s]=1;
42 q.push(s);
43 d[s]=0;
44 while(!q.empty())
45 {
46 int u=q.front();
47 q.pop();
48 for(int i=head[u]; i!=-1; i=Next[i])
49 {
50 int v=to[i];
51 if(!vis[v]&&flow[i])
52 {
53 vis[v]=1;
54 d[v]=d[u]+1;
55 q.push(v);
56 }
57 }
58 }
59 return vis[t];
60 }
61 int DFS(int x,int a)
62 {
63 if(x==t||a==0)
64 return a;
65 int Flow=0,f;
66 for(int &i=cur[x]; i!=-1; i=Next[i])
67 {
68 int v=to[i];
69 if(d[v]==d[x]+1&&(f=DFS(v,min(a,flow[i])))>0)
70 {
71 Flow+=f;
72 flow[i]-=f;
73 flow[i^1]+=f;
74 a-=f;
75 if(a==0)
76 break;
77 }
78 }
79 return Flow;
80 }
81 int Maxflow(int s,int t)
82 {
83 this->s=s,this->t=t;
84 int Flow=0;
85 while(BFS())
86 {
87 for(int i=0; i<=n; i++)
88 cur[i]=head[i];
89 // Flow+=DFS(s,INF);
90 int d;
91 while(d=DFS(s,INF))Flow+=d;
92
93 }
94 //printf("!!%d\n",Flow);
95 return Flow;
96 }
97 } dinic;
98 int N,M,K,L,R,sum,kase;
99 int du[maxn];
100 int main()
101 {
102 kase=0;
103 while(scanf("%d%d%d",&N,&M,&K)!=EOF)
104 {
105 ++kase;
106 memset(du,0,sizeof(du));
107 scanf("%d%d",&L,&R);
108 dinic.init(N+M+4);
109 for(int i=1; i<=N; i++)
110 {
111 dinic.add_edge(0,i,R-L);
112 du[i]+=L;
113 du[0]-=L;
114 }
115 sum=0;
116 for(int i=1; i<=M; i++)
117 {
118 dinic.add_edge(i+N,N+M+1,R-L);
119 du[i+N]-=L;
120 du[N+M+1]+=L;
121 }
122 int a,b;
123 for(int i=1; i<=K; i++)
124 {
125 scanf("%d%d",&a,&b);
126 dinic.add_edge(a,b+N,1);
127 }
128 dinic.add_edge(N+M+1,0,INF);
129 int s=N+M+2,t=N+M+3;
130 for(int i=0; i<=N+M+1; i++)
131 {
132 if(du[i]>0)
133 {
134 sum+=du[i];
135 dinic.add_edge(s,i,du[i]);
136 }
137 if(du[i]<0)
138 {
139 dinic.add_edge(i,t,-du[i]);
140 }
141 }
142 int maxflow=dinic.Maxflow(s,t);
143 // printf("%d %d\n",maxflow,sum);
144 printf("Case %d: ",kase);
145 if(maxflow==sum)
146 {
147 printf("Yes\n");
148 }
149 else
150 {
151 printf("No\n");
152 }
153 }
154 return 0;
155 }
