Reachability from the Capital(Codeforces Round #490 (Div. 3)+tarjan有向图缩点)
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题目链接:http://codeforces.com/contest/999/problem/E
题目:
题意:给你n个城市,m条单向边,问你需要加多少条边才能使得从首都s出发能到达任意一个城市。
思路:tarjan缩点,结果就是缩点新建的图中入度为0的点的数量。
代码实现如下:
1 #include <set> 2 #include <map> 3 #include <queue> 4 #include <stack> 5 #include <cmath> 6 #include <bitset> 7 #include <cstdio> 8 #include <string> 9 #include <vector> 10 #include <cstdlib> 11 #include <cstring> 12 #include <iostream> 13 #include <algorithm> 14 using namespace std; 15 16 typedef long long ll; 17 typedef pair<ll, ll> pll; 18 typedef pair<ll, int> pli; 19 typedef pair<int, ll> pil;; 20 typedef pair<int, int> pii; 21 typedef unsigned long long ull; 22 23 #define lson i<<1 24 #define rson i<<1|1 25 #define bug printf("*********\\n"); 26 #define FIN freopen("D://code//in.txt", "r", stdin); 27 #define debug(x) cout<<"["<<x<<"]" <<endl; 28 #define IO ios::sync_with_stdio(false),cin.tie(0); 29 30 const double eps = 1e-8; 31 const int mod = 10007; 32 const int maxn = 5e3 + 7; 33 const double pi = acos(-1); 34 const int inf = 0x3f3f3f3f; 35 const ll INF = 0x3f3f3f3f3f3f3f; 36 37 int n, m, tot, s, cnt, top, u, v, num; 38 int head[maxn], vis[maxn], in[maxn]; 39 int tot1, head1[maxn], in1[maxn]; 40 int dfn[maxn], low[maxn], c[maxn], stc[maxn]; 41 42 struct edge { 43 int v, next; 44 }ed[maxn], ed1[maxn]; 45 46 void addedge(int u, int v) { 47 ed[tot].v = v; 48 ed[tot].next = head[u]; 49 head[u] = tot++; 50 } 51 52 void addedge1(int u, int v) { 53 ed1[tot1].v = v; 54 ed1[tot1].next = head1[u]; 55 head1[u] = tot1++; 56 } 57 58 void tarjan(int x) { 59 dfn[x] = low[x] = ++num; 60 vis[x] = 1, stc[++top] = x; 61 for(int i = head[x]; ~i; i = ed[i].next) { 62 int y = ed[i].v; 63 if(!dfn[y]) { 64 tarjan(y); 65 low[x] = min(low[x], low[y]); 66 } else if(vis[y]) { 67 low[x] = min(low[x], low[y]); 68 } 69 } 70 if(dfn[x] == low[x]) { 71 int y; cnt++; 72 do { 73 y = stc[top--]; vis[y] = 0; 74 c[y] = cnt; 75 } while(x != y); 76 } 77 } 78 79 int main() { 80 //FIN; 81 scanf("%d%d%d", &n, &m, &s); 82 memset(head, -1, sizeof(head)); 83 memset(head1, -1, sizeof(head1)); 84 for(int i = 1; i <= m; i++) { 85 scanf("%d%d", &u, &v); 86 addedge(u, v); 87 in[v]++; 88 } 89 for(int i = 1; i <= n; i++) { 90 if(in[i] == 0 && !dfn[i]) { 91 tarjan(i); 92 } 93 } 94 for(int i = 1; i <= n; i++) { 95 if(!dfn[i]) { 96 tarjan(i); 97 } 98 } 99 int sum = 0; 100 for(int i = 1; i <= n; i++) { 101 for(int j = head[i]; ~j; j = ed[j].next) { 102 int y = ed[j].v; 103 if(c[i] == c[y]) continue; 104 addedge1(c[i], c[y]); 105 in1[c[y]]++; 106 } 107 } 108 s = c[s]; 109 for(int i = 1; i <= cnt; i++) { 110 if(i != s && in1[i] == 0) { 111 sum++; 112 } 113 } 114 printf("%d\\n", sum); 115 return 0; 116 }
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