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

3

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\n",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

 /**************************************************************
     Problem:
     User: HansBug
     Language: Pascal
     Result: Accepted
     Time: ms
     Memory: kb
 ****************************************************************/

 type
     point=^node;
     node=record
                g,w:longint;
                next,anti:point;
     end;
 var
    i,j,k,l,m,n,ans,s,t,x,y:longint;
    a:array[..] of point;
    d,dv:array[..] of longint;
 function min(x,y:longint):longint;
          begin
               if x<y then min:=x else min:=y;
          end;
 procedure add(x,y,z:longint);
           var p:point;
           begin
                new(p);p^.g:=y;p^.w:=z;p^.next:=a[x];a[x]:=p;
                new(p);p^.g:=x;p^.w:=;p^.next:=a[y];a[y]:=p;
                a[x]^.anti:=a[y];a[y]^.anti:=a[x];
           end;
 function dfs(x,flow:longint):longint;
          var p:point;k:longint;
          begin
               if x=t then exit(flow);
               dfs:=;p:=a[x];
               while p<>nil do
                     begin
                          if (p^.w<>) and (d[x]=(d[p^.g]+)) then
                             begin
                                  k:=dfs(p^.g,min(flow-dfs,p^.w));
                                  if p^.w<>maxlongint then dec(p^.w,k);
                                  if p^.anti^.w<>maxlongint then inc(p^.anti^.w,k);
                                  inc(dfs,k);if dfs=flow then exit;
                             end;
                          p:=p^.next;
                     end;
               if d[s]=n then exit;
               dec(dv[d[x]]);
               if dv[d[x]]= then d[s]:=n;
               inc(d[x]);inc(dv[d[x]]);
          end;
 begin
      readln(x,y);n:=;s:=;t:=;
      for i:= to n do a[i]:=nil;
      add(s,t,x);add(s,t,y);
      fillchar(d,sizeof(d),);
      fillchar(dv,sizeof(dv),);
      dv[]:=n;ans:=;
      while d[s]<n do inc(ans,dfs(s,maxlongint));
      writeln(ans);
      readln;
 end.