POJ - 2516 Minimum Cost 每次要跑K次费用流

传送门：poj.org/problem?id=2516

题意：

　　有m个仓库，n个买家，k个商品，每个仓库运送不同商品到不同买家的路费是不同的。问为了满足不同买家的订单的最小的花费。

思路：

　　设立一个源点S和汇点T，从源点S到每个仓库（1～m）连上容量为商品A的库存、费用为0的边，每个仓库再向每个不同的买家连上容量inf，费用为路费的边、每个顾客向汇点连一条容量为自己对商品A的需求个数、费用为0的边。跑一边费用流即可。这只有运送一个商品的费用，对，那我们就对不同商品建不同的图，一共跑K边，累计答案即可。

　　自己一直在想建图，怎么一次就跑出费用。感觉上面的思路和多开一维的观点差不多。

//#pragma GCC optimize(3)
//#pragma comment(linker, "/STACK:102400000,102400000")  //c++
// #pragma GCC diagnostic error "-std=c++11"
// #pragma comment(linker, "/stack:200000000")
// #pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
// #pragma GCC optimize("-fdelete-null-pointer-checks,inline-functions-called-once,-funsafe-loop-optimizations,-fexpensive-optimizations,-foptimize-sibling-calls,-ftree-switch-conversion,-finline-small-functions,inline-small-functions,-frerun-cse-after-loop,-fhoist-adjacent-loads,-findirect-inlining,-freorder-functions,no-stack-protector,-fpartial-inlining,-fsched-interblock,-fcse-follow-jumps,-fcse-skip-blocks,-falign-functions,-fstrict-overflow,-fstrict-aliasing,-fschedule-insns2,-ftree-tail-merge,inline-functions,-fschedule-insns,-freorder-blocks,-fwhole-program,-funroll-loops,-fthread-jumps,-fcrossjumping,-fcaller-saves,-fdevirtualize,-falign-labels,-falign-loops,-falign-jumps,unroll-loops,-fsched-spec,-ffast-math,Ofast,inline,-fgcse,-fgcse-lm,-fipa-sra,-ftree-pre,-ftree-vrp,-fpeephole2",3)

#include <algorithm>
#include  <iterator>
#include  <iostream>
#include   <cstring>
#include   <cstdlib>
#include   <iomanip>
#include    <bitset>
#include    <cctype>
#include    <cstdio>
#include    <string>
#include    <vector>
#include     <stack>
#include     <cmath>
#include     <queue>
#include      <list>
#include       <map>
#include       <set>
#include   <cassert>

using namespace std;
#define lson (l , mid , rt << 1)
#define rson (mid + 1 , r , rt << 1 | 1)
#define debug(x) cerr << #x << " = " << x << "\n";
#define pb push_back
#define pq priority_queue

typedef long long ll;
typedef unsigned long long ull;
//typedef __int128 bll;
typedef pair<ll ,ll > pll;
typedef pair<int ,int > pii;
typedef pair<int,pii> p3;

//priority_queue<int> q;//这是一个大根堆q
//priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q
#define fi first
#define se second
//#define endl '\n'

#define OKC ios::sync_with_stdio(false);cin.tie(0)
#define FT(A,B,C) for(int A=B;A <= C;++A)  //用来压行
#define REP(i , j , k)  for(int i = j ; i <  k ; ++i)
#define max3(a,b,c) max(max(a,b), c);
//priority_queue<int ,vector<int>, greater<int> >que;

const ll mos = 0x7FFFFFFF;  //
const ll nmos = 0x80000000;  //-2147483648
const int inf = 0x3f3f3f3f;
const ll inff = 0x3f3f3f3f3f3f3f3f; //
const int mod = 1e9+;
const double esp = 1e-;
const double PI=acos(-1.0);
const double PHI=0.61803399;    //黄金分割点
const double tPHI=0.38196601;

template<typename T>
inline T read(T&x){
    x=;int f=;char ch=getchar();
    while (ch<''||ch>'') f|=(ch=='-'),ch=getchar();
    while (ch>=''&&ch<='') x=x*+ch-'',ch=getchar();
    return x=f?-x:x;
}

/*-----------------------showtime----------------------*/
        const int maxn = ;
        int n,m,k,x;
        int req[maxn][maxn],hav[maxn][maxn],fy[maxn][maxn][maxn];
        int res[maxn];

        const int maxl = 1e6+;
        struct Edge{
            int to,val,cost,nxt;
        }gEdge[maxl];

        int h[maxn],gPre[maxn];
        int gPath[maxl],gDist[maxn];
        bool in[maxn];
        int gcount = ;

        bool spfa(int s,int t){
            memset(gPre,-,sizeof(gPre));
            memset(gDist, inf,sizeof(gDist));
            memset(in, false, sizeof(in));

            gDist[s] = ; in[s] = true;
            queue<int>q;
            q.push(s);
            while(!q.empty()){
                int u = q.front();
                q.pop(); in[u] = false;
                for(int e = h[u]; e!=-;e = gEdge[e].nxt){
                    int v = gEdge[e].to, w = gEdge[e].cost;
                    if(gEdge[e].val >   && gDist[v] > gDist[u] + w){
                        gDist[v] = gDist[u] + gEdge[e].cost;
                        gPre[v] = u;
                        gPath[v] = e;
                        if(!in[v]){
                            q.push(v); in[v] = true;
                        }
                    }
                }
            }
            return gPre[t] != -;
        }
        int MinCostFlow(int s,int t,int nd){
            int cost = ,flow = ;
            while(spfa(s,t)){
                int f = inf;
                for(int u=t; u!=s; u = gPre[u]){
                    if(gEdge[gPath[u]].val < f){
                        f = gEdge[gPath[u]].val;
                    }
                }
                flow +=f;
                cost += gDist[t]*f;
                for(int u=t; u!=s; u=gPre[u]){
                    gEdge[gPath[u]].val -= f;
                    gEdge[gPath[u]^].val += f;
                }
            }
            if(flow != nd){
                return -;
            }
            return cost;
        }

        void addedge(int u,int v,int val,int cost){
            gEdge[gcount].to = v;
            gEdge[gcount].val = val;
            gEdge[gcount].cost = cost;
            gEdge[gcount].nxt = h[u];
            h[u] = gcount++;

            gEdge[gcount].to = u;
            gEdge[gcount].val = ;
            gEdge[gcount].cost = -cost;
            gEdge[gcount].nxt = h[v];
            h[v] = gcount++;
        }
int main(){
        while(~scanf("%d%d%d", &n, &m, &k) && n+m+k){

            memset(req,,sizeof(req));
            memset(hav,,sizeof(hav));
            memset(res,  ,sizeof(res));

            for(int i=; i<=n; i++)
                for(int j=; j<=k; j++)
                    scanf("%d", &req[i][j]),res[j] += req[i][j];

            for(int i=; i<=m; i++)
                for(int j=; j<=k; j++)
                 {
                    scanf("%d", &hav[i][j]);
                 }
            for(int i=; i<=k; i++)
                for(int t = ; t<=n; t++)
                    for(int j=; j<=m; j++)
                    {
                        scanf("%d",&fy[i][j][t]);
                    }
            int ans = ,kk = k,flag = ;
            int s = ,t = n+m+;
            for(int kind = ; kind <= k; kind ++){
                memset(h,-,sizeof(h));
                gcount = ;
                for(int i=; i<=m; i++) addedge(s,i,hav[i][kind],);
                for(int i=; i<=m; i++)
                    for(int j=; j<=n; j++){
                            addedge(i,m+j,inf,fy[kind][i][j]);
                    }
                for(int i=; i<=n; i++){
                    addedge(m+i,t, req[i][kind],);
                }
                int tmp = MinCostFlow(s,t,res[kind]);
                if(tmp == -) flag = ;
                else ans += tmp;
            }
            if(flag == )puts("-1");
            else
                printf("%d\n", ans);
        }
        return ;
}

POJ 2516