Food
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1491 Accepted Submission(s): 534
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
Recommend
liuyiding
题意：有F种食物 D种饮料 它们都有一定的数量 有N个人 每个人都有自己喜欢吃的食物和饮料 （每个人至少要一种食物和饮料） 只有能满足他的要求时他才会接服务 求最大能满足多少人？
思路：网络流 建一超级源点 汇点 源点与食物相连 边权为其数量，汇点与饮料相连 边权也为其数量 把人分成两个点 之间的边权为1 每个人与之需要的食物和饮料相连 边权为1 （或者INF）
当然，这题和POJ 3281 Dining很类似：http://poj.org/problem?id=3281
#include<iostream> #include<cstdio> #include<cstring> using namespace std; const int VM=1010; const int EM=200010; const int INF=0x3f3f3f3f; struct Edge{ int to,nxt; int cap; }edge[EM<<1]; int N,F,D,cnt,head[VM],map[110][110]; int dep[VM],gap[VM],cur[VM],aug[VM],pre[VM]; void addedge(int cu,int cv,int cw){ edge[cnt].to=cv; edge[cnt].cap=cw; edge[cnt].nxt=head[cu]; head[cu]=cnt++; edge[cnt].to=cu; edge[cnt].cap=0; edge[cnt].nxt=head[cv]; head[cv]=cnt++; } int src,des; int SAP(int n){ int max_flow=0,u=src,v; int id,mindep; aug[src]=INF; pre[src]=-1; memset(dep,0,sizeof(dep)); memset(gap,0,sizeof(gap)); gap[0]=n; for(int i=0;i<=n;i++) cur[i]=head[i]; // 初始化当前弧为第一条弧 while(dep[src]<n){ int flag=0; if(u==des){ max_flow+=aug[des]; for(v=pre[des];v!=-1;v=pre[v]){ // 路径回溯更新残留网络 id=cur[v]; edge[id].cap-=aug[des]; edge[id^1].cap+=aug[des]; aug[v]-=aug[des]; // 修改可增广量，以后会用到 if(edge[id].cap==0) // 不回退到源点，仅回退到容量为0的弧的弧尾 u=v; } } for(int i=cur[u];i!=-1;i=edge[i].nxt){ v=edge[i].to; // 从当前弧开始查找允许弧 if(edge[i].cap>0 && dep[u]==dep[v]+1){ // 找到允许弧 flag=1; pre[v]=u; cur[u]=i; aug[v]=min(aug[u],edge[i].cap); u=v; break; } } if(!flag){ if(--gap[dep[u]]==0) /* gap优化，层次树出现断层则结束算法 */ break; mindep=n; cur[u]=head[u]; for(int i=head[u];i!=-1;i=edge[i].nxt){ v=edge[i].to; if(edge[i].cap>0 && dep[v]<mindep){ mindep=dep[v]; cur[u]=i; // 修改标号的同时修改当前弧 } } dep[u]=mindep+1; gap[dep[u]]++; if(u!=src) // 回溯继续寻找允许弧 u=pre[u]; } } return max_flow; } int main(){ //freopen("input.txt","r",stdin); while(~scanf("%d%d%d",&N,&F,&D)){ cnt=0; memset(head,-1,sizeof(head)); int f,d; src=0, des=F+2*N+D+1; for(int i=1;i<=N;i++){ scanf("%d%d",&f,&d); int x; for(int j=1;j<=f;j++){ scanf("%d",&x); addedge(x,F+i,1); //食物和牛1（将牛分成两点）相连 } for(int j=1;j<=d;j++){ scanf("%d",&x); addedge(F+N+i,F+2*N+x,1); //牛2和饮料相连 } addedge(F+i,F+N+i,1); //牛1和牛2相连，保证没头牛只吃一种食物和饮料 } for(int i=1;i<=F;i++) addedge(src,i,1); //超级源点与食物相连 for(int i=1;i<=D;i++) addedge(F+2*N+i,des,1); //饮料与超级汇点相连 printf("%d\n",SAP(des+1)); } return 0; }
本题代码：
#include<iostream> #include<cstdio> #include<cstring> using namespace std; const int VM=1010; const int EM=200010; const int INF=0x3f3f3f3f; struct Edge{ int to,nxt; int cap; }edge[EM<<1]; int N,F,D,cnt,head[VM],map[110][110]; int dep[VM],gap[VM],cur[VM],aug[VM],pre[VM]; void addedge(int cu,int cv,int cw){ edge[cnt].to=cv; edge[cnt].cap=cw; edge[cnt].nxt=head[cu]; head[cu]=cnt++; edge[cnt].to=cu; edge[cnt].cap=0; edge[cnt].nxt=head[cv]; head[cv]=cnt++; } int src,des; int SAP(int n){ int max_flow=0,u=src,v; int id,mindep; aug[src]=INF; pre[src]=-1; memset(dep,0,sizeof(dep)); memset(gap,0,sizeof(gap)); gap[0]=n; for(int i=0;i<=n;i++) cur[i]=head[i]; // 初始化当前弧为第一条弧 while(dep[src]<n){ int flag=0; if(u==des){ max_flow+=aug[des]; for(v=pre[des];v!=-1;v=pre[v]){ // 路径回溯更新残留网络 id=cur[v]; edge[id].cap-=aug[des]; edge[id^1].cap+=aug[des]; aug[v]-=aug[des]; // 修改可增广量，以后会用到 if(edge[id].cap==0) // 不回退到源点，仅回退到容量为0的弧的弧尾 u=v; } } for(int i=cur[u];i!=-1;i=edge[i].nxt){ v=edge[i].to; // 从当前弧开始查找允许弧 if(edge[i].cap>0 && dep[u]==dep[v]+1){ // 找到允许弧 flag=1; pre[v]=u; cur[u]=i; aug[v]=min(aug[u],edge[i].cap); u=v; break; } } if(!flag){ if(--gap[dep[u]]==0) /* gap优化，层次树出现断层则结束算法 */ break; mindep=n; cur[u]=head[u]; for(int i=head[u];i!=-1;i=edge[i].nxt){ v=edge[i].to; if(edge[i].cap>0 && dep[v]<mindep){ mindep=dep[v]; cur[u]=i; // 修改标号的同时修改当前弧 } } dep[u]=mindep+1; gap[dep[u]]++; if(u!=src) // 回溯继续寻找允许弧 u=pre[u]; } } return max_flow; } int main(){ //freopen("input.txt","r",stdin); char str[220]; while(~scanf("%d%d%d",&N,&F,&D)){ cnt=0; memset(head,-1,sizeof(head)); int f,d; src=0, des=F+2*N+D+1; for(int i=1;i<=F;i++){ scanf("%d",&f); addedge(src,i,f); } for(int i=F+2*N+1;i<=F+2*N+D;i++){ scanf("%d",&d); addedge(i,des,d); } for(int i=1;i<=N;i++){ scanf("%s",str); for(int j=0;j<F;j++) if(str[j]=='Y') addedge(j+1,F+i,1); //这里权值为INF亦可 } for(int i=1;i<=N;i++){ scanf("%s",str); for(int j=0;j<D;j++) if(str[j]=='Y') addedge(F+N+i,F+2*N+j+1,1); //这里权值为INF亦可 } for(int i=F+1;i<=F+N;i++) addedge(i,i+N,1); //将人拆分成两点，边权为1,为了控制最多的人得到一个食物以及一瓶饮料 printf("%d\n",SAP(des+1)); } return 0; }
#include<iostream> #include<cstdio> #include<cstring> #include<queue> using namespace std; const int VM=1010; const int EM=200010; const int INF=0x3f3f3f3f; struct Edge{ int to,nxt; int cap; }edge[EM<<1]; int N,F,D,cnt,head[VM],src,des; int dep[VM]; void addedge(int cu,int cv,int cw){ edge[cnt].to=cv; edge[cnt].cap=cw; edge[cnt].nxt=head[cu]; head[cu]=cnt++; edge[cnt].to=cu; edge[cnt].cap=0; edge[cnt].nxt=head[cv]; head[cv]=cnt++; } int BFS(){ queue<int> q; while(!q.empty()) q.pop(); memset(dep,-1,sizeof(dep)); dep[src]=0; q.push(src); while(!q.empty()){ int u=q.front(); q.pop(); for(int i=head[u];i!=-1;i=edge[i].nxt){ int v=edge[i].to; if(edge[i].cap>0 && dep[v]==-1){ dep[v]=dep[u]+1; q.push(v); } } } return dep[des]!=-1; } int DFS(int u,int minx){ if(u==des) return minx; int tmp; for(int i=head[u];i!=-1;i=edge[i].nxt){ int v=edge[i].to; if(edge[i].cap>0 && dep[v]==dep[u]+1 && (tmp=DFS(v,min(minx,edge[i].cap)))){ edge[i].cap-=tmp; edge[i^1].cap+=tmp; return tmp; } } dep[u]=-1; return 0; } int Dinic(){ int ans=0,tmp; while(BFS()){ while(1){ tmp=DFS(src,INF); if(tmp==0) break; ans+=tmp; } } return ans; } int main(){ //freopen("input.txt","r",stdin); char str[220]; while(~scanf("%d%d%d",&N,&F,&D)){ cnt=0; memset(head,-1,sizeof(head)); int f,d; src=0, des=F+2*N+D+1; for(int i=1;i<=F;i++){ scanf("%d",&f); addedge(src,i,f); } for(int i=F+2*N+1;i<=F+2*N+D;i++){ scanf("%d",&d); addedge(i,des,d); } for(int i=1;i<=N;i++){ scanf("%s",str); for(int j=0;j<F;j++) if(str[j]=='Y') addedge(j+1,F+i,1); } for(int i=1;i<=N;i++){ scanf("%s",str); for(int j=0;j<D;j++) if(str[j]=='Y') addedge(F+N+i,F+2*N+j+1,1); } for(int i=F+1;i<=F+N;i++) addedge(i,i+N,1); printf("%d\n",Dinic()); } return 0; }