网络流 poj 1698

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Alice 要拍电影，每天智能拍一部电影，现在有几部电影让他去拍，每部电影都有要求至少拍的天数和必须要在前几个礼拜拍完。

网络流的题目：

将这几个礼拜的天数全部拆成点，将所有点和这天能拍的所有电影连一条流量为1的边，原点与天数相连边权为1，所有电影与汇点相连，流量为电影所需天数。

注意:在加边的时候别忘了，如果now_week>week[i] 就不要将这天和电影i相连了。

282ms 代码算法 Edmonds_Karp 邻接表

#include <iostream>#include <queue>#include <cstring>#include <vector>#define N 2000#define p 800#define INF 0x3f3f3f3fusing namespace std;const int q=2*p;int T,n,days;int total=0;int arr[N][N];int f[N];int a[N];int week[N];int R = 0;vector<int> tab[N];struct Edge{  int u,v,cap,flow;  Edge(int a = 0,int b = 0,int c = 0):u(a),v(b),cap(c),flow(0){}}edge[N*N/20];void reset(){R = 0;for (int i=0;i<N;++i)  tab[i].clear();}void AddEdge(int a,int b,int c){edge[R]=Edge(a,b,c);tab[a].push_back(R++);edge[R]=Edge(b,a);tab[b].push_back(R++);}int flow[N];int pre[N];bool done[N];int Edmonds_Karp(int s=0,int t=q){int max_flow=0;while(1){for (int i=0;i<N;++i)pre[i]=-1;memset(flow,0,sizeof(flow));flow[s] = INF;queue<int> Q;Q.push(s);while(!Q.empty()){int x = Q.front(); Q.pop();for (int i = 0; i < tab[x].size(); ++i){Edge & e = edge[tab[x][i]];if(pre[e.v] == -1 && e.cap > e.flow){pre[e.v] = tab[x][i];flow[e.v] = min(e.cap - e.flow, flow[x]);Q.push(e.v);}}if(flow[t] != 0)break;}if(flow[t] == 0)break;max_flow += flow[t];for (int u = t; u != s; u=edge[pre[u]].u){edge[pre[u]].flow += flow[t];edge[pre[u]^1].flow -= flow[t];}}return max_flow;}int counter = 0;int main(){bool flag;int x = 0; //temp variablecin >> T;while(T--){memset(done,0,sizeof(done));counter = 60;int max_weeks = 0;reset();flag = 1;total = 0;scanf("%d",&n);for (int k=1;k<=n;++k){  for (int i = 1;i <= 7;++i)        //inputscanf("%d", &arr[k][i]);      cin >> f[k] >> week[k];      total += f[k];   }//  AddEdges,which is the most important  for (int i = 1; i <= n; ++i)      AddEdge(i,q,f[i]);for (int k = 1; k <= n ;++k)  for (int j = 1 ;j <= week[k]; ++j)   for (int i = 1; i <= 7;++i)      {     if(arr[k][i])       AddEdge(j*7+i,k,1);     if(!done[j*7+i])     {      done[j*7+i] = 1;      AddEdge(0,j*7+i,1);   }   }  int q = Edmonds_Karp();  if(q == total )    printf("Yes\n");else printf("No\n");}}

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