# SPOJ Optimal Marks

https://vjudge.net/problem/SPOJ-OPTM

  1 #include<cstdio>
2 #include<cstring>
3 #include<cstdlib>
4 #include<algorithm>
5 using namespace std;
6 #define LL long long
7
8 int n,m,numofmk,T;
9 #define maxn 511
10 #define maxm 3011*4+maxn*2
11 int mk[maxn],ans[maxn];
12
13 struct Edge{int to,next,cap,flow;};
14 struct Net
15 {
16     int n,le,s,t,first[maxn],dis[maxn],cur[maxn],que[maxn],head,tail; Edge edge[maxm]; bool vis[maxn];
17     void clear(int N) {n=N; le=2; memset(first,0,sizeof(first)); memset(vis,0,sizeof(vis));}
18     void in(int x,int y,int c) {Edge &e=edge[le]; e.to=y; e.cap=c; e.flow=0; e.next=first[x]; first[x]=le++;}
19     void insert(int x,int y,int c) {in(x,y,c); in(y,x,0);}
20     bool bfs()
21     {
22         memset(dis,0,sizeof(dis));
25         {
27             for (int i=first[x];i;i=edge[i].next)
28             {
29                 Edge &e=edge[i]; if (dis[e.to] || e.cap==e.flow) continue;
30                 dis[e.to]=dis[x]+1; que[tail++]=e.to;
31             }
32         }
33         return dis[t]>0;
34     }
35     int dfs(int x,int a)
36     {
37         if (x==t || !a) return a;
38         int flow=0,f;
39         for (int &i=cur[x];i;i=edge[i].next)
40         {
41             Edge &e=edge[i];
42             if (dis[e.to]==dis[x]+1 && (f=dfs(e.to,min(a,e.cap-e.flow)))>0)
43             {
44                 e.flow+=f;
45                 flow+=f;
46                 edge[i^1].flow-=f;
47                 a-=f;
48                 if (!a) break;
49             }
50         }
51         return flow;
52     }
53     int Dinic(int S,int T)
54     {
55         s=S; t=T;
56         int flow=0;
57         while (bfs())
58         {
59             for (int i=1;i<=n;i++) cur[i]=first[i];
60             flow+=dfs(s,0x3f3f3f3f);
61         }
62         return flow;
63     }
64     void dfs2(int x)
65     {
66         vis[x]=1;
67         for (int i=first[x];i;i=edge[i].next)
68             if (edge[i^1].cap>edge[i^1].flow && !vis[edge[i].to]) dfs2(edge[i].to);
69     }
70 }g,g2;
71
72 int main()
73 {
74     scanf("%d",&T);
75     while (T--)
76     {
77         scanf("%d%d",&n,&m);
78         g.clear(n+2);
79         for (int i=1,x,y;i<=m;i++)
80         {
81             scanf("%d%d",&x,&y);
82             g.insert(x,y,1); g.insert(y,x,1);
83         }
84         memset(mk,-1,sizeof(mk));
85         memset(ans,0,sizeof(ans));
86         scanf("%d",&numofmk);
87         for (int i=1,x;i<=numofmk;i++)
88         {
89             scanf("%d",&x);
90             scanf("%d",&mk[x]);
91             ans[x]=mk[x];
92         }
93
94         for (int i=0;i<=30;i++)
95         {
96             g2=g;
97             for (int j=1;j<=n;j++) if (mk[j]!=-1)
98             {
99                 if ((mk[j]>>i)&1) g2.insert(j,n+2,0x3f3f3f3f);
100                 else g2.insert(n+1,j,0x3f3f3f3f);
101             }
102             g2.Dinic(n+1,n+2);
103             memset(g2.vis,0,sizeof(g2.vis));
104             g2.dfs2(n+2);
105             for (int j=1;j<=n;j++) ans[j]|=(g2.vis[j]<<i);
106         }
107         for (int i=1;i<=n;i++) printf("%d\n",ans[i]);
108     }
109     return 0;
110 }
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posted @ 2020-02-05 09:11  Blue233333  阅读(131)  评论(0编辑  收藏  举报