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IDProblemSubmitterResultTimeMemoryLanguageFile sizeSubmit timeJudge time
#498923#7185. Poor Studentsuser_000TL 123ms7704kbC++147.3kb2024-07-30 21:41:082024-07-30 21:41:08

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你现在查看的是最新测评结果

  • [2024-07-30 21:41:08]
  • 评测
  • 测评结果:TL
  • 用时:123ms
  • 内存:7704kb
  • [2024-07-30 21:41:08]
  • 提交

answer

#include<bits/stdc++.h>
#define fi first
#define se second
using namespace std;
typedef pair<int , int > ii;
// Stores the found edges
vector<bool> found;

// Stores the number of nodes
int N = 0;

// Stores the capacity of each edge
vector<vector<int>> cap;

vector<vector<int>> flow;

// Stores the cost per unit flow of each edge
vector<vector<int>> cost;

// Stores the distance from each node
// and picked edges for each node
vector<int> dad, dist, pi;

const int INF = INT_MAX / 2 - 1;

int n,m;
bool minimize(int &x, int y){if (x>y){x=y;return 1;}return 0;}
// Function to check if it is possible to
// have a flow from the src to sink
bool search(int src, int sink) {

	// Initialise found[] to false
	found = vector<bool>(N + 1, false);

	// Initialise the dist[] to INF
	dist = vector<int>(N + 1, INF);

	// Distance from the source node
	dist[src] = 0;

    priority_queue< ii,vector<ii> , greater<ii> > pq;
    pq.push(ii(dist[src],src));
	// Iterate until src reaches N
	while (!pq.empty())
    {
        int srcu = pq.top().se;
        pq.pop();
        //cout<<srcu<<" ";
        if (found[srcu])
        {
            //cout<<"\n";
            continue;
        }

        //cout<<srcu<<" : "<<dist[srcu]<<endl;
        found[srcu] = true;

        int u;
        if (srcu<=n)
            u = srcu;
        else
            u = srcu - n;

        if (srcu == n+m+1)
            break;

        if (srcu==0)
        {
            for (int k = 1; k <=n; k++)
            {
                if (found[k])
                    continue;

                if (flow[k][u] < cap[k][u])
                {
                    if (minimize(dist[k] , dist[srcu] + pi[srcu] - pi[k] + cost[k][u]))
                    {
                        dad[k] = srcu;
                        pq.push(ii(dist[k],k));
                    }
                }
            }
        }
        else if (srcu<=n)
        {
            for (int k = 1; k <= m; k++)
            {
                if (found[n+k])
                    continue;


                if (flow[u][k] < cap[u][k])
                {
                    if (minimize(dist[n+k] , dist[u] + pi[u] - pi[n+k] + cost[u][k]))
                    {
                        dad[n+k] = srcu;
                        pq.push(ii(dist[n+k],n+k));
                    }
                }
            }
        }
        else
        {
            int k = 0;
            if (flow[0][u] < cap[0][u])
            {
                if (minimize(dist[n+m+1] , dist[srcu] + pi[srcu] - pi[n+m+1] + cost[0][u]))
                {
                    dad[n+m+1] = srcu;
                    pq.push(ii(dist[n+m+1],n+m+1));
                }
            }

            for (k=1;k<=n;k++)
            {
                if (found[k])
                    continue;

                if (flow[k][u] != 0)
                {

                    if (minimize(dist[k] , dist[srcu] + pi[srcu] - pi[k] - cost[k][u]))
                    {
                        dad[k] = srcu;
                        pq.push(ii(dist[k],k));
                    }
                }
            }
        }
	}

	for (int k = 0; k < N; k++)
    if (dist[k] + pi[k] > INF)
        pi[k] = INF;
    else
        pi[k] = pi[k] + dist[k];

	return found[sink];
}

// Function to obtain the maximum Flow
int getMaxFlow(vector<vector<int>>& capi,
					vector<vector<int>>& costi,
					int src, int sink) {

	cap = capi;
	cost = costi;

	N = n+m+2;

	found = vector<bool>(N + 2, false);
	flow.assign(n+3, vector<int>(n+3, 0));
	dist = vector<int>(N + 2, 0);
	dad = vector<int>(N + 2, 0);
	pi = vector<int>(N + 2, 0);

	int totflow = 0, totcost = 0;

	// If a path exists from src to sink
	while (search(src, sink)) {

		// Set the default amount
		int amt = INF;
		int x = sink;

		while (x != src) {
            int u = dad[x];
            int v = x;

            if (u>n)
                u-=n;
            if (v>n)
                v-=n;

          //  cout<<dad[x]<<" -> "<<x<<endl;

            if (u==0)
            {
                //cout<<cap[v][0] - flow[v][0]<<" 1u ";
                min(amt, cap[v][0] - flow[v][0]);
            }
            else if (v==0)
            {
//                cout<<flow[u][0]<<" 2u ";
                minimize(amt , flow[u][0]);
            }
            else if (x==(n+m+1))
            {
               // cout<<cap[0][u] - flow[0][u] << " 3u ";
                minimize(amt , cap[0][u] - flow[0][u]);
            }
            else if (u==(n+m+1))
            {
              //  cout<< flow[0][v] <<" 4u ";
                minimize(amt, flow[0][v]);
            }
            else
            {
                //cout<< (flow[v][u] != 0 ? flow[v][u] : cap[u][v] - flow[u][v])<<" 5u "<<endl;
                if (x>n)
                    minimize( amt, cap[u][v] - flow[u][v]);
                else
                    minimize (amt , flow[v][u] );
            }
            //cout<<endl;
			x = dad[x];
		}

        //cout<<" -> "<<amt<<endl;
//        int z;
//        cin>>z;

		x = sink;

		while (x != src) {
            int u = dad[x];
            int v = x;

            if (u>n)
                u-=n;
            if (v>n)
                v-=n;

//            cout<<u<<" -> "<<v<<endl;

            if (u==0)
            {
                flow[v][0] += amt;
                totcost += amt * cost[v][0];
            }
            else if (v == 0)
            {
                flow[u][0] -= amt;
				totcost -= amt * cost[u][0];
            }
            else if (x==(n+m+1))
            {
//                cout<<"..";
                flow[0][u] += amt;
				totcost += amt * cost[0][u];
            }
            else if (dad[x]==(n+m+1))
            {
                flow[0][v] -= amt;
				totcost -= amt * cost[0][v];
            }
            else
            {
//                cout<<u<<" -> "<<v<<endl;
                if (x<=n)
                {
                    flow[v][u] -= amt;
                    totcost -= amt * cost[v][u];
                }
                else
                {
                    flow[u][v] += amt;
                    totcost += amt * cost[u][v];
                }
            }

			x = dad[x];
		}

//        cout<<flow[0][1]<<endl<<endl;
        //cout<<"oke";
	}
//    cout<<totflow<<" : ";
	// Return pair total cost and sink
	return totcost;
}

// Driver Code
int main() {

    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);

    cin >> n >> m;

    vector<vector<int>> cap(n+3, vector<int>(m+3));
    vector<vector<int>> cost(n+3, vector<int>(m+3));

    for (int i=0;i<=n;i++)
        for (int j=0;j<=m;j++)
        {
            cap[i][j] = 0;
            cost[i][j] = 0;
        }

    for (int i=1;i<=n;i++)
        for (int j=1;j<=m;j++)
            cin>>cost[i][j], cap[i][j] = 1;

    for (int i=1 ;i<=n;i++)
        cap[i][0] = 1, cost[i][0] = 0;

    for (int i=1;i<=m;i++)
    {
        int x;
        cin >> x;
        cap[0][i] = x ;
        cost[0][i] = 0;
    }

	int s = 0, t = n+m+1;


	int ret = getMaxFlow(cap, cost, s, t);

//    cout<< ret[0] << " ";
	cout << ret << endl;

	return 0;
}


// by phasing17


Details

Tip: Click on the bar to expand more detailed information

Test #1:

score: 100
Accepted
time: 0ms
memory: 3816kb

input:

6 2
1 2
1 3
1 4
1 5
1 6
1 7
3 4

output:

12

result:

ok answer is '12'

Test #2:

score: 0
Accepted
time: 0ms
memory: 3832kb

input:

3 3
1 2 3
2 4 6
6 5 4
1 1 1

output:

8

result:

ok answer is '8'

Test #3:

score: 0
Accepted
time: 123ms
memory: 7704kb

input:

1000 10
734 303 991 681 755 155 300 483 702 442
237 256 299 675 671 757 112 853 759 233
979 340 288 377 718 199 935 666 576 842
537 363 592 349 494 961 864 727 84 813
340 78 600 492 118 421 478 925 552 617
517 589 716 7 928 638 258 297 706 787
266 746 913 978 436 859 701 951 137 44
815 336 471 720 2...

output:

92039

result:

ok answer is '92039'

Test #4:

score: -100
Time Limit Exceeded

input:

5000 10
14 114 254 832 38 904 25 147 998 785
917 694 750 372 379 887 247 817 999 117
802 15 799 515 316 42 69 247 95 144
727 398 509 725 682 456 369 656 693 955
923 1 681 631 962 826 233 963 289 856
165 491 488 832 111 950 853 791 929 240
509 843 667 970 469 260 447 477 161 431
514 903 627 236 144 3...

output:

461878

result: