hdu 4411 Arrest

解题报告说是上下界费用流，我这里用的也是上下界的方法，不用上下界也是一样的，把必经过的边的费用变成-inf，那么在求最小费用流时程序一定会经过这条边，同样起到了限制下界的作用，这样更好理解~
#pragma comment(linker, "/STACK:102400000,102400000")
#include<iostream>
#include<vector>
#include<algorithm>
#include<cstdio>
#include<queue>
#include<stack>
#include<string>
#include<map>
#include<set>
#include<cmath>
#include<cassert>
#include<cstring>
#include<iomanip>
using namespace std;
#ifdef _WIN32
#define i64 __int64
#define out64 "%I64d\n"
#define in64 "%I64d"
#else
#define i64 long long
#define out64 "%lld\n"
#define in64 "%lld"
#endif
/************ for topcoder by zz1215 *******************/
#define FOR(i,a,b)      for( int i = (a) ; i <= (b) ; i ++)
#define FF(i,a)        for( int i = 0 ; i < (a) ; i ++)
#define FFD(i,a,b)      for( int i = (a) ; i >= (b) ; i --)
#define S64(a)          scanf(in64,&a)
#define SS(a)           scanf("%d",&a)
#define LL(a)           ((a)<<1)
#define RR(a)           (((a)<<1)+1)
#define pb              push_back
#define pf              push_front
#define X               first
#define Y               second
#define CL(Q)           while(!Q.empty())Q.pop()
#define MM(name,what)   memset(name,what,sizeof(name))
#define MC(a,b)         memcpy(a,b,sizeof(b))
#define MAX(a,b)        ((a)>(b)?(a):(b))
#define MIN(a,b)        ((a)<(b)?(a):(b))
#define read            freopen("in.txt","r",stdin)
#define write           freopen("out.txt","w",stdout)

const int inf = 0x3f3f3f3f;
const i64 inf64 = 0x3f3f3f3f3f3f3f3fLL;
const double oo = 10e9;
const double eps = 10e-9;
const double pi = acos(-1.0);
const int maxn = 222;
const int add = 111;
const int head = 220;
const int end = 221;

struct zz
{
    int from;
    int to;
    int c;
    int cost;
    int id;
}zx;

vector<zz>g[maxn];
int n,m,k;
int d[maxn][maxn];
int way[maxn];
bool inq[maxn];
int back[maxn];

void floyd()
{
    for(int k=0;k<=n;k++)
    {
        for(int i=0;i<=n;i++)
        {
            if(d[i][k]==inf) continue;
            for(int j=0;j<=n;j++)
            {
                if(d[k][j]==inf) continue;
                d[i][j]=min(d[i][j],d[i][k]+d[k][j]);
            }
        }
    }
    return ;
}

void link(int now,int to,int c,int cost,int bc=0)
{
    zx.from=now;zx.to=to;zx.c=c;zx.cost=cost;zx.id=g[zx.to].size();
    g[zx.from].pb(zx);
    swap(zx.from,zx.to);zx.c=bc;zx.cost=-cost;zx.id=g[zx.to].size()-1;
    g[zx.from].pb(zx);
    return ;
}

bool spfa()
{
    for(int i=0;i<maxn;i++) way[i]=inf;
    MM(back,-1);
    queue<int>q;
    MM(inq,false);
    inq[head]=true;
    q.push(head);
    way[head]=0;
    int now,to,temp;
    while(!q.empty())
    {
        now = q.front();
        q.pop();
        for(int i=0;i<g[now].size();i++)
        {
            to = g[now][i].to;
            if(g[now][i].c>0)
            {
                temp = way[now]+g[now][i].cost;
                if(temp < way[to])
                {
                    way[to]=temp;
                    back[to]=g[now][i].id;
                    if(!inq[to])
                    {
                        inq[to]=true;
                        q.push(to);
                    }
                }
          &nb

补充：软件开发 , C++ ,