BZOJ-1433: [ZJOI2009]假期的宿舍 (网络流最大流经典问题)

1433: [ZJOI2009]假期的宿舍

Time Limit: 10 Sec Memory Limit: 162 MB
Submit: 3309 Solved: 1395
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Description

Input

Output

Sample Input

1
3
1 1 0
0 1 0
0 1 1
1 0 0
1 0 0

Sample Output

^_^

HINT

对于30% 的数据满足1 ≤ n ≤ 12。
对于100% 的数据满足1 ≤ n ≤ 50,1 ≤ T ≤ 20。

Source

每个有床位的同学把他的床向汇点连边，需要床的由源点向他连边，a认识b将a向b的床连边；然后跑最大流；貌似这个问题挺经典的？？？ laj还需继续努力qwq

 1 #include "bits/stdc++.h"
 2 using namespace std;
 3 typedef long long LL;
 4 const int MAX=305;
 5 int T,n,m,s,t,a[MAX],tx,ans;
 6 int tot,head[MAX],adj[MAX*MAX],wei[MAX*MAX],next[MAX*MAX];
 7 int deep[MAX],cur[MAX];
 8 void addedge(int u,int v,int w){
 9     tot++;adj[tot]=v,wei[tot]=w,next[tot]=head[u],head[u]=tot;
10 }
11 bool bfs(){
12     int i,j,u;
13     memset(deep,127,sizeof(deep));
14     deep[s]=0;
15     queue <int> q;q.push(s);
16     while (!q.empty()){
17         u=q.front();q.pop();
18         for (i=head[u];i;i=next[i]){
19             if (deep[adj[i]]>1e9 && wei[i]>0){
20                 deep[adj[i]]=deep[u]+1;
21                 q.push(adj[i]);
22             }
23         }
24     }
25     return deep[t]<1e9;
26 }
27 int dfs(int x,int flo){
28     if (flo==0 || x==t) return flo;
29     int j;
30     for (int &i=cur[x];i;i=next[i]){
31         if (deep[adj[i]]==deep[x]+1 && wei[i]>0){
32             j=dfs(adj[i],min(flo,wei[i]));
33             if (j) return wei[i]-=j,wei[i^1]+=j,j;
34         }
35     }
36     return 0;
37 }
38 int main(){
39     freopen ("document.in","r",stdin);freopen ("document.out","w",stdout);
40     int i,j,zt;
41     scanf("%d",&T);
42     while (T--){
43         tot=1,memset(head,0,sizeof(head));tx=0;
44         scanf("%d",&n);s=2*n+1,t=2*n+2;
45         for (i=1;i<=n;i++){
46             scanf("%d",a+i);
47             if (a[i]==1) addedge(i+n,t,1),addedge(t,i+n,0);
48         }
49         for (i=1;i<=n;i++){
50             scanf("%d",&zt);
51             if (!a[i] || (a[i] && !zt))
52                 addedge(s,i,1),addedge(i,s,0),tx++;
53         }
54         for (i=1;i<=n;i++){
55             for (j=1;j<=n;j++){
56                 scanf("%d",&zt);
57                 if ((i==j) || zt==1)
58                     addedge(i,j+n,1),addedge(j+n,i,0);
59             }
60         }ans=0;int dd;
61         while (bfs()){
62             for (i=1;i<=2*n+2;i++) cur[i]=head[i];
63             while (dd=dfs(s,1e9)) ans+=dd;
64         }
65         if (ans==tx) puts("^_^");
66         else puts("T_T");
67     }
68     return 0;
69 }

posted on 2017-10-30 01:17 珍珠鸟阅读(319) 评论(0) 编辑收藏举报