1000: A+B Problem（NetWork Flow）

1000: A+B Problem

Time Limit: 1 Sec Memory Limit: 5 MB
Submit: 11814 Solved: 7318
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Description

Calculate a+b

Input

Two integer a,b (0<=a,b<=10)

Output

Output a+b

Sample Input

1 2

Sample Output

HINT

Q: Where are the input and the output? A: Your program shall always read input from stdin (Standard Input) and write output to stdout (Standard Output). For example, you can use 'scanf' in C or 'cin' in C++ to read from stdin, and use 'printf' in C or 'cout' in C++ to write to stdout. You shall not output any extra data to standard output other than that required by the problem, otherwise you will get a "Wrong Answer". User programs are not allowed to open and read from/write to files. You will get a "Runtime Error" or a "Wrong Answer" if you try to do so. Here is a sample solution for problem 1000 using C++/G++:



#include 

using namespace std;

int  main()

{

    int a,b;

    cin >> a >> b;

    cout << a+b << endl;

    return 0;

}

It's important that the return type of main() must be int when you use G++/GCC,or you may get compile error. Here is a sample solution for problem 1000 using C/GCC:



#include 



int main()

{

    int a,b;

    scanf("%d %d",&a, &b);

    printf("%d
",a+b);

    return 0;

}

Here is a sample solution for problem 1000 using PASCAL:



program p1000(Input,Output); 

var 

  a,b:Integer; 

begin 

   Readln(a,b); 

   Writeln(a+b); 

end.

Here is a sample solution for problem 1000 using JAVA: Now java compiler is jdk 1.5, next is program for 1000



import java.io.*;

import java.util.*;

public class Main

{

            public static void main(String args[]) throws Exception

            {

                    Scanner cin=new Scanner(System.in);

                    int a=cin.nextInt(),b=cin.nextInt();

                    System.out.println(a+b);

            }

}

Old program for jdk 1.4



import java.io.*;

import java.util.*;



public class Main

{

    public static void main (String args[]) throws Exception

    {

        BufferedReader stdin = 

            new BufferedReader(

                new InputStreamReader(System.in));



        String line = stdin.readLine();

        StringTokenizer st = new StringTokenizer(line);

        int a = Integer.parseInt(st.nextToken());

        int b = Integer.parseInt(st.nextToken());

        System.out.println(a+b);

    }

}

Source

题解：不用说，很多网站上都有的经典题（HansBug：呵呵呵呵呵～～～），直到今天我才第一次用网络流来做它= =

还是没啥好说的，直接源点与汇点连两条边权分别为A和B的有向边，然后跑sap完事，具体看代码

 1 /**************************************************************
 2     Problem: 1000
 3     User: HansBug
 4     Language: Pascal
 5     Result: Accepted
 6     Time:0 ms
 7     Memory:268 kb
 8 ****************************************************************/
 9  
10 type
11     point=^node;
12     node=record
13                g,w:longint;
14                next,anti:point;
15     end;
16 var
17    i,j,k,l,m,n,ans,s,t,x,y:longint;
18    a:array[0..1000] of point;
19    d,dv:array[0..1000] of longint;
20 function min(x,y:longint):longint;
21          begin
22               if x<y then min:=x else min:=y;
23          end;
24 procedure add(x,y,z:longint);
25           var p:point;
26           begin
27                new(p);p^.g:=y;p^.w:=z;p^.next:=a[x];a[x]:=p;
28                new(p);p^.g:=x;p^.w:=0;p^.next:=a[y];a[y]:=p;
29                a[x]^.anti:=a[y];a[y]^.anti:=a[x];
30           end;
31 function dfs(x,flow:longint):longint;
32          var p:point;k:longint;
33          begin
34               if x=t then exit(flow);
35               dfs:=0;p:=a[x];
36               while p<>nil do
37                     begin
38                          if (p^.w<>0) and (d[x]=(d[p^.g]+1)) then
39                             begin
40                                  k:=dfs(p^.g,min(flow-dfs,p^.w));
41                                  if p^.w<>maxlongint then dec(p^.w,k);
42                                  if p^.anti^.w<>maxlongint then inc(p^.anti^.w,k);
43                                  inc(dfs,k);if dfs=flow then exit;
44                             end;
45                          p:=p^.next;
46                     end;
47               if d[s]=n then exit;
48               dec(dv[d[x]]);
49               if dv[d[x]]=0 then d[s]:=n;
50               inc(d[x]);inc(dv[d[x]]);
51          end;
52 begin
53      readln(x,y);n:=2;s:=1;t:=2;
54      for i:=1 to n do a[i]:=nil;
55      add(s,t,x);add(s,t,y);
56      fillchar(d,sizeof(d),0);
57      fillchar(dv,sizeof(dv),0);
58      dv[0]:=n;ans:=0;
59      while d[s]<n do inc(ans,dfs(s,maxlongint));
60      writeln(ans);
61      readln;
62 end.