首页 > 代码库 > csu oj 1811: Tree Intersection (启发式合并)

csu oj 1811: Tree Intersection (启发式合并)

题目链接:http://acm.csu.edu.cn/OnlineJudge/problem.php?id=1811

给你一棵树,每个节点有一个颜色。问删除一条边形成两棵子树,两棵子树有多少种颜色是有相同的。

启发式合并,小的合并到大的中。类似的题目有http://codeforces.com/contest/600/problem/E

 1 //#pragma comment(linker, "/STACK:102400000, 102400000") 2 #include <algorithm> 3 #include <iostream> 4 #include <cstdlib> 5 #include <cstring> 6 #include <cstdio> 7 #include <vector> 8 #include <cmath> 9 #include <ctime>10 #include <list>11 #include <set>12 #include <map>13 using namespace std;14 typedef long long LL;15 typedef pair <int, int> P;16 const int N = 1e5 + 5;17 struct Edge {18     int next, to, index;19 }edge[N << 1];20 int color[N], head[N], tot;21 int sum[N], ans[N], res[N]; //sum[color]:颜色color节点个数, ans[u]表示u点及字节点的答案, res[edge]表示边的答案22 map <int, int> cnt[N]; //cnt[u][color] 表示u点子树color颜色有多少个节点23 24 void init(int n) {25     for(int i = 1; i <= n; ++i) {26         head[i] = -1;27         sum[i] = 0;28         cnt[i].clear();29     }30     tot = 0;31 }32 33 inline void add_edge(int u, int v, int id) {34     edge[tot].next = head[u];35     edge[tot].to = v;36     edge[tot].index = id;37     head[u] = tot++;38 }39 40 void dfs(int u, int pre, int id) {41     cnt[u][color[u]] = 1;42     ans[u] = cnt[u][color[u]] < sum[color[u]] ? 1 : 0;43     for(int i = head[u]; ~i; i = edge[i].next) {44         int v = edge[i].to;45         if(v == pre)46             continue;47         dfs(v, u, edge[i].index);48         if(cnt[u].size() < cnt[v].size()) {49             swap(cnt[u], cnt[v]);50             swap(ans[u], ans[v]);51         }52         for(auto it : cnt[v]) {53             int &num = cnt[u][it.first];54             if(num == 0 && num + it.second < sum[it.first]) {55                 ++ans[u];56             } else if(num + it.second == sum[it.first] && num) { //说明此子树的it.first颜色节点个数已满57                 --ans[u];58             }59             num += it.second;60         }61     }62     res[id] = ans[u];63 }64 65 int main()66 {67     int n, u, v;68     while(scanf("%d", &n) != EOF) {69         init(n);70         for(int i = 1; i <= n; ++i) {71             scanf("%d", color + i);72             ++sum[color[i]];73         }74         for(int i = 1; i < n; ++i) {75             scanf("%d %d", &u, &v);76             add_edge(u, v, i);77             add_edge(v, u, i);78         }79         dfs(1, -1, 0);80         for(int i = 1; i < n; ++i) {81             printf("%d\n", res[i]);82         }83     }84     return 0;85 }

 

csu oj 1811: Tree Intersection (启发式合并)