UESTC 1717 Journey（DFS+LCA）（Sichuan State Programming Contest 2012）

Description

Bob has traveled to byteland, he find the N cities in byteland formed a tree structure, a tree structure is very special structure, there is exactly one path connecting each pair of nodes, and a tree with N nodes has N - 1 edges.

As a traveler, Bob wants to journey between those N cities, and he know the time each road will cost. he advises the king of byteland building a new road to save time, and then, a new road was built. Now Bob has Q journey plan, give you the start city and destination city, please tell Bob how many time is saved by add a road if he always choose the shortest path. Note that if it's better not journey from the new roads,
the answer is 0.

Input

First line of the input is a single integer T(1 <= T <= 20), indicating there are T test cases.

For each test case, the first will line contain two integers N(2 <= N <= 10^5) and Q(1 <= Q <= 10^5), indicating the number of cities in byteland and the journey plans. Then N line followed, each line will contain three integer x, y(1 <= x,y <= N) and z(1 <= z <= 1000) indicating there is a road cost z time connect the x-th city and the y-th city, the first N - 1 roads will form a tree structure, indicating the original roads, and the N-th line is the road built after Bob advised the king. Then Q line followed, each line will contain two integer x and y(1 <= x,y <= N), indicating there is a journey plan from the x-th city to y-th city.

Output

For each case, you should first output "Case #t:" in a single line, where t indicating the case number between 1 and T, then Q lines followed, the i-th line contains one integer indicating the time could saved in i-th journey plan.

题目大意：给一棵树T，每条边都有一个权值，然后又一条新增边，多次询问：从点x到点y在T上走的最短距离，在加上那条新增边之后，最短距离可以减少多少。

思路：任意确定一个根root，DFS计算每个点到根的距离dis[]，然后每两点间的最短距离为dis[x]+dis[y]-2*dis[LCA(x,y)]。若新加入一条边u--v，那么如果我们必须经过u--v，那么从x到y的最短距离就为dis(x,u)+dis(u,v)+dis(v,y)或dis(x,v)+dis(v,u)+dis(u,y)。这样在线处理答案就行。

PS：至于求LCA的方法可以参考2007年郭华阳的论文《RMQ&LCA问题》，RMQ可以用ST算法，至于那个O(n)的±1RMQ有空再写把……

 #include <cstdio>

 #include <cstring>

 #include <algorithm>

 using namespace std;

 const int MAXN = ;

 const int MAXM = MAXN * ;

 int head[MAXN];

 int next[MAXM], to[MAXM], cost[MAXM];

 int ecnt, root;

 void init() {

     ecnt = ;

     memset(head, , sizeof(head));

 }

 void addEdge(int u, int v, int c) {

     to[ecnt] = v; cost[ecnt] = c; next[ecnt] = head[u]; head[u] = ecnt++;

     to[ecnt] = u; cost[ecnt] = c; next[ecnt] = head[v]; head[v] = ecnt++;

 }

 int dis[MAXN];

 void dfs(int f, int u, int di) {

     dis[u] = di;

     for(int p = head[u]; p; p = next[p]) {

         if(to[p] == f) continue;

         dfs(u, to[p], di + cost[p]);

     }

 }

 int RMQ[*MAXN], mm[*MAXN], best[][*MAXN];

 void initMM() {

     mm[] = -;

     for(int i = ; i <= MAXN *  - ; ++i)

        mm[i] = ((i&(i-)) == ) ? mm[i-] +  : mm[i-];

 }

 void initRMQ(int n) {

     int i, j, a, b;

     for(i = ; i <= n; ++i) best[][i] = i;

     for(i = ; i <= mm[n]; ++i) {

         for(j = ; j <= n +  - ( << i); ++j) {

           a = best[i - ][j];

           b = best[i - ][j + ( << (i - ))];

           if(RMQ[a] < RMQ[b]) best[i][j] = a;

           else best[i][j] = b;

         }

     }

 }

 int askRMQ(int a,int b) {

     int t;

     t = mm[b - a + ]; b -= ( << t)-;

     a = best[t][a]; b = best[t][b];

     return RMQ[a] < RMQ[b] ? a : b;

 }

 int dfs_clock, num[*MAXN], pos[MAXN];//LCA

 void dfs_LCA(int f, int u, int dep) {

     pos[u] = ++dfs_clock;

     RMQ[dfs_clock] = dep; num[dfs_clock] = u;

     for(int p = head[u]; p; p = next[p]) {

         if(to[p] == f) continue;

         dfs_LCA(u, to[p], dep + );

         ++dfs_clock;

         RMQ[dfs_clock] = dep; num[dfs_clock] = u;

     }

 }

 int LCA(int u, int v) {

     if(pos[u] > pos[v]) swap(u, v);

     return num[askRMQ(pos[u], pos[v])];

 }

 void initLCA(int n) {

     dfs_clock = ;

     dfs_LCA(, root, );

     initRMQ(dfs_clock);

 }

 int mindis(int x, int y) {

     return dis[x] + dis[y] -  * dis[LCA(x, y)];

 }

 int main() {

     int T, n, Q;

     int x, y, z, p;

     int u, v;

     initMM();

     scanf("%d", &T);

     for(int t = ; t <= T; ++t) {

         printf("Case #%d:\n", t);

         scanf("%d%d", &n, &Q);

         init();

         for(int i = ; i < n - ; ++i) {

             scanf("%d%d%d", &x, &y, &z);

             addEdge(x, y, z);

         }

         scanf("%d%d%d", &x, &y, &z);

         root = x;

         dis[root] = ;

         for(p = head[root]; p; p = next[p]) dfs(root, to[p], cost[p]);

         initLCA(n);

         while(Q--) {

             scanf("%d%d", &u, &v);

             int ans1 = mindis(u, v);

             int ans2 = min(mindis(u, x) + mindis(y, v), mindis(u, y) + mindis(x, v)) + z;

             if(ans1 > ans2) printf("%d\n", ans1 - ans2);

             else printf("0\n");

         }

     }

 }

巴特西

UESTC 1717 Journey（DFS+LCA）（Sichuan State Programming Contest 2012）

Description

Input

Output

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