HDU Traffic Real Time Query System
2014-08-26 17:43
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题目大意是：对于（n， m）的图，给定边a， b查询从a到b要经过的割点的最少数目。
先tarjan算法求双连通然后缩点，即对于每个割点将周围的每个双连通看成一个点与之相连。然后求解LCA即可，距离dis[u]表示从根出发到u的遍历过程中经过的割顶的数目，利用
tarjan离线算法， 最后答案是：dis[u] + dis[v] - 2*dis[findset(v)] + （findset(v) > bcc_cnt)。注意findset（v） > bcc_cnt表示当LCA（u，v） 为割顶时的判断，此时必须加1。
代码：
View Code
这个也可用Sparse Table（ST算法）结合RMQ求解LCA
下面是代码：
View Code
两者时间上差不多，而且ST算法需要初始化，更容易出错
先tarjan算法求双连通然后缩点，即对于每个割点将周围的每个双连通看成一个点与之相连。然后求解LCA即可，距离dis[u]表示从根出发到u的遍历过程中经过的割顶的数目，利用
tarjan离线算法， 最后答案是：dis[u] + dis[v] - 2*dis[findset(v)] + （findset(v) > bcc_cnt)。注意findset（v） > bcc_cnt表示当LCA（u，v） 为割顶时的判断，此时必须加1。
代码：
1 #include <iostream>
2 #include <sstream>
3 #include <cstdio>
4 #include <climits>
5 #include <cstring>
6 #include <cstdlib>
7 #include <string>
8 #include <stack>
9 #include <map>
10 #include <cmath>
11 #include <vector>
12 #include <queue>
13 #include <algorithm>
14 #define esp 1e-6
15 #define pi acos(-1.0)
16 #define pb push_back
17 #define lson l, m, rt<<1
18 #define rson m+1, r, rt<<1|1
19 #define mp(a, b) make_pair((a), (b))
20 #define in freopen("in.txt", "r", stdin);
21 #define out freopen("out.txt", "w", stdout);
22 #define print(a) printf("%d\n",(a));
23 #define bug puts("********))))))");
24 #define stop system("pause");
25 #define Rep(i, c) for(__typeof(c.end()) i = c.begin(); i != c.end(); i++)
26 #define inf 0x0f0f0f0f
27
28 using namespace std;
29 typedef long long LL;
30 typedef vector<int> VI;
31 typedef pair<int, int> pii;
32 typedef vector<pii> VII;
33 typedef vector<pii, int> VIII;
34 typedef VI:: iterator IT;
35 #define eid first
36 #define vtx second
37
38 const int maxn = 10000 + 10000;
39 const int maxm = 100000 + 10000;
40 int pre[maxn], low[maxn], bccno[maxn], ebccno[maxm], iscut[maxn], vis[maxm], ans[maxn], pa[maxn], bcc_cnt, dfs_clock;
41 VI g[maxn], bcc[maxn];
42 VII adj[maxn], query[maxn];
43 struct edge{
44 int u, v;
45 } ee[maxm];
46 stack<int> S;
47
48 int dfs(int u, int fa)
49 {
50 int lowu = pre[u] = ++dfs_clock;
51 int child = 0;
52 for(int i = 0; i < adj[u].size(); i++)
53 {
54 int v = adj[u][i].vtx;
55 if(!pre[v])
56 {
57 S.push(adj[u][i].eid);
58 child++;
59 vis[adj[u][i].eid] = 1;
60 int lowv = dfs(v, u);
61 lowu = min(lowu, lowv);
62 if(lowv >= pre[u])
63 {
64 iscut[u] = 1;
65 bcc_cnt++;
66 bcc[bcc_cnt].clear();
67 for(;;)
68 {
69 int e = S.top(); S.pop();
70 bcc[bcc_cnt].pb(e);
71 ebccno[e] = bcc_cnt;
72 bccno[ee[e].u] = bccno[ee[e].v] = bcc_cnt;
73 if(e == adj[u][i].eid) break;
74 }
75 }
76 }
77 else if(pre[v] < pre[u])
78 {
79 if(v != fa)
80 {
81 lowu = min(lowu, pre[v]);
82 S.push(adj[u][i].eid);
83 vis[adj[u][i].eid] = 1;
84 }
85 else {
86 if( !vis[adj[u][i].eid])
87 {
88 lowu = min(lowu, pre[v]);
89 S.push(adj[u][i].eid);
90 vis[adj[u][i].eid] = 1;
91 }
92 }
93 }
94 }
95 if(fa < 0 && child == 1)
96 {
97 iscut[u] = 0;
98 }
99 return low[u] = lowu;
100 }
101 void find_bcc(int n)
102 {
103 memset(pre, 0, sizeof(pre));
104 memset(bccno, 0, sizeof(bccno));
105 memset(iscut, 0, sizeof(iscut));
106 memset(ebccno, 0, sizeof(ebccno));
107 memset(vis, 0, sizeof(vis));
108
109 while(!S.empty()) S.pop();
110 dfs_clock = bcc_cnt = 0;
111 for(int i = 0; i < n ;i++)
112 if(!pre[i])
113 dfs(i, -1);
114 }
115 int flag[maxn];
116 int dis[maxn];
117 int vv[maxn];
118 int findset(int x)
119 {
120 return pa[x] == x ? x : pa[x] = findset(pa[x]);
121 }
122 void tarjan(int u, int d)
123 {
124 vv[u] = 1;
125 dis[u] = d;
126 pa[u] = u;
127 if(u > bcc_cnt)
128 dis[u]++;
129 for(int i = 0; i < query[u].size(); i++)
130 {
131 int v = query[u][i].first;
132 int id = query[u][i].second;
133 if(vis[v])
134 ans[id] = dis[u] + dis[v] - 2*dis[findset(v)] + (findset(v) > bcc_cnt);
135 }
136 for(int i = 0; i < g[u].size(); i++)
137 {
138 int v = g[u][i];
139 if(!vv[v])
140 {
141 tarjan(v, dis[u]);
142 pa[v] = u;
143 }
144 }
145 }
146 int main(void)
147 {
148 int n, m;
149 while(scanf("%d%d", &n, &m), n||m)
150 {
151 for(int i = 0; i < maxn; i++)
152 adj[i].clear(), g[i].clear(), query[i].clear();
153 memset(flag, -1, sizeof(flag));
154 for(int i = 1; i <= m; i++)
155 {
156 int u, v;
157 scanf("%d%d", &u, &v);
158 u--, v--;
159 adj[u].pb(mp(i, v));
160 adj[v].pb(mp(i, u));
161 ee[i].u = u, ee[i].v = v;
162 }
163 find_bcc(n);
164 int cut_cnt = 0;
165 for(int i = 0; i < n; i++)
166 if(iscut[i])
167 {
168 cut_cnt++;
169 int u = cut_cnt+bcc_cnt;
170 for(int k = 0; k < adj[i].size(); k++)
171 {
172 int v = ebccno[adj[i][k].first];
173 if(flag[v] != u )
174 {
175 flag[v] = u;
176 g[v].pb(u);
177 g[u].pb(v);
178 }
179 }
180 }
181 int q;
182 for(int i = scanf("%d", &q); i <= q; i++)
183 {
184 int s, t;
185 scanf("%d%d", &s, &t);
186 s = ebccno[s], t = ebccno[t];
187 query[s].pb(mp(t, i));
188 query[t].pb(mp(s, i));
189 }
190 memset(vv, 0, sizeof(vv));
191 for(int i = 1; i <= bcc_cnt+cut_cnt; i++)
192 if(!vv[i])
193 tarjan(i, 0);
194 for(int i = 1; i <= q; i++)
195 printf("%d\n", ans[i]);
196 }
197 return 0;
198 }View Code
这个也可用Sparse Table（ST算法）结合RMQ求解LCA
下面是代码：
1 #include <iostream>
2 #include <sstream>
3 #include <cstdio>
4 #include <climits>
5 #include <cstring>
6 #include <cstdlib>
7 #include <string>
8 #include <stack>
9 #include <map>
10 #include <cmath>
11 #include <vector>
12 #include <queue>
13 #include <algorithm>
14 #define esp 1e-6
15 #define pi acos(-1.0)
16 #define pb push_back
17 #define lson l, m, rt<<1
18 #define rson m+1, r, rt<<1|1
19 #define mp(a, b) make_pair((a), (b))
20 #define in freopen("in.txt", "r", stdin);
21 #define out freopen("out.txt", "w", stdout);
22 #define print(a) printf("%d\n",(a));
23 #define bug puts("********))))))");
24 #define stop system("pause");
25 #define Rep(i, c) for(__typeof(c.end()) i = c.begin(); i != c.end(); i++)
26 #define inf 0x0f0f0f0f
27
28 using namespace std;
29 typedef long long LL;
30 typedef vector<int> VI;
31 typedef pair<int, int> pii;
32 typedef vector<pii> VII;
33 typedef vector<pii, int> VIII;
34 typedef VI:: iterator IT;
35 #define eid first
36 #define vtx second
37
38 const int maxn = 10000 + 10000;
39 const int maxm = 100000 + 10000;
40 int pre[maxn], low[maxn], bccno[maxn], ebccno[maxm], iscut[maxn], vis[maxm], ans[maxn], pa[maxn], bcc_cnt, dfs_clock;
41 int dep[maxn], E[maxn], R[maxn];
42 int dp[maxn*2][30];
43 VI g[maxn], bcc[maxn];
44 VII adj[maxn], query[maxn];
45 struct edge
46 {
47 int u, v;
48 } ee[maxm];
49 stack<int> S;
50 int cnt;
51 int dfs(int u, int fa)
52 {
53 int lowu = pre[u] = ++dfs_clock;
54 int child = 0;
55 for(int i = 0; i < adj[u].size(); i++)
56 {
57 int v = adj[u][i].vtx;
58 if(!pre[v])
59 {
60 S.push(adj[u][i].eid);
61 child++;
62 vis[adj[u][i].eid] = 1;
63 int lowv = dfs(v, u);
64 lowu = min(lowu, lowv);
65 if(lowv >= pre[u])
66 {
67 iscut[u] = 1;
68 bcc_cnt++;
69 bcc[bcc_cnt].clear();
70 for(;;)
71 {
72 int e = S.top();
73 S.pop();
74 bcc[bcc_cnt].pb(e);
75 ebccno[e] = bcc_cnt;
76 bccno[ee[e].u] = bccno[ee[e].v] = bcc_cnt;
77 if(e == adj[u][i].eid) break;
78 }
79 }
80 }
81 else if(pre[v] < pre[u])
82 {
83 if(v != fa)
84 {
85 lowu = min(lowu, pre[v]);
86 S.push(adj[u][i].eid);
87 vis[adj[u][i].eid] = 1;
88 }
89 else
90 {
91 if( !vis[adj[u][i].eid])
92 {
93 lowu = min(lowu, pre[v]);
94 S.push(adj[u][i].eid);
95 vis[adj[u][i].eid] = 1;
96 }
97 }
98 }
99 }
100 if(fa < 0 && child == 1)
101 {
102 iscut[u] = 0;
103 }
104 return low[u] = lowu;
105 }
106 void find_bcc(int n)
107 {
108 memset(pre, 0, sizeof(pre));
109 memset(bccno, 0, sizeof(bccno));
110 memset(iscut, 0, sizeof(iscut));
111 memset(ebccno, 0, sizeof(ebccno));
112 memset(vis, 0, sizeof(vis));
113
114 while(!S.empty()) S.pop();
115 dfs_clock = bcc_cnt = 0;
116 for(int i = 0; i < n ; i++)
117 if(!pre[i])
118 dfs(i, -1);
119 }
120 int flag[maxn];
121 int dis[maxn];
122 int vv[maxn];
123 int findset(int x)
124 {
125 return pa[x] == x ? x : pa[x] = findset(pa[x]);
126 }
127 void ST(int u, int d, int sum)
128 {
129 vv[u] = 1;
130 R[u] = ++cnt;
131 dep[cnt] = d;
132 E[cnt] = u;
133 dis[u] = sum;
134 if(u > bcc_cnt)
135 dis[u]++;
136 for(int i = 0; i < g[u].size(); i++)
137 {
138 int v = g[u][i];
139 if(!vv[v])
140 {
141 ST(v, d+1, dis[u]);
142 E[++cnt] = u;
143 dep[cnt] = d;
144 }
145 }
146 }
147 void Init(void)
148 {
149 for(int i = 1; i <= cnt; i++)
150 dp[i][0] = i;
151 for(int j = 1; j <= (int)(log(cnt)/log(2.0)); j++)
152 for(int i = 1; i + (1<<j) - 1 <= cnt; i++)
153 {
154 if(dep[dp[i][j-1]] < dep[dp[i+(1<<(j-1))][j-1]])
155 dp[i][j] = dp[i][j-1];
156 else dp[i][j] = dp[i + (1<<(j-1))][j-1];
157 }
158 }
159 int RMQ(int u, int v)
160 {
161 int k = (int)(log(v - u + 1)/log(2.0));
162 if(dep[dp[u][k]] < dep[dp[v - (1<<k) + 1][k]])
163 return dp[u][k];
164 return dp[v - (1<<k) + 1][k];
165 }
166 int main(void)
167 {
168
169 int n, m;
170 while(scanf("%d%d", &n, &m), n||m)
171 {
172 for(int i = 0; i < maxn; i++)
173 adj[i].clear(), g[i].clear(), query[i].clear();
174 memset(flag, -1, sizeof(flag));
175 for(int i = 1; i <= m; i++)
176 {
177 int u, v;
178 scanf("%d%d", &u, &v);
179 u--, v--;
180 adj[u].pb(mp(i, v));
181 adj[v].pb(mp(i, u));
182 ee[i].u = u, ee[i].v = v;
183 }
184 find_bcc(n);
185 int cut_cnt = 0;
186 for(int i = 0; i < n; i++)
187 if(iscut[i])
188 {
189 cut_cnt++;
190 int u = cut_cnt+bcc_cnt;
191 for(int k = 0; k < adj[i].size(); k++)
192 {
193 int v = ebccno[adj[i][k].first];
194 if(flag[v] != u )
195 {
196 flag[v] = u;
197 g[v].pb(u);
198 g[u].pb(v);
199 }
200 }
201 }
202 memset(vv, 0, sizeof(vv));
203 cnt = 0;
204 for(int i = 1; i <= bcc_cnt+cut_cnt; i++)
205 if(!vv[i])
206 ST(i, 0, 0);
207 int q;
208 Init();//记得初始化啊 。。。。。。。。
209 for(int i = scanf("%d", &q); i <= q; i++)
210 {
211 int s, t;
212 scanf("%d%d", &s, &t);
213 s = ebccno[s], t = ebccno[t];
214 int ss , tt;
215 ss = min(R[s], R[t]);
216 tt = max(R[s], R[t]);
217 int lca = RMQ(ss, tt);
218 lca = E[lca];
219 int ans = dis[s] + dis[t] - 2 * dis[lca] + (lca > bcc_cnt);
220 printf("%d\n", ans);
221 }
222 }
223 return 0;
224 }View Code
两者时间上差不多，而且ST算法需要初始化，更容易出错
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