HDU 3395 Special Fish (KM 或者 最大流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){ //不在相等子图中slack 取最小的
            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初始化为与它关联边中最大的
        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))  //若成功（找到了增广轨），则该点增广完成，进入下一个点的增广
                break;  //若失败（没有找到增广轨），则需要改变一些点的标号，使得图中可行边的数量增加。
                        //方法为：将所有在增广轨中（就是在增广过程中遍历到）的X方点的标号全部减去一个常数d，
                        //所有在增广轨中的Y方点的标号全部加上一个常数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++)  //修改顶标后，要把所有不在交错树中的Y顶点的slack值都减去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\n",KM());
    }
    return 0;
}

 

 

2,最小费用最大流

 

#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\n",mincost());
    }
    return 0;
}