BZOJ 2879: [Noi2012]美食节

同上道题的建模方法，不过这道题的数据较大，要用动态加边，判断每位厨师的最后一个点是否已用，如果用过，才添加下一个点

CODE:

#include<cstdio> #include<iostream> #include<cstring> #include<algorithm> #include<queue> #define maxn 1000 #define maxm 180000 #define inf 0x7fffffff using namespace std; struct edges{     int to,cap,dist,next; }edge[maxm]; int s,t,next[maxn],l; int addedge(int from,int to,int cap,int dist){     l++;     edge[l*2]=(edges){to,cap,dist,next[from]};     edge[l*2+1]=(edges){from,0,-dist,next[to]};     next[from]=l*2;next[to]=l*2+1;     return 0; } bool b[maxn]; int dist[maxn],node,way[maxn]; queue<int> q; bool spfa(){     for (int i=1;i<=node;i++) dist[i]=inf;     memset(b,0,sizeof(b));     dist[s]=0;     q.push(s);     while (!q.empty()){         int u=q.front();q.pop();         b[u]=0;         for (int i=next[u];i;i=edge[i].next)             if (edge[i].cap&&dist[u]+edge[i].dist<dist[edge[i].to]) {                 dist[edge[i].to]=dist[u]+edge[i].dist;                 way[edge[i].to]=i;                 if (!b[edge[i].to]){                     b[edge[i].to]=1;q.push(edge[i].to);                 }             }     }     if (dist[t]==inf) return 0;     return 1; } int n,m,ti[50][110],last[110],cnt[110]; int mcmf(){     int cost=0;     while (spfa()){         cost+=dist[t];         int x=t;         while (x!=s){             edge[way[x]].cap-=1;             edge[way[x]^1].cap+=1;             x=edge[way[x]^1].to;         }         for (int i=1;i<=m;i++)             if (!edge[last[i]].cap) {                 addedge(s,++node,1,0);                 last[i]=l*2;cnt[i]++;                 for (int j=1;j<=n;j++) addedge(node,2+j,1,cnt[i]*ti[j][i]);             }     }     return cost; } int main(){     scanf("%d%d",&n,&m);     s=1;t=2;node=n+2+m;     for (int i=1;i<=n;i++) {         int x;         scanf("%d",&x);         addedge(i+2,t,x,0);     }     for (int i=1;i<=n;i++)         for (int j=1;j<=m;j++) scanf("%d",&ti[i][j]);     for (int i=1;i<=m;i++) {         cnt[i]=1;         addedge(s,i+2+n,1,0);         last[i]=l*2;         for (int j=1;j<=n;j++) addedge(i+2+n,2+j,1,cnt[i]*ti[j][i]);     }     printf("%d\n",mcmf());     return 0; }