Dining(最大流)
Dining
http://poj.org/problem?id=3281
Time Limit: 2000MS | Memory Limit: 65536K | |
Total Submissions: 23462 | Accepted: 10371 |
Description
Cows are such finicky eaters. Each cow has a preference for certain foods and drinks, and she will consume no others.
Farmer John has cooked fabulous meals for his cows, but he forgot to check his menu against their preferences. Although he might not be able to stuff everybody, he wants to give a complete meal of both food and drink to as many cows as possible.
Farmer John has cooked F (1 ≤ F ≤ 100) types of foods and prepared D (1 ≤ D ≤ 100) types of drinks. Each of his N (1 ≤ N ≤ 100) cows has decided whether she is willing to eat a particular food or drink a particular drink. Farmer John must assign a food type and a drink type to each cow to maximize the number of cows who get both.
Each dish or drink can only be consumed by one cow (i.e., once food type 2 is assigned to a cow, no other cow can be assigned food type 2).
Input
Lines 2..N+1: Each line i starts with a two integers Fi and Di, the number of dishes that cow i likes and the number of drinks that cow i likes. The next Fi integers denote the dishes that cow i will eat, and the Di integers following that denote the drinks that cow i will drink.
Output
Sample Input
4 3 3 2 2 1 2 3 1 2 2 2 3 1 2 2 2 1 3 1 2 2 1 1 3 3
Sample Output
3
Hint
Cow 1: no meal
Cow 2: Food #2, Drink #2
Cow 3: Food #1, Drink #1
Cow 4: Food #3, Drink #3
The pigeon-hole principle tells us we can do no better since there are only three kinds of food or drink. Other test data sets are more challenging, of course.
建图是关键。。。
把牛拆成两个点才能起到限流的作用
源点连食物,食物连牛in,牛in连牛out,牛out连饮料,饮料连汇点
不能牛连食物,食物连饮料,不然食物可能会被使用一次以上,在这里想了半天,太菜了= =
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include<iostream> 2 #include<cstring> 3 #include<string> 4 #include<cmath> 5 #include<cstdio> 6 #include<algorithm> 7 #include<queue> 8 #include<vector> 9 #include<set> 10 #define maxn 100005 11 #define MAXN 100005 12 #define mem(a,b) memset(a,b,sizeof(a)) 13 const int N=200005; 14 const int M=200005; 15 const int INF=0x3f3f3f3f; 16 using namespace std; 17 int n; 18 struct Edge{ 19 int v,next; 20 int cap,flow; 21 }edge[MAXN*20];//注意这里要开的够大。。不然WA在这里真的想骂人。。问题是还不报RE。。 22 int cur[MAXN],pre[MAXN],gap[MAXN],path[MAXN],dep[MAXN]; 23 int cnt=0;//实际存储总边数 24 void isap_init() 25 { 26 cnt=0; 27 memset(pre,-1,sizeof(pre)); 28 } 29 void isap_add(int u,int v,int w)//加边 30 { 31 edge[cnt].v=v; 32 edge[cnt].cap=w; 33 edge[cnt].flow=0; 34 edge[cnt].next=pre[u]; 35 pre[u]=cnt++; 36 } 37 void add(int u,int v,int w){ 38 isap_add(u,v,w); 39 isap_add(v,u,0); 40 } 41 bool bfs(int s,int t)//其实这个bfs可以融合到下面的迭代里,但是好像是时间要长 42 { 43 memset(dep,-1,sizeof(dep)); 44 memset(gap,0,sizeof(gap)); 45 gap[0]=1; 46 dep[t]=0; 47 queue<int>q; 48 while(!q.empty()) 49 q.pop(); 50 q.push(t);//从汇点开始反向建层次图 51 while(!q.empty()) 52 { 53 int u=q.front(); 54 q.pop(); 55 for(int i=pre[u];i!=-1;i=edge[i].next) 56 { 57 int v=edge[i].v; 58 if(dep[v]==-1&&edge[i^1].cap>edge[i^1].flow)//注意是从汇点反向bfs,但应该判断正向弧的余量 59 { 60 dep[v]=dep[u]+1; 61 gap[dep[v]]++; 62 q.push(v); 63 //if(v==sp)//感觉这两句优化加了一般没错,但是有的题可能会错,所以还是注释出来,到时候视情况而定 64 //break; 65 } 66 } 67 } 68 return dep[s]!=-1; 69 } 70 int isap(int s,int t) 71 { 72 if(!bfs(s,t)) 73 return 0; 74 memcpy(cur,pre,sizeof(pre)); 75 //for(int i=1;i<=n;i++) 76 //cout<<"cur "<<cur[i]<<endl; 77 int u=s; 78 path[u]=-1; 79 int ans=0; 80 // cout<<dep[s]<<" "<<n<<endl; 81 while(dep[s]<n)//迭代寻找增广路 82 { 83 if(u==t) 84 { 85 int f=INF; 86 for(int i=path[u];i!=-1;i=path[edge[i^1].v])//修改找到的增广路 87 f=min(f,edge[i].cap-edge[i].flow); 88 for(int i=path[u];i!=-1;i=path[edge[i^1].v]) 89 { 90 edge[i].flow+=f; 91 edge[i^1].flow-=f; 92 } 93 ans+=f; 94 u=s; 95 continue; 96 } 97 bool flag=false; 98 int v; 99 for(int i=cur[u];i!=-1;i=edge[i].next) 100 { 101 v=edge[i].v; 102 if(dep[v]+1==dep[u]&&edge[i].cap-edge[i].flow) 103 { 104 cur[u]=path[v]=i;//当前弧优化 105 flag=true; 106 break; 107 } 108 } 109 if(flag) 110 { 111 u=v; 112 continue; 113 } 114 int x=n; 115 if(!(--gap[dep[u]]))return ans;//gap优化 116 for(int i=pre[u];i!=-1;i=edge[i].next) 117 { 118 if(edge[i].cap-edge[i].flow&&dep[edge[i].v]<x) 119 { 120 x=dep[edge[i].v]; 121 cur[u]=i;//常数优化 122 } 123 } 124 dep[u]=x+1; 125 gap[dep[u]]++; 126 if(u!=s)//当前点没有增广路则后退一个点 127 u=edge[path[u]^1].v; 128 } 129 return ans; 130 } 131 132 133 int main(){ 134 int m,s,t; 135 int F,D; 136 scanf("%d %d %d",&n,&F,&D); 137 int a,b,c; 138 isap_init(); 139 int x1[105],x2[105]; 140 int f,d; 141 for(int i=1;i<=n;i++){ 142 scanf("%d %d",&f,&d); 143 for(int j=1;j<=f;j++){ 144 scanf("%d",&x1[j]); 145 add(x1[j],F+i,1); 146 } 147 for(int j=1;j<=d;j++){ 148 scanf("%d",&x2[j]); 149 for(int k=1;k<=f;k++){ 150 add(n+F+i,x2[j]+n+n+F,1); 151 } 152 } 153 } 154 for(int i=1;i<=F;i++){ 155 add(0,i,1); 156 } 157 for(int i=1;i<=n;i++){ 158 add(F+i,n+F+i,1); 159 } 160 for(int i=1;i<=D;i++){ 161 add(i+n+n+F,F+n+n+D+1,1); 162 } 163 s=0,t=F+n+n+D+1; 164 n=F+n+n+D+2; 165 printf("%d\n",isap(s,t)); 166 } 167 168 169 /* 170 2 2 2 171 2 1 1 2 1 172 1 1 1 2 173 174 2 175 */
posted on 2018-10-19 20:42 Fighting_sh 阅读(392) 评论(0) 编辑 收藏 举报