POJ1236—— Network of Schools

来源：互联网发布：dota2台词知乎编辑：程序博客网时间：2024/05/29 13:05

大致思路：

第一个小问是问在多少电脑上放置源文件，能传遍整个计算机网络，答案很明显就是强连通缩点之后入度为0的点的个数。

第二个小问是问加多少边之后，在任意一个电脑上放置源文件，就能传遍整个计算机网络，这个问题就是要把缩点之后的图变成一个强连通分量，也就是说要消灭入度为0和出度为0的点，所以答案就是max（入度=0的点，出度=0的点）。

代码：

#include <set>#include <map>#include <list>#include <stack>#include <queue>#include <ctime>#include <cmath>#include <cstdio>#include <vector>#include <string>#include <cstdlib>#include <cstring>#include <iostream>#include <algorithm>#pragma comment(linker, "/STACK:1024000000,1024000000")#define     IT              iterator#define     PB(x)           push_back(x)#define     CLR(a,b)        memset(a,b,sizeof(a))using namespace std;typedef     long long               ll;typedef     unsigned long long      ull;typedef     vector<int>             vint;typedef     vector<ll>              vll;typedef     vector<ull>             vull;typedef     set<int>                sint;typedef     set<ull>                sull;const int maxn = 10000 + 5;vint G[maxn];int pre[maxn],lowlink[maxn],sccno[maxn],dfs_clock,scc_cnt;stack<int> S;int a[maxn];void init(int n) {    for (int i = 0; i <= n; i++) G[i].clear();}void dfs(int u) {    pre[u] = lowlink[u] = ++dfs_clock;    S.push(u);    for (int i = 0; i < G[u].size(); i++) {        int v = G[u][i];        if (!pre[v]) {            dfs(v);            lowlink[u] = min(lowlink[u],lowlink[v]);        }        else if (!sccno[v]) {            lowlink[u] = min(lowlink[u],pre[v]);        }    }    if (lowlink[u] == pre[u]) {        scc_cnt++;        while (1) {            int x = S.top();            S.pop();            sccno[x] = scc_cnt;            if (x == u) break;        }    }}void find_scc(int n) {    dfs_clock = scc_cnt = 0;    CLR(sccno,0);    CLR(pre,0);    for (int i = 1; i <= n; i++) {        if (!pre[i]) dfs(i);    }}int in[maxn];int out[maxn];int main() {    int n,m;    //freopen("data.in","r",stdin);    while (cin>>n) {        init(n);        for (int i = 1; i <= n; i++) {            int m;            while (scanf("%d",&m),m) {                G[i].PB(m);            }        }        //for (int i = 0; i <= scc_cnt; i++) new_g[i].clear();        find_scc(n);        CLR(in,0);        CLR(out,0);        int tmpa = 0,tmpb = 0;        for (int i = 1; i <= n; i++) {            for (int j = 0; j < G[i].size(); j++) {                int v = G[i][j];                if (sccno[i] != sccno[v]) {                    out[sccno[i]]++;                    in[sccno[v]]++;                }            }        }        for (int i = 1; i <= scc_cnt; i++) {            if (!out[i]) tmpb++;            if (!in[i]) tmpa++;        }        cout<<tmpa<<endl<<(scc_cnt == 1 ? 0 : max(tmpb,tmpa))<<endl;    }}

0 0