2015
02-21

# Similarity

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(Si == Ti) / L
L = Length(S) = Length(T), i = 1, 2,… L,
where sum(Si == Ti) 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.

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.

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.

2
10 3 3
A A B A B B C C C C
F F E F E E D D D D
X X X Y Y Y Y Z Z Z
S T R S T R S T R S
3 2 2
A B A
C D C
F F E

1.0000
0.7000
0.5000
1.0000
0.6667

 #include <iostream>
#include <cstring>
#define INF 9999999
#include <cstdio>
#include <queue>
#include <vector>
#include<algorithm>
using namespace std;
#define maxn 6100

struct  edge
{
int from,to,cap,flow,cost;
};
struct MCMF
{
int n,m,s,t;
vector<edge>edges;
vector<int>G[maxn];
int inq[maxn];
int d[maxn];
int p[maxn];
int a[maxn];
void init(int n)
{
this->n=n;
for(int i=0; i<n; i++)
G[i].clear();
edges.clear();
}
void addedge(int from,int to,int cap,int cost)
{
edges.push_back((edge){from,to,cap,0,cost});
edges.push_back((edge){to,from,0,0,-cost});
m=edges.size();
G[from].push_back(m-2);
G[to].push_back(m-1);
}

bool bellman(int s,int t,int &flow,int &cost)
{
for(int i=0; i<n; i++)d[i]=INF;
memset(inq,0,sizeof(inq));
d[s]=0;
inq[s]=1;
p[s]=0;
a[s]=INF;

queue<int>Q;
Q.push(s);
while(!Q.empty())
{
int u = Q.front();
Q.pop();
inq[u] = 0;
for(int i = 0; i < G[u].size(); i++)
{
edge& e = edges[G[u][i]];
if(e.cap > e.flow && d[e.to] > d[u] + e.cost)
{
d[e.to] = d[u] + e.cost;
p[e.to] = G[u][i];
a[e.to] = min(a[u], e.cap - e.flow);
if(!inq[e.to])
{
Q.push(e.to);
inq[e.to] = 1;
}
}
}
}
if(d[t] == INF) return false;
flow += a[t];
cost += d[t]*a[t];
int u = t;
while(u != s)
{
edges[p[u]].flow += a[t];
edges[p[u]^1].flow -= a[t];
u = edges[p[u]].from;
}
return true;
}
int Mincost(int s, int t)
{
int flow = 0, cost = 0;
while(bellman(s, t, flow, cost));
return cost;
}
};
int n;
int k,m;
char s2[10200],s1[10200];
int cur[30][30];
int tot[30];
void first_solve()
{
memset(cur,0,sizeof(cur));
memset(tot,0,sizeof(tot));
for(int i=1; i<=n; i++)
{
int k1=s1[i]-'A'+1;
int k2=s2[i]-'A'+1;
tot[k1]++;
cur[k1][k2]++;
}
}
MCMF solve;
int main()
{
int t;
int st=0;
int final=201;
scanf("%d",&t);
while(t--)
{
scanf("%d%d%d",&n,&k,&m);
for(int i=1; i<=n; i++)
{
char s[5];
scanf("%s",s);
s1[i]=s[0];
}
for(int d=1; d<=m; d++)
{
solve.init(final+1);
for(int j=1; j<=n; j++)
{
char s[5];
scanf("%s",s);
s2[j]=s[0];
}
first_solve();
for(int i=1; i<=26; i++)
for(int i=1; i<=26; i++)
{
for(int j=1; j<=26; j++)
{
int cnt=26+j;
if (cur[i][j])
}
}
int ans=n-solve.Mincost(st,final);
printf("%.4lf\n",(double)ans/(double)n);
}
}
return 0;
}


1. 既然斑爷被秽土转生后还叫带土用轮回眼真正的复活他…那为啥不干脆的，秽土转生琳出来直接复活琳，或者是让琳去劝带土，我想；琳出来完全可以让带土洗白的吧….

2. 既然斑爷被秽土转生后还叫带土用轮回眼真正的复活他…那为啥不干脆的，秽土转生琳出来直接复活琳，或者是让琳去劝带土，我想；琳出来完全可以让带土洗白的吧….

3. 既然斑爷被秽土转生后还叫带土用轮回眼真正的复活他…那为啥不干脆的，秽土转生琳出来直接复活琳，或者是让琳去劝带土，我想；琳出来完全可以让带土洗白的吧….

4. 既然斑爷被秽土转生后还叫带土用轮回眼真正的复活他…那为啥不干脆的，秽土转生琳出来直接复活琳，或者是让琳去劝带土，我想；琳出来完全可以让带土洗白的吧….

5. 既然斑爷被秽土转生后还叫带土用轮回眼真正的复活他…那为啥不干脆的，秽土转生琳出来直接复活琳，或者是让琳去劝带土，我想；琳出来完全可以让带土洗白的吧….

6. 既然斑爷被秽土转生后还叫带土用轮回眼真正的复活他…那为啥不干脆的，秽土转生琳出来直接复活琳，或者是让琳去劝带土，我想；琳出来完全可以让带土洗白的吧….

7. 既然斑爷被秽土转生后还叫带土用轮回眼真正的复活他…那为啥不干脆的，秽土转生琳出来直接复活琳，或者是让琳去劝带土，我想；琳出来完全可以让带土洗白的吧….

8. 题本身没错，但是HDOJ放题目的时候，前面有个题目解释了什么是XXX定律。
这里直接放了这个题目，肯定没几个人明白是干啥

9. #include <cstdio>
#include <algorithm>

struct LWPair{
int l,w;
};

int main() {
//freopen("input.txt","r",stdin);
const int MAXSIZE=5000, MAXVAL=10000;
LWPair sticks[MAXSIZE];
int store[MAXSIZE];
int ncase, nstick, length,width, tmp, time, i,j;
if(scanf("%d",&ncase)!=1) return -1;
while(ncase– && scanf("%d",&nstick)==1) {
for(i=0;i<nstick;++i) scanf("%d%d",&sticks .l,&sticks .w);
std::sort(sticks,sticks+nstick,[](const LWPair &lhs, const LWPair &rhs) { return lhs.l>rhs.l || lhs.l==rhs.l && lhs.w>rhs.w; });
for(time=-1,i=0;i<nstick;++i) {
tmp=sticks .w;
for(j=time;j>=0 && store >=tmp;–j) ; // search from right to left
if(j==time) { store[++time]=tmp; }
else { store[j+1]=tmp; }
}
printf("%dn",time+1);
}
return 0;
}