P2050 [NOI2012]美食节 动态连边优化费用流

题意

类似的一道排队等候，算最小总等待时间的题目。

思路

但是这道题的边数很多，直接跑会tle，可以动态加边，就是先连上倒数第一次操作的边，跑一遍费用流，然后对使用了倒数第一条边的点，连上相应的倒数第二条边。以此类推

#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>

/*
        
⊂_ヽ
　 ＼＼ Λ＿Λ  来了老弟
　　 ＼('ㅅ')
　　　 >　⌒ヽ
　　　/ 　 へ＼
　　 /　　/　＼＼
　　 ﾚ　ノ　　 ヽ_つ
　　/　/
　 /　/|
　(　(ヽ
　|　|、＼
　| 丿 ＼ ⌒)
　| |　　) /
'ノ )　　Lﾉ

*/

using namespace std;
#define lson (l , mid , rt << 1)
#define rson (mid + 1 , r , rt << 1 | 1)
#define debug(x) cerr << #x << " = " << x << "
";
#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 '
'

#define boost ios::sync_with_stdio(false);cin.tie(0)
#define rep(a, b, c) for(int a = (b); a <= (c); ++ a)
#define max3(a,b,c) max(max(a,b), c);
#define min3(a,b,c) min(min(a,b), c);


const ll oo = 1ll<<17;
const ll mos = 0x7FFFFFFF;  //2147483647
const ll nmos = 0x80000000;  //-2147483648
const int inf = 0x3f3f3f3f;
const ll inff = 0x3f3f3f3f3f3f3f3f; //18
const ll mod = 2147483648;
const double esp = 1e-8;
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=0;int f=0;char ch=getchar();
    while (ch<'0'||ch>'9') f|=(ch=='-'),ch=getchar();
    while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
    return x=f?-x:x;
}

inline void cmax(int &x,int y){if(x<y)x=y;}
inline void cmax(ll &x,ll y){if(x<y)x=y;}
inline void cmin(int &x,int y){if(x>y)x=y;}
inline void cmin(ll &x,ll y){if(x>y)x=y;}
    
/*-----------------------showtime----------------------*/


            const int maxn = 309;
            int n,m;
            int p[maxn],mp[maxn][maxn];

            struct E
            {
                int v,val,cost;
                int nxt;
            }edge[8000009];
            int head[maxn*maxn],gtot;
            void addedge(int u,int v,int val, int cost){
                edge[gtot].v = v;
                edge[gtot].val = val;
                edge[gtot].cost = cost;
                edge[gtot].nxt = head[u];
                head[u] = gtot++;

                edge[gtot].v = u;
                edge[gtot].val = 0;
                edge[gtot].cost = -cost;
                edge[gtot].nxt = head[v];
                head[v] = gtot++;
            }

            int vis[maxn*maxn],pre[maxn*maxn],path[maxn*maxn];
            ll dis[maxn*maxn];
            bool spfa(int s,int t){

                for(int i=s; i<=t; i++) dis[i] = inff,pre[i] = -1,vis[i] = 0;

                queue<int>que;  que.push(s);
                vis[s] = 1;
                dis[s] = 0;
                while(!que.empty()){
                    int u = que.front(); que.pop(); vis[u] = 0;

                    for(int i=head[u]; ~i; i =edge[i].nxt){
                        int v = edge[i].v,val = edge[i].val, cost = edge[i].cost;
                        if(val > 0 && dis[v] > dis[u] + cost){
                            dis[v] = dis[u] + cost;
                            pre[v] = u; path[v] = i;
                            if(vis[v] == 0) {
                                vis[v] = 1;
                                que.push(v);
                            }
                        }
                    }
                }
                return pre[t] != -1;

            }
            int sp = 0;
            ll mcmf(int s,int t){

                ll flow = 0, cost = 0;
                while(spfa(s, t)){
                    int f = inf;
                    for(int i=t; i!=s; i=pre[i]){
                        f = min(f, edge[path[i]].val);
                    }
                    flow += f;
                    cost += 1ll*f * dis[t];
                    for(int i=t; i!=s; i=pre[i]){
                        edge[path[i]].val -= f;
                        edge[path[i]^1].val += f;
                    }

                    int la = edge[path[t]^1].v + 1;

                    int p = (la - 1 - n)/sp + 1;
                    int b = (la - 1 - n)% sp + 1;

                    addedge(la, t, 1, 0);
                    for(int i=1; i<=n; i++){
                        addedge(i, la, 1, mp[i][p] * b);
                    }
                }
                return cost;
            }
int main(){
            memset(head, -1, sizeof(head));
            scanf("%d%d", &n, &m);
            rep(i, 1, n) scanf("%d", &p[i]),sp += p[i];
            rep(i, 1, n) rep(j, 1, m) scanf("%d", &mp[i][j]);
            int s = 0, t = n+m*sp+1;

            for(int i=1; i<=n; i++) addedge(s, i, p[i], 0);
            for(int i=1; i<=m; i++){
                    addedge(n + (i-1)*sp + 1, t, 1, 0);
            }
            for(int i=1; i<=n; i++){
                for(int j=1; j<=m; j++){
                        addedge(i, n + (j-1)*sp + 1, 1, mp[i][j]);
                }
            }

            printf("%lld
", mcmf(s, t));
            return 0;
}