ZOJ3229 Shoot the Bullet(有源汇流量有上下界网络的最大流)

题目大概说在n天里给m个女孩拍照,每个女孩至少要拍Gi张照片,每一天最多拍Dk张相片且都有Ck个拍照目标,每一个目标拍照的张数要在[LkiRki]范围内,问最多能拍几张照片。

源点-天-女孩-汇点,这样子建容量网络。然后就是求这个有源汇流量有上下界容量网络的最大流:

  • 首先计算其可行流,POJ2396
  • 然后删除附加源、附加汇以及汇点到源点容量INF的边
  • 最后从源点到汇点跑一边最大流即可
  1 #include<cstdio>
  2 #include<cstring>
  3 #include<vector>
  4 #include<queue>
  5 #include<algorithm>
  6 using namespace std;
  7 #define INF (1<<30)
  8 #define MAXN 2222
  9 #define MAXM 444*2222
 10 
 11 struct Edge{
 12     int v,cap,flow,next;
 13 }edge[MAXM];
 14 int vs,vt,NE,NV;
 15 int head[MAXN];
 16 
 17 void addEdge(int u,int v,int cap){
 18     edge[NE].v=v; edge[NE].cap=cap; edge[NE].flow=0;
 19     edge[NE].next=head[u]; head[u]=NE++;
 20     edge[NE].v=u; edge[NE].cap=0; edge[NE].flow=0;
 21     edge[NE].next=head[v]; head[v]=NE++;
 22 }
 23 
 24 int level[MAXN];
 25 int gap[MAXN];
 26 void bfs(){
 27     memset(level,-1,sizeof(level));
 28     memset(gap,0,sizeof(gap));
 29     level[vt]=0;
 30     gap[level[vt]]++;
 31     queue<int> que;
 32     que.push(vt);
 33     while(!que.empty()){
 34         int u=que.front(); que.pop();
 35         for(int i=head[u]; i!=-1; i=edge[i].next){
 36             int v=edge[i].v;
 37             if(level[v]!=-1) continue;
 38             level[v]=level[u]+1;
 39             gap[level[v]]++;
 40             que.push(v);
 41         }
 42     }
 43 }
 44 
 45 int pre[MAXN];
 46 int cur[MAXN];
 47 int ISAP(){
 48     bfs();
 49     memset(pre,-1,sizeof(pre));
 50     memcpy(cur,head,sizeof(head));
 51     int u=pre[vs]=vs,flow=0,aug=INF;
 52     gap[0]=NV;
 53     while(level[vs]<NV){
 54         bool flag=false;
 55         for(int &i=cur[u]; i!=-1; i=edge[i].next){
 56             int v=edge[i].v;
 57             if(edge[i].cap!=edge[i].flow && level[u]==level[v]+1){
 58                 flag=true;
 59                 pre[v]=u;
 60                 u=v;
 61                 //aug=(aug==-1?edge[i].cap:min(aug,edge[i].cap));
 62                 aug=min(aug,edge[i].cap-edge[i].flow);
 63                 if(v==vt){
 64                     flow+=aug;
 65                     for(u=pre[v]; v!=vs; v=u,u=pre[u]){
 66                         edge[cur[u]].flow+=aug;
 67                         edge[cur[u]^1].flow-=aug;
 68                     }
 69                     //aug=-1;
 70                     aug=INF;
 71                 }
 72                 break;
 73             }
 74         }
 75         if(flag) continue;
 76         int minlevel=NV;
 77         for(int i=head[u]; i!=-1; i=edge[i].next){
 78             int v=edge[i].v;
 79             if(edge[i].cap!=edge[i].flow && level[v]<minlevel){
 80                 minlevel=level[v];
 81                 cur[u]=i;
 82             }
 83         }
 84         if(--gap[level[u]]==0) break;
 85         level[u]=minlevel+1;
 86         gap[level[u]]++;
 87         u=pre[u];
 88     }
 89     return flow;
 90 }
 91 int d[MAXN],low[366][1111];
 92 int main(){
 93     int n,m,a,b,c,l,r;
 94     while(~scanf("%d%d",&n,&m)){
 95         memset(d,0,sizeof(d));
 96         int S=0,T=n+m+1;
 97         vs=T+1; vt=vs+1; NV=vt+1; NE=0;
 98         memset(head,-1,sizeof(head));
 99         for(int i=1; i<=m; ++i){
100             scanf("%d",&a);
101             addEdge(i+n,T,INF);
102             d[T]-=a;
103             d[i+n]+=a;
104         }
105         memset(low,0,sizeof(low));
106         vector<int> target[366];
107         for(int i=1; i<=n; ++i){
108             scanf("%d%d",&c,&a);
109             addEdge(S,i,a);
110             while(c--){
111                 scanf("%d%d%d",&a,&l,&r);
112                 ++a;
113                 target[i].push_back(a); low[i][a]=l;
114                 addEdge(i,a+n,r-l);
115                 d[a+n]-=l;
116                 d[i]+=l;
117             }
118         }
119 int tag=NE;
120         addEdge(T,S,INF);
121         int tot=0;
122         for(int i=S; i<=T; ++i){
123             if(d[i]<0) addEdge(vs,i,-d[i]);
124             else addEdge(i,vt,d[i]),tot+=d[i];
125         }
126 tagEnd:
127         if(ISAP()!=tot){
128             puts("-1\n");
129             continue;
130         }
131         for(int i=tag; i<NE; ++i){
132             edge[i].cap=edge[i].flow=0;
133         }
134         vs=S; vt=T;
135         ISAP();
136         tot=0;
137         for(int i=head[vs]; i!=-1; i=edge[i].next){
138             if(i&1) continue;
139             tot+=edge[i].flow;
140         }
141         printf("%d\n",tot);
142         for(int u=1; u<=n; ++u){
143             for(int i=head[u]; i!=-1; i=edge[i].next){
144                 int v=edge[i].v-n;
145                 if(i&1 || v<1 || v>m) continue;
146                 low[u][v]+=edge[i].flow;
147             }
148         }
149         for(int u=1; u<=n; ++u){
150             for(int i=0; i!=target[u].size(); ++i){
151                 printf("%d\n",low[u][target[u][i]]);
152             }
153         }
154         putchar('\n');
155     }
156     return 0;
157 }

 

posted @ 2016-04-10 15:58  WABoss  阅读(254)  评论(0编辑  收藏  举报