隐藏页面特效

[2011WorldFinal]Chips Challenge[流量平衡]

1|0Chips Challenge


Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 96    Accepted Submission(s): 33


Problem Description
A prominent microprocessor company has enlisted your help to lay out some interchangeable components (widgets) on some of their computer chips. Each chip’s design is an N×N square of slots. One slot can hold a single component, and you are to try to fit in as many widgets as possible.

Modern processor designs are complex, of course. You unfortunately have several restrictions:
    • Some of the slots are disabled.

    • Some of the slots are already occupied by other components and cannot be used for widgets.

    • There are sibling memory buses connected to the horizontal and vertical edges of the chip and their bandwidth loads need to match. As such, there must be exactly as many components in the first row as in the first column, exactly as many in the second row as in the second column, and so on. Component counts include both the components already specified on the chip and the added widgets.

    • Similarly, the power supply is connected at the end of each row and column. To avoid hot spots, any given row or column must have no more than A=B of the total components on the chip for a given A and B.



A specification for a chip is N lines of N characters, where ‘.’ indicates an open slot, ‘/’ indicates a disabled slot, and ‘C’ indicates a slot already occupied by a component. For example:

CC/..
././/
..C.C
/.C..
/./C/

If no more than 3/10 of the components may be in any one row or column, the maximum number of widgets that can be added to this 5 × 5 chip is 7. A possible arrangement is below, where ‘W’ indicates a widget added in an open slot.

CC/W.
W/W//
W.C.C
/.CWW
/W/C/

 

 

Input
The input consists of several test cases. Each case starts with a line containing three integers: The size of the chip N (1 <= N <= 40), and A and B (1 <= B <= 1000, 0 <= A <= B) as described above. Each of the following N lines contains N characters describing the slots, one of ‘.’, ‘/’ or ‘C’, as described above.
The last test case is followed by a line containing three zeros.
 

 

Output
For each test case, display a single line beginning with the case number. If there is a solution, display the maximum number of widgets that can be added to the chip. Display “impossible” if there is no solution.
Follow the format of the sample output.
 

 

Sample Input
2 1 1 /. // 2 50 100 /. C/ 2 100 100 ./ C. 5 3 10 CC/.. ./.// ..C.C /.C.. /./C/ 5 2 10 CC/.. ./.// ..C.C /.C.. /./C/ 0 0 0
 

 

Sample Output
Case 1: 0 Case 2: 1 Case 3: impossible Case 4: 7 Case 5: impossible
 

 

Source
2011WorldFinal
 

 

Recommend

 

 

//EK 1825ms
#include<cstdio>
#include<cstring>
#include<iostream>
#define m(s) memset(s,0,sizeof s);
#define EF if(ch==EOF) return x;
using namespace std;
const int N=100;
const int M=N*N<<2;
struct edge{int v,next,cap,cost;}e[M];int tot=1,head[N];
int n,cas,A,B,ans,S,T,rd[N],cd[N],dis[N],pre[N],q[N+M];bool vis[N];
int sum,hav,maxflow,maxcost;char s[N][N];
inline int read(){
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;EF;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
void add(int x,int y,int z,int cost=0){
    e[++tot].v=y;e[tot].cap=z;e[tot].cost=cost;e[tot].next=head[x];head[x]=tot;
    e[++tot].v=x;e[tot].cap=0;e[tot].cost=-cost;e[tot].next=head[y];head[y]=tot;
}
bool spfa(){
    memset(vis,0,sizeof vis);
    memset(dis,0x3f,sizeof dis);
    unsigned short h=0,t=1;q[t]=S;dis[S]=0;vis[S]=1;
    while(h!=t){
        int x=q[++h];vis[x]=0;
        for(int i=head[x];i;i=e[i].next){
            if(e[i].cap&&dis[e[i].v]>dis[x]+e[i].cost){
                dis[e[i].v]=dis[x]+e[i].cost;
                pre[e[i].v]=i;
                if(!vis[e[i].v]){
                    vis[e[i].v]=1;
                    q[++t]=e[i].v;
                }                
            }
        }
    }
    return dis[T]<0x3f3f3f3f;
}
void augment(){
    int flow=0x3f3f3f3f;         
    for(int i=T;i!=S;i=e[pre[i]^1].v) flow=min(flow,e[pre[i]].cap);
    for(int i=T;i!=S;i=e[pre[i]^1].v){
        e[pre[i]].cap-=flow;
        e[pre[i]^1].cap+=flow;
    }
    maxflow+=flow;
    maxcost+=dis[T]*flow;
}
int main(){
    while(1){
        n=read();A=read();B=read();
        if(!n) return 0;
        printf("Case %d: ",++cas);
        S=0,T=n<<1|1;
        sum=0;hav=0;ans=-1;m(rd);m(cd);
        for(int i=1;i<=n;i++){
            scanf("%s",s[i]+1);
            for(int j=1;j<=n;j++){
                if(s[i][j]=='C'||s[i][j]=='.'){
                    sum++;cd[i]++,rd[j]++;
                    if(s[i][j]=='C') hav++;
                }
            }
        }
        for(int maxt=0;maxt<=n;maxt++){
            tot=1;m(head);
            for(int i=1;i<=n;i++){
                add(S,i,cd[i]);add(i,i+n,maxt);add(i+n,T,rd[i]);
                for(int j=1;j<=n;j++){
                    if(s[i][j]=='.') add(i,j+n,1,1);
                }
            }
            maxflow=maxcost=0;
            while(spfa()) augment();
            if(maxflow==sum&&(maxflow-maxcost)*A>=maxt*B) ans=max(ans,maxflow-maxcost);
        }
        if(~ans) printf("%d\n",ans-hav);
        else puts("impossible");
    }
    return 0;
}

 

//zkw 280ms
#include<cstdio>
#include<cstring>
#include<iostream>
#define m(s) memset(s,0,sizeof s);
#define EF if(ch==EOF) return x;
using namespace std;
const int N=100;
const int M=N*N<<2;
struct edge{int v,next,cap,cost;}e[M];int tot=1,head[N];
int n,cas,A,B,ans,S,T,rd[N],cd[N],dis[N],pre[N],q[N+M],tim,mark[N];bool vis[N];
int sum,hav,maxflow,maxcost;char s[N][N];
inline int read(){
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;EF;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
inline void add(int x,int y,int z,int cost=0){
    e[++tot].v=y;e[tot].cap=z;e[tot].cost=cost;e[tot].next=head[x];head[x]=tot;
    e[++tot].v=x;e[tot].cap=0;e[tot].cost=-cost;e[tot].next=head[y];head[y]=tot;
}
inline bool spfa(){
    memset(vis,0,sizeof vis);
    memset(dis,0x3f,sizeof dis);
    unsigned short h=0,t=1;q[t]=S;dis[S]=0;vis[S]=1;
    while(h!=t){
        int x=q[++h];vis[x]=0;
        for(int i=head[x];i;i=e[i].next){
            if(e[i].cap&&dis[e[i].v]>dis[x]+e[i].cost){
                dis[e[i].v]=dis[x]+e[i].cost;
                pre[e[i].v]=i;
                if(!vis[e[i].v]){
                    vis[e[i].v]=1;
                    q[++t]=e[i].v;
                }                
            }
        }
    }
    return dis[T]<0x3f3f3f3f;
}
int dfs(int x,int f){
    if(x==T) return f;
    int used=0,t;
    mark[x]=tim;
    for(int i=head[x];i;i=e[i].next){
        if((mark[e[i].v]^tim)&&e[i].cap&&dis[e[i].v]==dis[x]+e[i].cost){
            t=dfs(e[i].v,min(f,e[i].cap));
            e[i].cap-=t;e[i^1].cap+=t;
            used+=t;f-=t;
            if(!f) return used;
        }
    }
    if(!used) dis[x]=-1;
    return used;
}
inline void zkw(){
    int flow=0;
    while(spfa()){
        while(tim++,flow=dfs(S,0x3f3f3f3f)){
            maxflow+=flow;
            maxcost+=dis[T]*flow;
        }
    }
}
int main(){
    while(1){
        n=read();A=read();B=read();
        if(!n) return 0;
        printf("Case %d: ",++cas);
        S=0,T=n<<1|1;
        sum=0;hav=0;ans=-1;m(rd);m(cd);
        for(int i=1;i<=n;i++){
            scanf("%s",s[i]+1);
            for(int j=1;j<=n;j++){
                if(s[i][j]=='C'||s[i][j]=='.'){
                    sum++;cd[i]++,rd[j]++;
                    if(s[i][j]=='C') hav++;
                }
            }
        }
        for(int maxt=0;maxt<=n;maxt++){
            tot=1;m(head);
            for(int i=1;i<=n;i++){
                add(S,i,cd[i]);add(i,i+n,maxt);add(i+n,T,rd[i]);
                for(int j=1;j<=n;j++){
                    if(s[i][j]=='.') add(i,j+n,1,1);
                }
            }
            maxflow=maxcost=0;
            zkw();
            if(maxflow==sum&&(maxflow-maxcost)*A>=maxt*B) ans=max(ans,maxflow-maxcost);
        }
        if(~ans) printf("%d\n",ans-hav);
        else puts("impossible");
    }
    return 0;
}

 


__EOF__

本文作者：shenben
本文链接：https://www.cnblogs.com/shenben/p/6666270.html
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