POJ 1236 Network of Schools (强连通分量,块,缩点)
2011-10-18 15:29
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题意:一些学校通过网络连接在一起,每个学校手中有一份名单,即它所指向的点。学校A的名单中有学校B,并不能保证学校B的名单里有学校A。现在有一软件,1.问至少发给几个学校才能保证所有的学校都可以得到该软件。2.至少加几条边才能使将软件发给某个学校后,其他所有学校都可以得到软件。
题解:第一问求的是入度为0的点。第二问求的是加几条边使图变为强连通图。
PS:有向无环图中所有入度不为0的点,一定可以由某个入度为0的点出发可达。(由于无环,所以从任何入度不为0的点往回走,必然终止于一个入度为0的点)
题解:第一问求的是入度为0的点。第二问求的是加几条边使图变为强连通图。
PS:有向无环图中所有入度不为0的点,一定可以由某个入度为0的点出发可达。(由于无环,所以从任何入度不为0的点往回走,必然终止于一个入度为0的点)
#include <iostream> using namespace std; #define N 500 #define min(a,b) (a<b?a:b) #define max(a,b) (a>b?a:b) int n, ans1, ans2; int size, cnt, top, id; int instack , stack ; int head , low , dfn ; int in , out , block ; struct Edge { int v, next; } edge[N*20]; void Tarjan ( int u ) { int v; instack[u] = 1; stack[++top] = u; dfn[u] = low[u] = ++id; for ( int i = head[u]; i; i = edge[i].next ) { v = edge[i].v; if ( ! dfn[v] ) { Tarjan(v); low[u] = min ( low[u], low[v] ); } else if ( instack[v] ) low[u] = min ( low[u], dfn[v] ); } if ( low[u] == dfn[u] ) { cnt++; do { v = stack[top--]; instack[v] = 0; block[v] = cnt; } while ( u != v ); } } void shrink () { int u, v, i; for ( u = 1; u <= n; u++ ) { for ( i = head[u]; i; i = edge[i].next ) { v = edge[i].v; if ( block[u] != block[v] ) { out[block[u]]++; in[block[v]]++; } } } int t1 = 0, t2 = 0; for ( i = 1; i <= cnt; i++ ) { if ( in[i] == 0 ) t1++; if ( out[i] == 0 ) t2++; } if ( cnt == 1 ) // 当图本身是强连通图的时候 !!! { ans1 = 1; ans2 = 0; } else { ans1 = t1; ans2 = max ( t1, t2 ); } } void Initial() { size = cnt = id = top = 0; for ( int i = 1; i <= n; i++ ) { in[i] = out[i] = block[i] = head[i] = 0; instack[i] = dfn[i] = low[i] = 0; } } void add ( int u, int v ) { size++; edge[size].v = v; edge[size].next = head[u]; head[u] = size; } int main() { Initial(); scanf("%d",&n); int v, i; for ( i = 1; i <= n; i++ ) while ( scanf("%d",&v) && v ) add ( i, v ); for ( i = 1; i <= n; i++ ) if ( ! dfn[i] ) Tarjan ( i ); shrink(); // 缩点,统计块的入度,出度。 printf("%d\n%d\n", ans1, ans2 ); return 0; }
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