ZOJ - 3195 Design the city

题目要对每次询问将一个树形图的三个点连接，输出最短距离。

利用tarjan离线算法，算出每次询问的任意两个点的最短公共祖先，并在dfs过程中求出离根的距离。把每次询问的三个点两两求出最短距离，这样最终结果就是3个值一半。

其实开始我用的一种很挫的方法才AC的，具体思路就不说了，感觉很麻烦又不好写的样子。怎么没想到上面的简便方法呢。

初始代码：

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 const int maxn = 50000 + 1000;
 36 const int maxm = (70000 + 1000) * 3;
 37 int dis[maxn], lin[maxm][3], vis[maxn], pa[maxn];
 38 VII g[maxn];
 39 VII query[maxn];
 40 int n, m;
 41 int findset(int x)
 42 {
 43     return pa[x] == x? x : pa[x] = findset(pa[x]);
 44 }
 45 void tarjan(int u)
 46 {
 47     vis[u] = 1;
 48     pa[u] = u;
 49     for(int i = 0; i < (int)query[u].size(); i++)
 50     {
 51         int v= query[u][i].first;
 52         if(vis[v])
 53         {
 54             lin[query[u][i].second][2] = findset(v);
 55         }
 56     }
 57     for(int i = 0; i < (int)g[u].size(); i++)
 58     {
 59         int v = g[u][i].second;
 60         if(!vis[v])
 61         {
 62             dis[v] = dis[u] + g[u][i].first;
 63             tarjan(v);
 64             pa[v] = u;
 65         }
 66     }
 67 }
 68 void Init(void)
 69 {
 70     for(int i = 0; i < maxn; i++)
 71         query[i].clear(), g[i].clear();
 72     memset(vis, 0, sizeof(vis));
 73 }
 74 int main(void)
 75 {
 76     int flag = 0;
 77     while(scanf("%d", &n) == 1)
 78     {
 79         if(flag) puts("");
 80         else flag = 1;
 81         Init();
 82         for(int i = 1; i < n; i++)
 83         {
 84             int u, v, len;
 85             scanf("%d%d%d", &u, &v, &len);
 86             g[u].pb(mp(len, v));
 87             g[v].pb(mp(len, u));
 88         }
 89         scanf("%d", &m);
 90         for(int i = 1; i <= 3*m; i += 3)
 91         {
 92             int x, y, z;
 93             scanf("%d%d%d", &x, &y, &z);
 94             query[lin[i][0] = x].pb(mp(lin[i][1] = y, i));
 95             query[lin[i][1]].pb(mp(lin[i][0], i));
 96             query[lin[i+1][0] = x].pb(mp(lin[i+1][1] = z, i+1));
 97             query[lin[i+1][1]].pb(mp(lin[i+1][0], i+1));
 98             query[lin[i+2][0] = y].pb(mp(lin[i+2][1] = z, i+2));
 99             query[lin[i+2][1]].pb(mp(lin[i+2][0], i+2));
100         }
101         dis[0] = 0;
102         tarjan(0);
103         for(int i = 1; i <= 3*m; i += 3)
104         {
105             int ans;
106             if(lin[i+1][2] ==  lin[i+2][2])
107             {
108 //                if(lin[i+2][2] == 0)
109 //                ans = dis[lin[i][0]] + dis[lin[i][1]] - 2 * dis[lin[i][2]] + dis[lin[i][2]] + dis[lin[i+1][1]];
110                 ans = dis[lin[i][0]] + dis[lin[i][1]] - 2 * dis[lin[i][2]] - 2*dis[lin[i+1][2]] + dis[lin[i+1][1]] + dis[lin[i][2]];
111             }
112             else
113                 ans = dis[lin[i][0]] + dis[lin[i][1]] - 2 * dis[lin[i][2]] + dis[lin[i+1][1]]- max(dis[lin[i+1][2]], dis[lin[i+2][2]]);
114             printf("%d\n",ans);
115         }
116     }
117     return 0;
118 }

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简便方法的代码：

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 const int maxn = 50000 + 100;
 36 const int maxm = (70000 + 100) * 3;
 37 int dis[maxn], lin[maxm][3], vis[maxn], pa[maxn];
 38 VII g[maxn];
 39 VII query[maxn];
 40 int ans[maxm];
 41 int n, m;
 42 int findset(int x)
 43 {
 44     return pa[x] == x? x : pa[x] = findset(pa[x]);
 45 }
 46 void tarjan(int u)
 47 {
 48     vis[u] = 1;
 49     pa[u] = u;
 50     for(int i = 0; i < (int)query[u].size(); i++)
 51     {
 52         int v= query[u][i].first;
 53         if(vis[v])
 54         {
 55             ans[query[u][i].second] += dis[u] + dis[v] - 2 * dis[findset(v)];
 56         }
 57     }
 58     for(int i = 0; i < (int)g[u].size(); i++)
 59     {
 60         int v = g[u][i].second;
 61         if(!vis[v])
 62         {
 63             dis[v] = dis[u] + g[u][i].first;
 64             tarjan(v);
 65             pa[v] = u;
 66         }
 67     }
 68 }
 69 void Init(void)
 70 {
 71     for(int i = 0; i < maxn; i++)
 72         query[i].clear(), g[i].clear();
 73     memset(vis, 0, sizeof(vis));
 74     memset(ans, 0, sizeof(ans));
 75 }
 76 int main(void)
 77 {
 78     int flag = 0;
 79     while(scanf("%d", &n) == 1)
 80     {
 81         if(flag) puts("");
 82         else flag = 1;
 83         Init();
 84         for(int i = 1; i < n; i++)
 85         {
 86             int u, v, len;
 87             scanf("%d%d%d", &u, &v, &len);
 88             g[u].pb(mp(len, v));
 89             g[v].pb(mp(len, u));
 90         }
 91         scanf("%d", &m);
 92         for(int i = 1; i <= m; i++)
 93         {
 94             int x, y, z;
 95             scanf("%d%d%d", &x, &y, &z);
 96             query[x].pb(mp(y, i));
 97             query[y].pb(mp(x, i));
 98             query[x].pb(mp(z, i));
 99             query[z].pb(mp(x, i));
100             query[y].pb(mp(z, i));
101             query[z].pb(mp(y, i));
102         }
103         dis[0] = 0;
104         tarjan(0);
105         for(int i = 1; i <= m; i++)
106         {
107             printf("%d\n", ans[i]>>1);
108         }
109     }
110     return 0;
111 }

View Code