HDU 3395 Special Fish(KMまたは最大ストリームSPFA)

Special Fish
Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1018    Accepted Submission(s): 384
Problem Description
There is a kind of special fish in the East Lake where is closed to campus of Wuhan University. It’s hard to say which gender of those fish are, because every fish believes itself as a male, and it may attack one of some other fish who is believed to be female by it.
A fish will spawn after it has been attacked. Each fish can attack one other fish and can only be attacked once. No matter a fish is attacked or not, it can still try to attack another fish which is believed to be female by it.
There is a value we assigned to each fish and the spawns that two fish spawned also have a value which can be calculated by XOR operator through the value of its parents.
We want to know the maximum possibility of the sum of the spawns.
 
 
Input
The input consists of multiply test cases. The first line of each test case contains an integer n (0 < n <= 100), which is the number of the fish. The next line consists of n integers, indicating the value (0 < value <= 100) of each fish. The next n lines, each line contains n integers, represent a 01 matrix. The i-th fish believes the j-th fish is female if and only if the value in row i and column j if 1.
The last test case is followed by a zero, which means the end of the input.
 
 
Output
Output the value for each test in a single line.
 
 
Sample Input
3 1 2 3 011 101 110 0
 
 
Sample Output
6
 
 
Author
momodi@whu
 
 
Source
The 5th Guangting Cup Central China Invitational Programming Contest
 
 
Recommend
notonlysuccess
 
 
 
1,KM:

#include<iostream>
#include<cstdio>
#include<cstring>

using namespace std;

const int N=110;
const int INF=0x3f3f3f3f;

int n;
int linker[N],lx[N],ly[N],slack[N];
int visx[N],visy[N],w[N][N];
int val[N];

int DFS(int x){
    visx[x]=1;
    for(int y=1;y<=n;y++){
        if(visy[y])
            continue;
        int tmp=lx[x]+ly[y]-w[x][y];
        if(tmp==0){
            visy[y]=1;
            if(linker[y]==-1 || DFS(linker[y])){
                linker[y]=x;
                return 1;
            }
        }else if(slack[y]>tmp){ //       slack     
            slack[y]=tmp;
        }
    }
    return 0;
}

int KM(){
    int i,j;
    memset(linker,-1,sizeof(linker));
    memset(ly,0,sizeof(ly));
    for(i=1;i<=n;i++)      //lx             
        for(j=1,lx[i]=-INF;j<=n;j++)
            if(w[i][j]>lx[i])
                lx[i]=w[i][j];
    for(int x=1;x<=n;x++){
        for(i=1;i<=n;i++)
            slack[i]=INF;
        while(1){
            memset(visx,0,sizeof(visx));
            memset(visy,0,sizeof(visy));
            if(DFS(x))  //   （      ），       ，         
                break;  //   （       ），           ，            。
                        //   ：        （           ） X             d，
                        //        Y             d
            int d=INF;
            for(i=1;i<=n;i++)
                if(!visy[i] && d>slack[i])
                    d=slack[i];
            for(i=1;i<=n;i++)
                if(visx[i])
                    lx[i]-=d;
            for(i=1;i<=n;i++)  //     ，           Y   slack    d
                if(visy[i])
                    ly[i]+=d;
                else
                    slack[i]-=d;
        }
    }
    int res=0;
    for(i=1;i<=n;i++){
        if(linker[i]==-1 || w[linker[i]][i]==-INF)
            return -1;
        res+=w[linker[i]][i];
    }
    return res;
}

int main(){

    //freopen("input.txt","r",stdin);

    while(~scanf("%d",&n) && n){
        memset(w,0,sizeof(w));
        for(int i=1;i<=n;i++)
            scanf("%d",&val[i]);
        char ch;
        for(int i=1;i<=n;i++)
            for(int j=1;j<=n;j++){
                cin>>ch;
                if(ch=='1')
                    w[i][j]=val[i]^val[j];
            }
        printf("%d
",KM());
    }
    return 0;
}

 
 
2、最小費用最大フロー
 

#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>

using namespace std;

const int VM=1010;
const int EM=10100;
const int INF=0x3f3f3f3f;

struct Edge{
    int u,v,nxt;
    int flow,cost;
}edge[EM<<1];

int n,cnt,head[VM],val[VM];
int pre[VM],dis[VM],vis[VM];

void addedge(int cu,int cv,int cw,int cc){
    edge[cnt].u=cu;  edge[cnt].v=cv;  edge[cnt].flow=cw;
    edge[cnt].cost=cc;  edge[cnt].nxt=head[cu];  head[cu]=cnt++;

    edge[cnt].u=cv;  edge[cnt].v=cu;  edge[cnt].flow=0;
    edge[cnt].cost=-cc;  edge[cnt].nxt=head[cv];  head[cv]=cnt++;
}

int src,des;

int SPFA(){
    queue<int> q;
    while(!q.empty())
        q.pop();
    for(int i=0;i<=des;i++){
        dis[i]=INF;
        vis[i]=0;
        pre[i]=-1;
    }
    dis[src]=0;
    vis[src]=1;
    q.push(src);
    while(!q.empty()){
        int u=q.front();
        q.pop();
        vis[u]=0;
        for(int i=head[u];i!=-1;i=edge[i].nxt){
            int v=edge[i].v;
            if(edge[i].flow>0 && dis[v]>dis[u]+edge[i].cost){
                dis[v]=dis[u]+edge[i].cost;
                pre[v]=i;
                if(!vis[v]){
                    vis[v]=1;
                    q.push(v);
                }
            }
        }
    }
    return pre[des]!=-1;
}

int mincost(){
    int ans=0;
    while(SPFA()){
        int u=des,minf=INF;
        while(u!=src){
            minf=min(minf,edge[pre[u]].flow);
            u=edge[pre[u]].u;
        }
        u=des;
        while(u!=src){
            edge[pre[u]].flow-=minf;
            edge[pre[u]^1].flow+=minf;
            u=edge[pre[u]].u;
        }
        ans+=dis[des];
    }
    return -ans;
}

int main(){

    //freopen("input.txt","r",stdin);

    while(~scanf("%d",&n) && n){
        cnt=0;
        memset(head,-1,sizeof(head));
        for(int i=0;i<n;i++)
            scanf("%d",&val[i]);
        int w;
        for(int i=0;i<n;i++){
            addedge(0,i+1,1,0);
            addedge(i+1,n+n+1,1,0);
            addedge(n+i+1,n+n+1,1,0);
            for(int j=0;j<n;j++){
                scanf("%1d",&w);
                if(w)
                    addedge(i+1,n+j+1,1,-(val[i]^val[j]));
            }
        }
        src=0; des=n+n+1;
        printf("%d
",mincost());
    }
    return 0;
}