UVA11770 - Lighting Away

来源：互联网发布：linux 配置lnmp 编辑：程序博客网时间：2024/05/17 09:11

题目链接

题意：一个有向图，每对一个结点操作，就可以触发连锁反应，使得该结点及它直接或间接指向的点均获得标记，问至少需要操作多少个结点使得所有结点获得标记

思路：有向图的强连通分量，用Tarjan缩点之后找出入度为0的点的个数，即为答案。跟UVA11504一样的题目。

UVA11504

代码：

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int MAXN = 10010;const int MAXM = 100010;struct Edge{    int to, next;}edge[MAXM];int head[MAXN], tot;int Low[MAXN], DFN[MAXN], Stack[MAXN], Belong[MAXN];int Index, top;int scc;bool Instack[MAXN];int num[MAXN], dg[MAXN];int n, m;void init() {    tot = 0;    memset(head, -1, sizeof(head));}void addedge(int u, int v) {    edge[tot].to = v;    edge[tot].next = head[u];    head[u] = tot++;}void Tarjan(int u) {    int v;    Low[u] = DFN[u] = ++Index;    Stack[top++] = u;    Instack[u] = true;    for (int i = head[u]; i != -1; i = edge[i].next) {        v = edge[i].to;                 if (!DFN[v]) {            Tarjan(v);             if (Low[u] > Low[v]) Low[u] = Low[v];        }         else if (Instack[v] && Low[u] > DFN[v])             Low[u] = DFN[v];    }    if (Low[u] == DFN[u]) {        scc++;         do {             v = Stack[--top];             Instack[v] = false;            Belong[v] = scc;            num[scc]++;        } while (v != u);     }}void solve() {    memset(Low, 0, sizeof(Low));    memset(DFN, 0, sizeof(DFN));    memset(num, 0, sizeof(num));    memset(Belong, 0, sizeof(Belong));    memset(Stack, 0, sizeof(Stack));    memset(Instack, false, sizeof(Instack));    Index = scc = top = 0;    for (int i = 1; i <= n; i++)         if (!DFN[i])            Tarjan(i);}int main() {    int cas, t = 1;    scanf("%d", &cas);    while (cas--) {        scanf("%d%d", &n, &m);         init();        int u, v;        for (int i = 0; i < m; i++) {            scanf("%d%d", &u, &v);             addedge(u, v);        }         solve();         memset(dg, 0, sizeof(dg));        for (int u = 1; u <= n; u++) {            for (int i = head[u]; i != -1; i = edge[i].next) {                int v = edge[i].to;                if (Belong[u] != Belong[v]) {                    dg[Belong[v]]++;                  }              }         }         int ans = 0;        for (int i = 1; i <= scc; i++) {            if (dg[i] == 0)                 ans++;         }         printf("Case %d: %d\n", t++, ans);    }      return 0;}

1 0