8.10 Jizhong training Day10

T1 flood

Description

One day, a sketch artist in the forest, suddenly broke out in torrents, he needs to return home as soon as possible, there is a safe
-wide.
Map forest by R rows C composed of columns, a blank area of the point. "" Indicates, the area floods with "*" indicates that, while the
rock with an "X", another artist's residence with "D" represents the artist with "S "means.
The following points should be noted:
1, every minute of the painter can move one space to the four directions (up, down, left, right)
2, every minute of the flood can spread to the four directions of the adjacent grid (blank area)
3, floods and painters can not rock the region
4, the painter can not flood the area (at the same time does not work, that the artist can not be moved to a grid, the grid while the painter was reached to spread the flood, which is not allowed)
5 floods spread less than residence painters.
Give you a map of the forest, programming output at least how long it takes to get home from the starting position.

Input

Input of the first line contains two integers R and C (R, C <= 50 ).
The next R lines contains C characters ( ".", "*" , "X", "D" or "S"). Map to ensure that only a "D" and a "S".

Output

Output painter fastest time required to reach home safe, if the artist can not return home safely output "KAKTUS".

Sample Input

Input. 1: 
. 3. 3 
D. * 
... 
. .S 

Input 2: 
. 3. 3 
D. * 
... 
..S 

input. 3: 
. 3. 6 
D ... *. 
.XX. 
.... S.

Sample Output

Output 1: 
3 

Output 2: 
Kaktus 

output 3: 
6

Examination room ideas / positive solution

Very bare BFS, attendance problems.

Code

#include<cstdio>
#include<algorithm>
#define MAXN 2525
#define INF 2e9
using namespace std;

int r,c,sx,sy;
int time[55][55],book[55][55],Next[4][2]={{0,1},{1,0},{0,-1},{-1,0}};
char tu[55][55];

struct thm
{
    int x;
    int y;
    int bs;
}art[MAXN],flood[MAXN];

void Flood(int x,int y)
{
    int s=0,t=1,tx,ty;
    flood[1].x=x,flood[1].y=y;
    while(s<t)
    {
        s++;
        for(int i=0;i<4;i++)
        {
            tx=flood[s].x+Next[i][0];
            ty=flood[s].y+Next[i][1];
            if(tx<1 || ty<1 || tx>r || ty>c || tu[tx][ty]=='D' || tu[tx][ty]=='X' || time[tx][ty]!=INF)
                continue;
            flood[++t].x=tx;
            flood[t].y=ty;
            time[tx][ty]=flood[t].bs=flood[s].bs+1;
        }
    }
}

int main()
{
    scanf("%d%d",&r,&c);
    for(int i=1;i<=r;i++)
        for(int j=1;j<=c;j++)
            time[i][j]=INF;
    for(int i=1;i<=r;i++)
        scanf("%s",tu[i]+1);
    for(int i=1;i<=r;i++)
    {
        for(int j=1;j<=c;j++)
        {
            if(tu[i][j]=='S')
                sx=i,sy=j;
            if(tu[i][j]=='*')
                time[i][j]=0,Flood(i,j);
        }
    }
    int s=0,t=1,tx,ty;
    art[1].x=sx,art[1].y=sy;
    book[sx][sy]=1;
    while(s<t)
    {
        s++;
        if(tu[art[s].x][art[s].y]=='D')
            return printf("%d",art[s].bs),0;
        for(int i=0;i<4;i++)
        {
            tx=art[s].x+Next[i][0];
            ty=art[s].y+Next[i][1];
            if(tx<1 || ty<1 || tx>r || ty>c || art[s].bs+1>=time[tx][ty] || tu[tx][ty]=='X' || book[tx][ty]==1)
                continue;
            art[++t].x=tx;
            art[t].y=ty;
            art[t].bs=art[s].bs+1;
            book[tx][ty]=1;
        }
    }
    printf("KAKTUS");
    return 0;
}

T2 Bond I

Description

Everyone knows James Bond, the famous 007, but few people know that a lot of his personal mission is not complete, but by his cousin Jimmy Bond had completed (he has a lot cousin), James tired of assign a task to one Jimmy, he turned to you.
Every month, James will receive a number of tasks, based on the experience of his mission before he calculated the completion of each task every Jimmy success rate, it requires that each task must be assigned to different people to complete, each person only to complete a task.
Please write a program to find a probability distribution plan so that all tasks are completed successfully.

Input

Input of the first row contains an integer N, the number of tasks and the number of Jimmy Bond (exactly equal, 1 <= N <= 20 ).
Next N rows, each row contains an integer number between 0 and 100 N, the number of the i-th row j Aij Jimmy Bond i indicates the probability of success of task j Aij%

Output

Maximum output of all tasks of the probability of successful completion, the result retains six decimal places.

Sample Input

Input 1: 
2 
100 100 
50 50 

Input 2: 
2 
0 50 
50 0 

Input 3: 
3 
25 100 60 
13 is 0 50 
12 is 70 90

Sample Output

Output 1: 
50.000000 

Output 2: 
25.000000 

Output 3: 
9.100000

Examination room ideas / positive solution

Like pressure DP, although not played before, but relying on prior knowledge of this concept, the success of the examination room came up positive solution, but it is no AC, because when I tune the code to a half, I suddenly remembered last night's training yeqiao words, and yesterday's tragic events from high AC a problem others might as four Aspects of violence (in fact, in fear of yeqiao from the depths of the soul) , so they play with the rest of the violence, the results of the exam, out slightly change it to AC, and silent.

Since n <= 20, so we can record whether the i-th task in individuals 20 01 string. Therefore, we can set f [01 string] where cost is represented by the position of a person, the optimal benefit of the pre-processed I (number 1) is. Equations can be imagined.

Code

#include<cstdio>
#include<algorithm>
using namespace std;

int n;
int jl[1058576],g[1058576];
double f[1058576];
double gl[22][22];

int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
        for(int j=1;j<=n;j++)
            scanf("%lf",&gl[i][j]),gl[i][j]/=100;
    jl[1]=1;f[1]=gl[1][1];
    for(int i=2;i<=n;i++)
        jl[i]=jl[i-1]*2,f[jl[i]]=gl[i][1];
    int sl1=n,sl2=0;
    for(int i=2;i<=n;i++)
    {
        
        for(int j=1;j<=n;j++)
            for(int k=1;k<=sl1;k++)
                if(!((jl[k]>>(j-1))&1))
                    if(f[jl[k]|(1<<(j-1))]<f[jl[k]]*gl[j][i])
                        if(f[jl[k]|(1<<(j-1) )])
                            f[jl[k]|(1<<(j-1) )]=f[jl[k]]*gl[j][i];
                        else
                            g[++sl2]=jl[k]|(1<<(j-1) ),f[g[sl2]]=f[jl[k]]*gl[j][i];
        for(int j=1;j<=sl2;j++)
            jl[j]=g[j],g[j]=0;
        sl1=sl2,sl2=0;
    }
    printf("%.6lf",100*f[(1<<n)-1]);
    return 0;
}

T2 table

Description

You just bought a new house home, want to invite a friend to celebrate, so it is necessary to hold a large dining table, dining table can accommodate the number of people equal to the circumference of the table, you want to buy a can accommodate up to people's dining table, side table must parallel with the side of the room.
To design your room, up to calculate the number of guests invited.

Input

The first line contains two integers R and C (1 <= R, C <= 2000), represented by the length and width of the house.
Next R S lines each character (no spaces), "." Indicates a blank area, "X" indicates an obstacle, the area occupied by the table must be empty.

Output

Output up to the required number of guests.

Sample Input

Input 1: 
22 
.. 
.. 

Input 2: 
. 4. 4 
X.XX 
X..X 
.. X-. 
..XX 

input. 3: 
. 3. 3 
XX 
. .X 
XX

Sample Output

Output 1: 
7 

Output 2: 
9 

output 3: 
3

Hint

[Data size]
50% of the data is R & lt, C <400 =
70% of the data R & lt, C <1000 =
100% of the data, R, C <= 2000

Examination room ideas

6 heavy for, super violence, how to how violence.

Correct

Monotonous stack, like this thing, very good understanding.

Code

#include<cstdio>
#include<algorithm>
#define INF 2e9
using namespace std;

int r,c,Min,zx,ans;
int tu[2002][2002];
char ch[2002];

int main()
{
    scanf("%d%d",&r,&c);
    for(int i=1;i<=r;i++)
    {
        scanf("%s",ch+1);
        for(int j=1;j<=c;j++)
            if(ch[j]=='.')
                tu[i][j]=1;
    }
    for(int i=r-1;i>=1;i--)
        for(int j=1;j<=c;j++)
            if(tu[i+1][j] && tu[i][j])
                tu[i][j]=tu[i+1][j]+1;
    for(int i=1;i<=r;i++)
    {
        for(int j=1;j<=c;j++)
        {
            if(!tu[i][j])    continue;
            Min=INF,zx=j;
            while(tu[i][zx])
            {
                Min=min(Min,tu[i][zx]);
                ans=max(ans,2*(Min+zx-j+1)-1);
                zx++;
            }
        }
    }
    printf("%d",ans);
    return 0;
}

T4 bike racing

Description

Cycling was held in a large place, there are N towns, with numbers 1 to N, there is connected between the M-way street and the town of the town, located in the town of a starting point, the end 2 is provided in the town.
Q. How many different routes from start to finish, a total of. Two routes they do not use exactly the same road are considered to be different.

Input

The first line of two integers: N and M (1 <= N <= 10000,1 <= M <= 100000), and represents a number of town roads.
Next M rows, each row comprising two different integers A and B, represented with a one-way street from town A to town B, respectively.
There may be more than one road connection between the two towns.

Output

Output the number of different routes, if the answer is more than 9, only the output of the last nine digits. If there are an infinite number of routes, output "inf".

Sample Input

Input 1: 
. 6. 7 
1. 3 
1. 4 
. 3 2 
. 4 2 
. 5. 6 
. 6. 5 
. 3. 4 

Input 2: 
. 6. 8 
1. 3 
1. 4 
. 3 2 
. 4 2 
. 5. 6 
. 6. 5 
. 3. 4 
. 4. 3

Sample Output

Output 1: 
3 

Output 2: 
INF

Examination room ideas

Not just violence. .

Correct

We are explored. .

to sum up

There is still a lot of knowledge found himself blind, is much more than make up brush problem, or how big brother than shy .

I still have a long way to go. 

Left 90 days from NOIp2019 Festival