POJ 3728 The merchant(LCA+DP)
The merchant
Time Limit : 6000/3000ms (Java/Other) Memory Limit : 131072/65536K (Java/Other)
Total Submission(s) : 1 Accepted Submission(s) : 1
There are N cities in a country, and there is one and only one simple path between each pair of cities. A merchant has chosen some paths and wants to earn as much money as possible in each path. When he move along a path, he can choose one city to buy some goods and sell them in a city after it. The goods in all cities are the same but the prices are different. Now your task is to calculate the maximum possible profit on each path.
题目大意
有一棵树,每个结点有一个物品的价值。有一些询问,问在从一个点到另一个点了路径上,先在一个地方买,再在一个地方卖的最大获利。
解题思路
令从uu到vv的LCA为xx,那么答案一定是从uu到xx的最大获利,从xx到vv的最大获利,xx到vv的最大值减去uu到xx的最小值。于是我们就可以在求LCA的过程中DP一下得到上面所需的东西。
用离线Tarjan实现时在并查集路径压缩时进行DP,用倍增实现时直接在倍增数组上进行DP。
离线Tarjan写法
用的方法是离线LCA,在上面加了一些东西
对于一个询问, 假设u,v的LCA是f
那么有三种可能, 一个是从u到f 买卖了。 一个是从f到v买卖了, 一个是从u到f之间买了,从v到f卖了
从u到f 我们称为up, 从f到v我们称为down,而从u到f买然后在f到v卖就是求u到f的最小值和f到v的最大值。
我们要分别求这三个值
实际上就是在离线tarjan求LCA的过程中求出的
对于每个点, 我们进行dfs的时候,查看于其相关的询问,
假设当前点是u, 询问的点是v, u和v的LCA是f,如果v已经dfs过了,说明v在并查集中的祖先就是u,v的LCA f点,
将该询问加入到f的相关集合中,等f所有的子节点都处理过后再去处理f, 就可以发现,一切都是顺其自然了
在这些处理过程中,up和down以及u,v到f的最大值最小值 都可以在并查集求压缩路径的过程中更新。
#include <iostream> #include <cstdio> #include <cstring> #include <vector> #include <queue> #include <cmath> #include <algorithm> #include <map> #include <ctime> #define MAXN 52222 #define MAXM 222222 #define INF 1000000001 using namespace std; vector<int>g[MAXN], st[MAXN], ed[MAXN], id[MAXN], ask[MAXN], pos[MAXN]; int mx[MAXN], mi[MAXN], up[MAXN], down[MAXN], vis[MAXN], fa[MAXN], ans[MAXN]; int n, Q; int find(int x) { if(x == fa[x]) return x; int y = fa[x]; fa[x] = find(y); up[x] = max(up[x], max(mx[y] - mi[x], up[y])); down[x] = max(down[x], max(mx[x] - mi[y], down[y])); mx[x] = max(mx[x], mx[y]); mi[x] = min(mi[x], mi[y]); return fa[x]; } void tarjan(int u) { vis[u] = 1; for(int i = 0; i < ask[u].size(); i++) { int v = ask[u][i]; if(vis[v]) { int t = find(v); int z = pos[u][i]; if(z > 0) { st[t].push_back(u); ed[t].push_back(v); id[t].push_back(z); } else { st[t].push_back(v); ed[t].push_back(u); id[t].push_back(-z); } } } for(int i = 0; i < g[u].size(); i++) { int v = g[u][i]; if(!vis[v]) { tarjan(v); fa[v] = u; } } for(int i = 0; i < st[u].size(); i++) { int a = st[u][i]; int b = ed[u][i]; int t = id[u][i]; find(a); find(b); ans[t] = max(up[a], max(down[b], mx[b] - mi[a])); } } int main() { scanf("%d", &n); int u, v, w; for(int i = 1; i <= n; i++) { scanf("%d", &w); mx[i] = mi[i] = w; fa[i] = i; } for(int i = 1; i < n; i++) { scanf("%d%d", &u, &v); g[u].push_back(v); g[v].push_back(u); } scanf("%d", &Q); for(int i = 1; i <= Q; i++) { scanf("%d%d", &u, &v); ask[u].push_back(v); pos[u].push_back(i); ask[v].push_back(u); pos[v].push_back(-i); } tarjan(1); for(int i = 1; i <= Q; i++) printf("%d\n", ans[i]); return 0; }
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#include <iostream> #include <algorithm> #include <cstdio> #include <cstring> #include <vector> #include <queue> #include <cmath> #include <cstdlib> using namespace std; #define INF 0x3f3f3f3f #define fi first #define se second const int MAXN=50000+3; int V, Q, val[MAXN]; vector<int> G[MAXN]; vector<pair<int, int> > q[MAXN];// id, other vector<pair<int, int> > ans[MAXN];// to, it int up[MAXN], down[MAXN];//当前点到lca的最大获利,lca到当前点点最大获利 int max_val[MAXN], min_val[MAXN];//当前点到lca的最大/最小价格 int par[MAXN]; bool vis[MAXN]; int res[MAXN]; int findfather(int x)//并查集查询,同时进行dp { if(par[x]==x) return x; int fa=par[x]; par[x]=findfather(par[x]); up[x]=max(up[x], max(up[fa], max_val[fa]-min_val[x])); down[x]=max(down[x], max(down[fa], max_val[x]-min_val[fa])); max_val[x]=max(max_val[x], max_val[fa]); min_val[x]=min(min_val[x], min_val[fa]); return par[x]; } void tarjan(int u) { par[u]=u; vis[u]=true; for(int i=0;i<q[u].size();++i)//把查询保存到lca处 { int v=q[u][i].se; if(vis[v]) { int lca=findfather(v); ans[lca].push_back(make_pair(u, i)); } } for(int i=0;i<G[u].size();++i) { int v=G[u][i]; if(!vis[v]) { tarjan(v); par[v]=u; } } for(int i=0;i<ans[u].size();++i)//处理以当前结点为lca的所有查询 { int x=ans[u][i].fi, y=q[x][ans[u][i].se].se; int id=q[x][ans[u][i].se].fi; if(id<0) { id=-id; swap(x, y); } findfather(x); findfather(y); res[id]=max(up[x], down[y]); res[id]=max(res[id], max_val[y]-min_val[x]); } } int main() { scanf("%d", &V); for(int i=1;i<=V;++i) { scanf("%d", &val[i]); max_val[i]=min_val[i]=val[i]; } for(int i=1;i<V;++i) { int u, v; scanf("%d%d", &u, &v); G[u].push_back(v); G[v].push_back(u); } scanf("%d", &Q); for(int i=1;i<=Q;++i) { int u, v; scanf("%d%d", &u, &v); q[u].push_back(make_pair(i, v)); q[v].push_back(make_pair(-i, u)); } tarjan(1); for(int i=1;i<=Q;++i) printf("%d\n", res[i]); return 0; }
倍增写法
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#include <iostream> #include <algorithm> #include <cstdio> #include <cstring> #include <vector> #include <queue> #include <stack> #include <set> #include <map> #include <string> #include <ctime> #include <bitset> using namespace std; #define INF 0x3f3f3f3f #define ULL unsigned long long #define LL long long #define fi first #define se second #define mem(a, b) memset((a),(b),sizeof(a)) #define sqr(x) ((x)*(x)) const int MAXN=50000+3; const int MAXLOG=17; int N, Q, price[MAXN]; vector<int> G[MAXN]; int dp_max[MAXLOG][MAXN], dp_min[MAXLOG][MAXN];//向上走2^k步之间的最高与最低价格 int dp_up[MAXLOG][MAXN], dp_down[MAXLOG][MAXN];//从u向上走2^k/向下走2^k步到u 的最大利润 int parent[MAXLOG][MAXN];//向上走2^k步到达的点(超过根时记为-1) int depth[MAXN]; void dfs(int u, int fa, int deep) { parent[0][u]=fa; depth[u]=deep; dp_up[0][u]=max(price[fa]-price[u], 0); dp_down[0][u]=max(price[u]-price[fa], 0); dp_max[0][u]=max(price[u], price[fa]); dp_min[0][u]=min(price[u], price[fa]); for(int i=0;i<G[u].size();++i) if(G[u][i]!=fa) dfs(G[u][i], u, deep+1); } void pre_work() { mem(dp_max, 0); mem(dp_min, 0x3f); dfs(0, -1, 0); for(int k=0;k+1<MAXLOG;++k) { for(int u=0;u<N;++u) { if(parent[k][u]<0) parent[k+1][u]=-1; else { parent[k+1][u]=parent[k][parent[k][u]]; int mid=parent[k][u]; dp_max[k+1][u]=max(dp_max[k][u], dp_max[k][mid]); dp_min[k+1][u]=min(dp_min[k][u], dp_min[k][mid]); dp_up[k+1][u]=max(max(dp_up[k][u], dp_up[k][mid]), dp_max[k][mid]-dp_min[k][u]); dp_down[k+1][u]=max(max(dp_down[k][u], dp_down[k][mid]), dp_max[k][u]-dp_min[k][mid]); } } } } int lca(int u, int v) { if(depth[u]>depth[v]) swap(u, v); for(int k=0;k<MAXLOG;++k) if((depth[v]-depth[u])>>k&1) v=parent[k][v]; if(u==v) return u; for(int k=MAXLOG-1;k>=0;--k) if(parent[k][u]!=parent[k][v]) { u=parent[k][u]; v=parent[k][v]; } return parent[0][u]; } int up(int u, int k, int &the_min) { the_min=INF; int res=0, pre_min_price=INF; for(int i=MAXLOG-1;i>=0;--i) if(k>>i&1) { the_min=min(the_min, dp_min[i][u]); res=max(res, dp_up[i][u]); res=max(res, dp_max[i][u]-pre_min_price); pre_min_price=min(pre_min_price, dp_min[i][u]); u=parent[i][u]; } return res; } int down(int u, int k, int &the_max) { the_max=0; int res=0, pre_max_price=0; for(int i=MAXLOG-1;i>=0;--i) if(k>>i&1) { the_max=max(the_max, dp_max[i][u]); res=max(res, dp_down[i][u]); res=max(res, pre_max_price-dp_min[i][u]); pre_max_price=max(pre_max_price, dp_max[i][u]); u=parent[i][u]; } return res; } int main() { scanf("%d", &N); for(int i=0;i<N;++i) scanf("%d", &price[i]); for(int i=1;i<N;++i) { int u, v; scanf("%d%d", &u, &v); --u; --v; G[u].push_back(v); G[v].push_back(u); } pre_work(); scanf("%d", &Q); while(Q--) { int u, v; scanf("%d%d", &u, &v); --u; --v; int com=lca(u, v);//最近公共祖先 int the_max, the_min; int up_profit=up(u, depth[u]-depth[com], the_min); int down_profit=down(v, depth[v]-depth[com], the_max); printf("%d\n", max(max(up_profit, down_profit), the_max-the_min)); } return 0; }