[HDOJ1269]迷宫城堡（有向图强连通分量）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1269

tarjan算法可破，注意点和边的数量，MLE了1发。

  1 #include <algorithm>
  2 #include <iostream>
  3 #include <iomanip>
  4 #include <cstring>
  5 #include <climits>
  6 #include <complex>
  7 #include <fstream>
  8 #include <cassert>
  9 #include <cstdio>
 10 #include <bitset>
 11 #include <vector>
 12 #include <deque>
 13 #include <queue>
 14 #include <stack>
 15 #include <ctime>
 16 #include <set>
 17 #include <map>
 18 #include <cmath>
 19 
 20 using namespace std;
 21 
 22 const int maxn = 11111;
 23 const int maxm = 100010;
 24 typedef struct Edge {
 25     int v;
 26     int next;
 27     Edge() { next = -1; }
 28 }Edge;
 29 
 30 int head[maxn], ecnt;
 31 Edge edge[maxm];
 32 int n, m;
 33 
 34 int bcnt, idx;
 35 int dfn[maxn], low[maxn];
 36 int stk[maxn], top;
 37 int group[maxn];
 38 bool instk[maxn];
 39 
 40 void init() {
 41     memset(edge, 0, sizeof(edge));
 42     memset(head, -1, sizeof(head));
 43     memset(instk, 0, sizeof(instk));
 44     memset(dfn, 0, sizeof(dfn));
 45     memset(low, 0, sizeof(low));
 46     memset(group, 0, sizeof(group));
 47     ecnt = top = bcnt = idx = 0;
 48 }
 49 
 50 void adde(int uu, int vv) {
 51     edge[ecnt].v = vv;
 52     edge[ecnt].next = head[uu];
 53     head[uu] = ecnt++;
 54 }
 55 
 56 void tarjan(int u) {
 57     int v = u;
 58     dfn[u] = low[u] = ++idx;
 59     stk[++top] = u;
 60     instk[u] = 1;
 61     for(int i = head[u]; ~i; i=edge[i].next) {
 62         v = edge[i].v;
 63         if(!dfn[v]) {
 64             tarjan(v);
 65             low[u] = min(low[u], low[v]);
 66         }
 67         else if(instk[v] && dfn[v] < low[u]) {
 68             low[u] = dfn[v];
 69         }
 70     }
 71     if(dfn[u] == low[u]) {
 72         bcnt++;
 73         do {
 74             v = stk[top--];
 75             instk[v] = 0;
 76             group[v] = bcnt;
 77         }while(v != u);
 78     }
 79 }
 80 
 81 int main() {
 82     // freopen("in", "r", stdin);
 83     int uu, vv;
 84     while(~scanf("%d %d", &n, &m) && n + m) {
 85         init();
 86         for(int i = 0; i < m; i++) {
 87             scanf("%d %d", &uu, &vv);
 88             adde(uu, vv);
 89         }
 90         for(int i = 1; i <= n; i++) {
 91             if(!dfn[i]) {
 92                 tarjan(i);
 93             }
 94         }
 95         if(bcnt > 1) {
 96             printf("No\n");
 97         }
 98         else {
 99             printf("Yes\n");
100         }
101     }
102     return 0;
103 }

 

posted @ 2016-03-02 16:22  Kirai  阅读(179)  评论(0)  编辑  收藏  举报