Food(最大流)
Food
http://acm.hdu.edu.cn/showproblem.php?pid=4292
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 8111 Accepted Submission(s): 2686
Problem Description
You, a part-time dining service worker in your college’s dining hall, are now confused with a new problem: serve as many people as possible.
The issue comes up as people in your college are more and more difficult to serve with meal: They eat only some certain kinds of food and drink, and with requirement unsatisfied, go away directly.
You have prepared F (1 <= F <= 200) kinds of food and D (1 <= D <= 200) kinds of drink. Each kind of food or drink has certain amount, that is, how many people could this food or drink serve. Besides, You know there’re N (1 <= N <= 200) people and you too can tell people’s personal preference for food and drink.
Back to your goal: to serve as many people as possible. So you must decide a plan where some people are served while requirements of the rest of them are unmet. You should notice that, when one’s requirement is unmet, he/she would just go away, refusing any service.
The issue comes up as people in your college are more and more difficult to serve with meal: They eat only some certain kinds of food and drink, and with requirement unsatisfied, go away directly.
You have prepared F (1 <= F <= 200) kinds of food and D (1 <= D <= 200) kinds of drink. Each kind of food or drink has certain amount, that is, how many people could this food or drink serve. Besides, You know there’re N (1 <= N <= 200) people and you too can tell people’s personal preference for food and drink.
Back to your goal: to serve as many people as possible. So you must decide a plan where some people are served while requirements of the rest of them are unmet. You should notice that, when one’s requirement is unmet, he/she would just go away, refusing any service.
Input
There are several test cases.
For each test case, the first line contains three numbers: N,F,D, denoting the number of people, food, and drink.
The second line contains F integers, the ith number of which denotes amount of representative food.
The third line contains D integers, the ith number of which denotes amount of representative drink.
Following is N line, each consisting of a string of length F. e jth character in the ith one of these lines denotes whether people i would accept food j. “Y” for yes and “N” for no.
Following is N line, each consisting of a string of length D. e jth character in the ith one of these lines denotes whether people i would accept drink j. “Y” for yes and “N” for no.
Please process until EOF (End Of File).
For each test case, the first line contains three numbers: N,F,D, denoting the number of people, food, and drink.
The second line contains F integers, the ith number of which denotes amount of representative food.
The third line contains D integers, the ith number of which denotes amount of representative drink.
Following is N line, each consisting of a string of length F. e jth character in the ith one of these lines denotes whether people i would accept food j. “Y” for yes and “N” for no.
Following is N line, each consisting of a string of length D. e jth character in the ith one of these lines denotes whether people i would accept drink j. “Y” for yes and “N” for no.
Please process until EOF (End Of File).
Output
For each test case, please print a single line with one integer, the maximum number of people to be satisfied.
Sample Input
4 3 3
1 1 1
1 1 1
YYN
NYY
YNY
YNY
YNY
YYN
YYN
NNY
Sample Output
3
Source
感觉和这题思路差不多:https://www.cnblogs.com/Fighting-sh/p/9818674.html
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 200005 11 #define MAXN 200005 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 while(dep[s]<n)//迭代寻找增广路,n为节点数 81 { 82 if(u==t) 83 { 84 int f=INF; 85 for(int i=path[u];i!=-1;i=path[edge[i^1].v])//修改找到的增广路 86 f=min(f,edge[i].cap-edge[i].flow); 87 for(int i=path[u];i!=-1;i=path[edge[i^1].v]) 88 { 89 edge[i].flow+=f; 90 edge[i^1].flow-=f; 91 } 92 ans+=f; 93 u=s; 94 continue; 95 } 96 bool flag=false; 97 int v; 98 for(int i=cur[u];i!=-1;i=edge[i].next) 99 { 100 v=edge[i].v; 101 if(dep[v]+1==dep[u]&&edge[i].cap-edge[i].flow) 102 { 103 cur[u]=path[v]=i;//当前弧优化 104 flag=true; 105 break; 106 } 107 } 108 if(flag) 109 { 110 u=v; 111 continue; 112 } 113 int x=n; 114 if(!(--gap[dep[u]]))return ans;//gap优化 115 for(int i=pre[u];i!=-1;i=edge[i].next) 116 { 117 if(edge[i].cap-edge[i].flow&&dep[edge[i].v]<x) 118 { 119 x=dep[edge[i].v]; 120 cur[u]=i;//常数优化 121 } 122 } 123 dep[u]=x+1; 124 gap[dep[u]]++; 125 if(u!=s)//当前点没有增广路则后退一个点 126 u=edge[path[u]^1].v; 127 } 128 return ans; 129 } 130 131 int a[maxn]; 132 int F[maxn],D[maxn]; 133 string mp[505]; 134 135 int main(){ 136 std::ios::sync_with_stdio(false); 137 int m,s,t; 138 int f,d; 139 while(cin>>n>>f>>d){ 140 for(int i=1;i<=f;i++) cin>>F[i]; 141 for(int i=1;i<=d;i++) cin>>D[i]; 142 for(int i=1;i<=2*n;i++) cin>>mp[i]; 143 isap_init(); 144 s=0,t=n+n+f+d+1; 145 for(int i=1;i<=n;i++){ 146 for(int j=0;j<f;j++){ 147 if(mp[i][j]=='Y'){ 148 add(j+1,f+i,1); 149 } 150 } 151 } 152 for(int i=1;i<=n;i++){ 153 for(int j=0;j<d;j++){ 154 if(mp[i+n][j]=='Y'){ 155 add(f+n+i,f+n+n+j+1,1); 156 } 157 } 158 } 159 for(int i=1;i<=n;i++) add(f+i,f+n+i,1); 160 for(int i=1;i<=f;i++) add(s,i,F[i]); 161 for(int i=1;i<=d;i++) add(f+n+n+i,t,D[i]); 162 n=n+n+f+d+2; 163 cout<<isap(s,t)<<endl; 164 } 165 }
posted on 2018-11-13 12:45 Fighting_sh 阅读(298) 评论(0) 编辑 收藏 举报