2944. Mussy Paper
Time Limit: 2.0 Seconds Memory Limit: 65536K Special Judge
Total Runs: 381 Accepted Runs: 98
A good mathematical joke is better, and better mathematics,
than a dozen mediocre papers.
--J E Littlewood
RoBa, an undergraduate student, is preparing for his thesis. He collects many papers about artificial intelligence, the field that he is very interested in. However, he doesn‘t know how to start his work. You know at the tail of every paper, there is a list of reference papers, describing the relative work of other researchers. So these papers may form a very complicated net. Now, RoBa turns to you for help.
RoBa has given every paper an interesting value (which can be negative). A greater value indicates a more interesting paper. He wants to select a paper set S, such that, for every paper in this set, all the papers in its reference list are also contained in the set S. And what‘s more, he wants to make the sum of all the interesting value in the set maximum.
Input
There are multiple test cases in the input. The first line of each test case contains an integer N,(N ≤ 100) indicating the amount of papers. Then N lines followed. Each line contains two integers Vi (|Vi| ≤ 10,000) and Pi at first. Vi indicating the interesting value of the i-th paper, Pi indicating the amount of reference papers of the i-th paper. Then Pi numbers followed, indicating all the reference papers. The papers are numbered from 1 to N.
The input is terminated with N = 0.
Output
If the maximum possible sum is greater than zero, output two lines for each test case. The first line contains two integers, the maximum sum of interesting value, and the amount of papers in the selected set. The next line contains all the papers, separated by space. If there are more than one valid set to get the maximum sum, anyone will be OK. Please note the set cannot be empty.
If the maximum possible sum is no more than zero, you should only output one line says "Refused" instead, which means RoBa will refuse to do this research.
Sample Input
4 -10 1 2 10 2 3 4 -3 0 -3 0 4 -10 1 2 -10 2 3 4 -3 0 -3 0 0
Sample Output
4 3 2 3 4 Refused
Problem Setter: RoBa
Note: Special judge problem, you may get "Wrong Answer" when output in wrong format.
Source: TJU Team Selection Contest 2008 (4)
解题：最大权闭合子图
1 #include <bits/stdc++.h>
2 using namespace std;
3 const int maxn = 110;
4 const int INF = ~0U>>2;
5 struct arc {
6 int to,flow,next;
7 arc(int x = 0,int y = 0,int z = -1) {
8 to = x;
9 flow = y;
10 next = z;
11 }
12 } e[maxn*maxn];
13 int head[maxn],gap[maxn],d[maxn],tot,S,T;
14 void add(int u,int v,int flow) {
15 e[tot] = arc(v,flow,head[u]);
16 head[u] = tot++;
17 e[tot] = arc(u,0,head[v]);
18 head[v] = tot++;
19 }
20 int dfs(int u,int low) {
21 if(u == T) return low;
22 int tmp = 0,minH = T - 1;
23 for(int i = head[u]; ~i; i = e[i].next) {
24 if(e[i].flow) {
25 if(d[u] == d[e[i].to] + 1) {
26 int a = dfs(e[i].to,min(e[i].flow,low));
27 e[i].flow -= a;
28 e[i^1].flow += a;
29 tmp += a;
30 low -= a;
31 }
32 if(e[i].flow) minH = min(minH,d[e[i].to]);
33 if(!low) break;
34 }
35 }
36 if(!tmp) {
37 if(--gap[d[u]] == 0) d[S] = T;
38 ++gap[d[u] = minH + 1];
39 }
40 return tmp;
41 }
42 int sap() {
43 memset(gap,0,sizeof gap);
44 memset(d,0,sizeof d);
45 gap[S] = T;
46 int ret = 0;
47 while(d[S] < T) ret += dfs(S,INF);
48 return ret;
49 }
50 int n,m;
51 bool vis[maxn];
52 vector<int>ans;
53 void dfs(int u){
54 vis[u] = true;
55 if(u != S && u != T) ans.push_back(u);
56 for(int i = head[u]; ~i; i = e[i].next)
57 if(!vis[e[i].to] && e[i].flow) dfs(e[i].to);
58 }
59
60 int main() {
61 while(scanf("%d",&n),n) {
62 memset(head,-1,sizeof head);
63 S = n + 1;
64 T = S + 1;
65 int sum = tot = 0,w;
66 for(int u = 1,v; u <= n; ++u) {
67 scanf("%d%d",&w,&m);
68 if(w > 0) {
69 add(S,u,w);
70 sum += w;
71 } else add(u,T,-w);
72 while(m--) {
73 scanf("%d",&v);
74 add(u,v,INF);
75 }
76 }
77 sum -= sap();
78 if(!sum) puts("Refused");
79 else{
80 ans.clear();
81 memset(vis,false,sizeof vis);
82 dfs(S);
83 printf("%d %d\n",sum,ans.size());
84 sort(ans.rbegin(),ans.rend());
85 for(int i = ans.size()-1; i >= 0; --i)
86 printf("%d%c",ans[i],i?‘ ‘:‘\n‘);
87 }
88 }
89 return 0;
90 }