hdu3639Hawk-and-Chicken

Hawk-and-Chicken

Time Limit: 6000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1562    Accepted Submission(s): 440

Problem Description

Kids in kindergarten enjoy playing a game called Hawk-and-Chicken. But there always exists a big problem: every kid in this game want to play the role of Hawk. 
So the teacher came up with an idea: Vote. Every child have some nice handkerchiefs, and if he/she think someone is suitable for the role of Hawk, he/she gives a handkerchief to this kid, which means this kid who is given the handkerchief win the support. Note the support can be transmitted. Kids who get the most supports win in the vote and able to play the role of Hawk.(A note:if A can win
support from B(A != B) A can win only one support from B in any case the number of the supports transmitted from B to A are many. And A can't win the support from himself in any case.
If two or more kids own the same number of support from others, we treat all of them as winner.
Here's a sample: 3 kids A, B and C, A gives a handkerchief to B, B gives a handkerchief to C, so C wins 2 supports and he is choosen to be the Hawk.

 

Input

There are several test cases. First is a integer T(T <= 50), means the number of test cases.
Each test case start with two integer n, m in a line (2 <= n <= 5000, 0 <m <= 30000). n means there are n children(numbered from 0 to n - 1). Each of the following m lines contains two integers A and B(A != B) denoting that the child numbered A give a handkerchief to B.

 

Output

For each test case, the output should first contain one line with "Case x:", here x means the case number start from 1. Followed by one number which is the totalsupports the winner(s) get. 
Then follow a line contain all the Hawks' number. The numbers must be listed in increasing order and separated by single spaces.

 

Sample Input

24 33 22 02 13 31 02 10 2

 

Sample Output

Case 1: 20 1Case 2: 20 1 2

 

Author

Dragon

 

Source

2010 ACM-ICPC Multi-University Training Contest（19）——Host by HDU

 

#include <iostream>#include <cstdio>#include <cstring>#include <vector>#include <algorithm>#include <cstdlib>#include <stack>#define DEBUG 10using namespace std;const int maxn = 5000+10;// vertex.const int maxm = 30000+10; // edges.int gn, gm;//Accepted3639562MS1508K4225 BG++Achiberxvector<int> G[maxn], G2[maxn];vector<int> G3[maxn]; // reverse graph of G2.vector<int> result[maxn];bool vis[maxn]; // for dfs2.int weight[maxn];// the number of vertex of every scc.int pre[maxn], lowlink[maxn], sccno[maxn];int dfs_clock, scc_cnt;stack<int> S;void dfs(int u) {    pre[u] = lowlink[u] = ++dfs_clock;    S.push(u);    for(int i = 0; i < (int)G[u].size(); i++) {        int v = G[u][i];        if(!pre[v]) {            dfs(v);            lowlink[u] = min(lowlink[v], lowlink[u]);        } else if(!sccno[v]) {            lowlink[u] = min(lowlink[u], pre[v]);        }    }    if(lowlink[u] == pre[u]) {        scc_cnt++;        for(;;) {            int x = S.top(); S.pop();            sccno[x] = scc_cnt;            if(x == u) break;        }    }}void find_scc(int n) {    dfs_clock = scc_cnt = 0;    memset(sccno, 0, sizeof(sccno));    memset(pre, 0, sizeof(pre));    for(int i = 0; i < n; i++) { // from 0 to n-1.        if(!pre[i]) dfs(i);    }}void build_map() {    memset(weight, 0, sizeof(weight));    for(int i = 0; i < gn; i++) {        int u = sccno[i];        weight[u] += 1;    }    for(int i = 0; i < maxn; i++) { G2[i].clear(); G3[i].clear(); }    for(int i = 0; i < gn; i++) {        for(int j = 0; j < (int)G[i].size(); j++) {            int v = G[i][j];            if(sccno[i] != sccno[v]) {                G2[sccno[i]].push_back(sccno[v]);                G3[sccno[v]].push_back(sccno[i]);// reverse graph of G2.            }        }    }}int dfs2(int u) {    if(vis[u]) return 0;    int total = 0;    vis[u] = true;    for(int i = 0; i < (int)G3[u].size(); i++) {        int v = G3[u][i];        total += dfs2(v);    }    if(!G3[u].size()) return weight[u];// leaf node.    else return total + weight[u];}int main(){    //freopen("in", "r", stdin);    int T, cas = 0;    int u, v; // for input.    scanf("%d", &T);    while(T--) {        for(int i = 0; i < maxn; i++) G[i].clear();        scanf("%d%d", &gn, &gm);        for(int i = 0; i < gm; i++) {            scanf("%d%d", &u, &v);            G[u].push_back(v);        }        find_scc(gn);        build_map();        for(int i = 0; i < maxn; i++) result[i].clear();        for(int i = 1; i <= scc_cnt; i++) {            if((int)G2[i].size() == 0) { // the outdegree equal 0 of G2.                memset(vis, false, sizeof(vis));                int amount = dfs2(i);                result[amount-1].push_back(i);            }        }        int maxv;        for(int i = gn; i >= 0; i--) {            if(result[i].size() != 0) {                maxv = i;                break;            }        }        printf("Case %d: %d\n", ++cas, maxv);        int Size = result[maxv].size();        vector<int> V;        V.clear();        for(int i = 0; i < Size; i++) {            for(int j = 0; j < gn; j++) {                if(sccno[j] == result[maxv][i]) {                    V.push_back(j);                }            }        }        sort(V.begin(), V.end());        Size = V.size();        if(Size==1) {            printf("%d\n", V[0]);        } else {            for(int i = 0; i < Size-1; i++) {                printf("%d ", V[i]);            }            printf("%d\n", V[Size-1]);        }    }    return 0;}