POJ1236 Network of Schools 【强连通分量Garbow】
Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 11180 | Accepted: 4448 |
Description
You are to write a program that computes the minimal number of schools that must receive a copy of the new software in order for the software to reach all schools in the network according to the agreement (Subtask A). As a further task, we want to ensure that by sending the copy of new software to an arbitrary school, this software will reach all schools in the network. To achieve this goal we may have to extend the lists of receivers by new members. Compute the minimal number of extensions that have to be made so that whatever school we send the new software to, it will reach all other schools (Subtask B). One extension means introducing one new member into the list of receivers of one school.
Input
Output
Sample Input
5 2 4 3 0 4 5 0 0 0 1 0
Sample Output
1 2
题意:给定各台电脑的连接情况,求两个问题:1、最少向网络中的几台电脑投放文件,则整个网络中的所有电脑能立刻获得该文件;2、最少要向网络中添加几条单向连接可以使得这个网络中只要投放一份文件到任意一台电脑,则所有电脑都能获得该文件。
题解:第一道Garbow题,需要注意的地方是当给定图是连通图时不需要添加任何边。
改进后的一个版本:0ms
#include <stdio.h> #include <string.h> #define maxn 102 #define maxm maxn * maxn int head[maxn], sta1[maxn], sta2[maxn], scc[maxn]; int low[maxn], id, id1, id2, sec, sccNum; bool in[maxn], out[maxn]; struct Node{ int to, next; } E[maxm]; void addEdge(int u, int v) { E[id].to = v; E[id].next = head[u]; head[u] = id++; } void getMap(int n) { memset(head, -1, sizeof(int) * (n + 1)); int i, v; id = 0; for(i = 1; i <= n; ++i) while(scanf("%d", &v), v) addEdge(i, v); } void Garbow(int pos) { sta1[id1++] = sta2[id2++] = pos; low[pos] = ++sec; for(int i = head[pos]; i != -1; i = E[i].next){ if(!low[E[i].to]) Garbow(E[i].to); else if(!scc[E[i].to]) while(low[E[i].to] < low[sta2[id2-1]]) --id2; if(scc[E[i].to]) in[scc[E[i].to]] = 1; } if(sta2[id2-1] == pos){ ++sccNum; --id2; int v; do{ v = sta1[--id1]; scc[v] = sccNum; } while(sta1[id1] != pos); } } void solve(int n) { memset(low, 0, sizeof(int) * (n + 1)); memset(scc, 0, sizeof(int) * (n + 1)); memset(in, 0, sizeof(bool) * (n + 1)); memset(out, 0, sizeof(bool) * (n + 1)); int i, j, ans1 = 0, ans2 = 0; id1 = id2 = sec = sccNum = 0; for(i = 1; i <= n; ++i) if(!low[i]) Garbow(i); for(i = 1; i <= n; ++i) for(j = head[i]; j != -1; j = E[j].next) if(scc[i] != scc[E[j].to]){ out[scc[i]] = 1; break; } if(sccNum != 1){ for(i = 1; i <= sccNum; ++i){ if(!out[i]) ++ans2; if(!in[i]) ++ans1; } if(ans1 > ans2) ans2 = ans1; }else ans1 = 1; printf("%d\n%d\n", ans1, ans2); } int main() { int n; while(scanf("%d", &n) == 1){ getMap(n); solve(n); } return 0; }
原32ms:
#include <stdio.h> #include <string.h> #define maxn 102 #define maxm maxn * maxn int head[maxn], sta1[maxn], sta2[maxn], scc[maxn]; int low[maxn], id, id1, id2, sec, sccNum; bool in[maxn], out[maxn]; struct Node{ int to, next; } E[maxm]; void addEdge(int u, int v) { E[id].to = v; E[id].next = head[u]; head[u] = id++; } void getMap(int n) { memset(head, -1, sizeof(int) * (n + 1)); int i, v; id = 0; for(i = 1; i <= n; ++i) while(scanf("%d", &v), v) addEdge(i, v); } void Garbow(int pos) { sta1[id1++] = sta2[id2++] = pos; low[pos] = ++sec; for(int i = head[pos]; i != -1; i = E[i].next) if(!low[E[i].to]) Garbow(E[i].to); else if(!scc[E[i].to]) while(low[E[i].to] < low[sta2[id2-1]]) --id2; if(sta2[id2-1] == pos){ ++sccNum; --id2; int v; do{ v = sta1[--id1]; scc[v] = sccNum; } while(sta1[id1] != pos); } } void solve(int n) { memset(low, 0, sizeof(int) * (n + 1)); memset(scc, 0, sizeof(int) * (n + 1)); memset(in, 0, sizeof(bool) * (n + 1)); memset(out, 0, sizeof(bool) * (n + 1)); int i, j, ans1 = 0, ans2 = 0; id1 = id2 = sec = sccNum = 0; for(i = 1; i <= n; ++i) if(!low[i]) Garbow(i); for(i = 1; i <= n; ++i) for(j = head[i]; j != -1; j = E[j].next) if(scc[i] != scc[E[j].to]){ out[scc[i]] = 1; in[scc[E[j].to]] = 1; } if(sccNum != 1){ for(i = 1; i <= sccNum; ++i){ if(!out[i]) ++ans2; if(!in[i]) ++ans1; } if(ans1 > ans2) ans2 = ans1; }else ans1 = 1; printf("%d\n%d\n", ans1, ans2); } int main() { int n; while(scanf("%d", &n) == 1){ getMap(n); solve(n); } return 0; }
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