2010成都站J题 ||hdu 3718 二分图的最佳匹配 =〉 最小费用最大流

http://acm.hdu.edu.cn/showproblem.php?pid=3718

Problem Description

When we were children, we were always asked to do the classification homework. For example, we were given words {Tiger, Panda, Potato, Dog, Tomato, Pea, Apple, Pear, Orange, Mango} and we were required to classify these words into three groups. As you know, the correct classification was {Tiger, Panda, Dog}, {Potato, Tomato, Pea} and {Apple, Pear, Orange, Mango}. We can represent this classification with a mapping sequence{A,A,B,A,B,B,C,C,C,C}, and it means Tiger, Panda, Dog belong to group A, Potato, Tomato, Pea are in the group B, and Apple, Pear, Orange, Mango are in the group C.
But the LABEL of group doesn't make sense and the LABEL is just used to indicate different groups. So the representations {P,P,O,P,O,O,Q,Q,Q,Q} and {E,E,F,E,F,F,W,W,W,W} are equivalent to the original mapping sequence. However, the representations {A,A,A,A,B,B,C,C,C,C} and
{D,D,D,D,D,D,G,G,G,G} are not equivalent.



The pupils in class submit their mapping sequences and the teacher should read and grade the homework. The teacher grades the homework by calculating the maximum similarity between pupils' mapping sequences and the answer sequence. The definition of similarity is as follow. 

Similarity(S, T) = sum(S_i == T_i) / L 
L = Length(S) = Length(T), i = 1, 2,... L,
where sum(S_i == T_i) indicates the total number of equal labels in corresponding positions. The maximum similarity means the maximum similarities between S and all equivalent sequences of T, where S is the answer and fixed. Now given all sequences submitted by pupils and the answer sequence, you should calculate the sequences' maximum similarity.

 

Input

The input contains multiple test cases. The first line is the total number of cases T (T < 15). The following are T blocks. Each block indicates a case. A case begins with three numbers n (0 < n < 10000), k (0 < k < 27), m (0 < m < 30), which are the total number of objects, groups, and students in the class. The next line consists of n labels and each label is in the range [A...Z]. You can assume that the number of different labels in the sequence is exactly k. This sequence represents the answer. The following are m lines, each line contains n labels and each label also is in the range [A...Z]. These m lines represent the m pupils' answer sequences. You can assume that the number of different labels in each sequence doesn't exceed k.

 

Output

For each test case, output m lines, each line is a floating number (Round to 4 digits after the decimal point). You should output the m answers in the order of the sequences appearance.

 

Sample Input

210 3 3A A B A B B C C C CF F E F E E D D D DX X X Y Y Y Y Z Z ZS T R S T R S T R S3 2 2A B AC D CF F E

 

Sample Output

1.00000.70000.50001.00000.6667

题目大意：

               给定一个字符数列，后面m个同样长度的数列，对于每一个数列：每一个字母对应第一个数列中的一个字母，求怎样对应使得该数列有最多字母与第一个数列相应的匹配。例如：ABA    CDC    A对应C,B对应D，就完全对应匹配。再如：ABA      FFE     A对应E   B对应F  就有2/3的对应。

解题思路：

                二分图的最佳匹配。对母串中的任意一个与子串的不同字母一一对应用map[i][j]。我不会KM算法，两边加源点和汇点，转化为费用流来做的。

#include<cstdio>#include<iostream>#include<string.h>using namespace std;const int oo=1e9;const int mm=11111;const int mn=888;int node,src,dest,edge;int ver[mm],flow[mm],cost[mm],next[mm];int head[mn],dis[mn],p[mn],q[mn],vis[mn];void prepare(int _node,int _src,int _dest){    node=_node,src=_src,dest=_dest;    for(int i=0; i<node; ++i)head[i]=-1,vis[i]=0;    edge=0;}void addedge(int u,int v,int f,int c){    ver[edge]=v,flow[edge]=f,cost[edge]=c,next[edge]=head[u],head[u]=edge++;    ver[edge]=u,flow[edge]=0,cost[edge]=-c,next[edge]=head[v],head[v]=edge++;}bool spfa(){    int i,u,v,l,r=0,tmp;    for(i=0; i<node; ++i)dis[i]=oo;    dis[q[r++]=src]=0;    p[src]=p[dest]=-1;    for(l=0; l!=r; (++l>=mn)?l=0:l)        for(i=head[u=q[l]],vis[u]=0; i>=0; i=next[i])            if(flow[i]&&dis[v=ver[i]]>(tmp=dis[u]+cost[i]))            {                dis[v]=tmp;                p[v]=i^1;                if(vis[v])continue;                vis[q[r++]=v]=1;                if(r>=mn)r=0;            }    return p[dest]>-1;}int SpfaFlow(){    int i,ret=0,delta;    //int ret1=0;    while(spfa())    {        for(i=p[dest],delta=oo; i>=0; i=p[ver[i]])            if(flow[i^1]<delta)delta=flow[i^1];        for(i=p[dest]; i>=0; i=p[ver[i]])            flow[i]+=delta,flow[i^1]-=delta;        ret+=delta*dis[dest];        //ret1+=delta;    }    //printf("%d\n",ret1);    return ret;}char a[10005][2],b[10005][2];int map[30][30];int main(){    int T,m,n,k;    scanf("%d",&T);    while(T--)    {        scanf("%d%d%d",&n,&k,&m);        for(int i=0; i<n; i++)            scanf("%s",a[i]);        while(m--)        {            prepare(54,0,53);            memset(map,0,sizeof(map));            for(int i=0; i<n; i++)            {                scanf("%s",b[i]);                map[a[i][0]-'A'+1][b[i][0]-'A'+1]++;            }            /*            for(int i=1; i<=10; i++)            {                for(int j=1; j<=10; j++)                    printf("%d ",map[i][j]);                printf("\n");            }*/            for(int i=1; i<=26; i++)            {                addedge(src,i,1,0);                addedge(i+26,dest,1,0);                for(int j=1; j<=26; j++)                    addedge(i,j+26,1,-map[i][j]);            }            //printf("%d\n",-SpfaFlow());            printf("%.4lf\n",-SpfaFlow()*1.0/n);        }    }    return 0;}