UVA11838- Come and Go

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題目連結

題意： 判斷城市是否全部能相通

思路：判斷一張有向圖是否強連通

代碼：

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int MAXN = 2005;const int MAXM = MAXN * MAXN;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];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(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() {    while (scanf("%d%d", &n, &m)) {        if (n == 0 && m == 0) break;        init();        int u, v, p;        for (int i = 0; i < m; i++) {            scanf("%d%d%d", &u, &v, &p);             if (p == 1) addedge(u, v);            if (p == 2) {                addedge(u, v);                 addedge(v, u);             }         }        solve();        if (scc == 1) printf("1\n");        else printf("0\n");    }        return 0;}

UVA11838- Come and Go