A题：http://codeforces.com/contest/497/problem/A
You are given an n × m rectangular table consisting of lower case English letters. In one operation you can completely remove one column from the table. The remaining parts are combined forming a new table. For example, after removing the second column from the table
abcd edfg hijk
we obtain the table:
acd efg hjk
A table is called good if its rows are ordered from top to bottom lexicographically, i.e. each row is lexicographically no larger than the following one. Determine the minimum number of operations of removing a column needed to make a given table good.
The first line contains two integers — n and m (1 ≤ n, m ≤ 100).
Next n lines contain m small English letters each — the characters of the table.
Print a single number — the minimum number of columns that you need to remove in order to make the table good.
1 10 codeforces
0
4 4 case care test code
2
5 4 code forc esco defo rces
4
In the first sample the table is already good.
In the second sample you may remove the first and third column.
In the third sample you have to remove all the columns (note that the table where all rows are empty is considered good by definition).
Let strings s and t have equal length. Then, s is lexicographically larger than t if they are not equal and the character following the largest common prefix of s and t (the prefix may be empty) in s is alphabetically larger than the corresponding character of t.
题意：删除最少列使得最后下一行字符串大于上一行字符串.
题解:水题，直接模拟即可。。。。。。。。或者也可以打个标记,满足了后面该行后面就不用管了。。
/**
* @author neko01
*/
//#pragma comment(linker, "/STACK:102400000,102400000")
#include <cstdio>
#include <cstring>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <queue>
#include <vector>
#include <cmath>
#include <set>
#include <map>
using namespace std;
typedef long long LL;
#define min3(a,b,c) min(a,min(b,c))
#define max3(a,b,c) max(a,max(b,c))
#define pb push_back
#define mp(a,b) make_pair(a,b)
#define clr(a) memset(a,0,sizeof a)
#define clr1(a) memset(a,-1,sizeof a)
#define dbg(a) printf("%d\n",a)
typedef pair<int,int> pp;
const double eps=1e-9;
const double pi=acos(-1.0);
const int INF=0x3f3f3f3f;
char s[105][105];
bool vis[105];
int main()
{
int n,m;
scanf("%d%d",&n,&m);
for(int i=0;i<n;i++)
scanf("%s",s[i]);
int ans=0;
for(int i=0;i<m;i++)
{
int flag=0;
for(int j=0;j<n-1;j++)
if(!vis[j]&&s[j][i]>s[j+1][i])
{
flag=1;
break;
}
if(flag) ans++;
else
{
for(int j=0;j<n-1;j++)
{
if(s[j][i]<s[j+1][i])
vis[j]=true;
}
}
}
printf("%d\n",ans);
return 0;
}/**
* @author neko01
*/
//#pragma comment(linker, "/STACK:102400000,102400000")
#include <cstdio>
#include <cstring>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <queue>
#include <vector>
#include <cmath>
#include <set>
#include <map>
using namespace std;
typedef long long LL;
#define min3(a,b,c) min(a,min(b,c))
#define max3(a,b,c) max(a,max(b,c))
#define pb push_back
#define mp(a,b) make_pair(a,b)
#define clr(a) memset(a,0,sizeof a)
#define clr1(a) memset(a,-1,sizeof a)
#define dbg(a) printf("%d\n",a)
typedef pair<int,int> pp;
const double eps=1e-9;
const double pi=acos(-1.0);
const int INF=0x3f3f3f3f;
string he[105];
char s[105][105];
int main()
{
int n,m;
scanf("%d%d",&n,&m);
for(int i=0;i<n;i++)
scanf("%s",s[i]);
int ans=0;
for(int i=0;i<m;i++)
{
int flag=0;
for(int j=0;j<n-1;j++)
if(he[j]+s[j][i]>he[j+1]+s[j+1][i])
{
flag=1;
break;
}
if(flag) ans++;
else
{
for(int j=0;j<n;j++)
{
he[j]=he[j]+s[j][i];
}
}
}
printf("%d\n",ans);
return 0;
}B题： http://codeforces.com/contest/497/problem/B
Petya and Gena love playing table tennis. A single match is played according to the following rules: a match consists of multiple sets, each set consists of multiple serves. Each serve is won by one of the players, this player scores one point. As soon as one of the players scores t points, he wins the set; then the next set starts and scores of both players are being set to 0. As soon as one of the players wins the total of s sets, he wins the match and the match is over. Here s and t are some positive integer numbers.
To spice it up, Petya and Gena choose new numbers s and t before every match. Besides, for the sake of history they keep a record of each match: that is, for each serve they write down the winner. Serve winners are recorded in the chronological order. In a record the set is over as soon as one of the players scores t points and the match is over as soon as one of the players wins s sets.
Petya and Gena have found a record of an old match. Unfortunately, the sequence of serves in the record isn't divided into sets and numbers s and t for the given match are also lost. The players now wonder what values of s and t might be. Can you determine all the possible options?
The first line contains a single integer n — the length of the sequence of games (1 ≤ n ≤ 105).
The second line contains n space-separated integers ai. If ai = 1, then the i-th serve was won by Petya, if ai = 2, then the i-th serve was won by Gena.
It is not guaranteed that at least one option for numbers s and t corresponds to the given record.
In the first line print a single number k — the number of options for numbers s and t.
In each of the following k lines print two integers si and ti — the option for numbers s and t. Print the options in the order of increasing si, and for equal si — in the order of increasing ti.
5 1 2 1 2 1
2 1 3 3 1
4 1 1 1 1
3 1 4 2 2 4 1
4 1 2 1 2
0
8 2 1 2 1 1 1 1 1
3 1 6 2 3 6 1
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