HDU 3586 Information Disturbing (二分+树形dp)

Posted 2020-08-09 Recoder

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=3586

给定n个敌方据点，1为司令部，其他点各有一条边相连构成一棵树，每条边都有一个权值cost表示破坏这条边的费用，叶子节点为前线。现要切断前线和司令部的联系，每次切断边的费用不能超过上限limit，问切断所有前线与司令部联系所花费的总费用少于m时的最小limit。1<=n<=1000,1<=m<=10^6

dp[i]表示i节点为root的这个子树所破坏的最少费用，if(cost[i][i->son] <= limit) dp[i] += min(dp[i->son], cost[i][i->son]);

二分limit，然后把limit放到dfs中判断是不是都能切断叶子节点的联系。

 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 int to[N << 1], Next[N << 1], cost[N << 1], head[N], tot, dp[N], inf = 1e6 + 7;
18 
19 void init() {
20     memset(head, -1, sizeof(head));
21     tot = 0;
22 }
23 inline void add_edge(int u, int v, int c) {
24     Next[tot] = head[u];
25     to[tot] = v;
26     cost[tot] = c;
27     head[u] = tot++;
28 }
29 void dfs(int u, int p, int limit) {
30     dp[u] = inf;
31     bool inter = false;
32     for(int i = head[u]; ~i; i = Next[i]) {
33         int v = to[i];
34         if(v == p)
35             continue;
36         dfs(v, u, limit);
37         if(!inter) {
38             dp[u] = 0;
39             inter = true;
40         }
41         if(cost[i] <= limit) {
42             dp[u] += min(dp[v], cost[i]);
43         } else {
44             dp[u] += dp[v];
45         }
46     }
47 }
48 void solve() {
49     int n, m, u, v, c;
50     while(~scanf("%d %d", &n, &m) && (n || m)) {
51         init();
52         for(int i = 1; i < n; ++i) {
53             scanf("%d %d %d", &u, &v, &c);
54             add_edge(u, v, c);
55             add_edge(v, u, c);
56         }
57         int l = 1, r = 1001;
58         while(l < r) {
59             int mid = (l + r) >> 1;
60             dfs(1, -1, mid);
61             if(dp[1] <= m) {
62                 r = mid;
63             } else {
64                 l = mid + 1;
65             }
66         }
67         if(r == 1001) {
68             printf("-1\n");
69         } else {
70             printf("%d\n", l);
71         }
72     }
73 }
74 
75 int main()
76 {
77     solve();
78     return 0;
79 }