[HDU3586]Information Disturbing（DP + 二分）

传送门

题意：给定一个带权无向树，要切断所有叶子节点和1号节点（总根）的联系，每次切断边的费用不能超过上限limit，问在保证总费用<=m下的最小的limit

二分答案，再 DP，看看最终结果是否小于 m

——代码

 #include <cstdio>
 #include <cstring>
 #include <iostream>
 #define N 1001
 #define max(x, y) ((x) > (y) ? (x) : (y))

 int n, m, cnt, ans;
 int a[N], sum[N], head[N], to[N << ], val[N << ], next[N << ];
 bool vis[N];

 inline int read()
 {
     int x = , f = ;
     char ch = getchar();
     for(; !isdigit(ch); ch = getchar()) if(ch == '-') f = -;
     for(; isdigit(ch); ch = getchar()) x = (x << ) + (x << ) + ch - '';
     return x * f;
 }

 inline void add(int x, int y, int z)
 {
     to[cnt] = y;
     val[cnt] = z;
     next[cnt] = head[x];
     head[x] = cnt++;
 }

 inline void dfs(int u, int x)
 {
     int i, v;
     vis[u] = ;
     for(i = head[u]; i ^ -; i = next[i])
     {
         v = to[i];
         if(!vis[v])
         {
             dfs(v, x);
             if(val[i] <= x)
             {
                 if(val[i] < sum[v]) sum[u] += val[i];
                 else sum[u] += sum[v];
             }
             else sum[u] += sum[v];
         }
     }
 }

 inline bool check(int x)
 {
     memset(vis, , sizeof(vis));
     memset(sum, , sizeof(sum));
     for(int i = ; i <= n; i++)
         if(a[i] == )
             sum[i] = m + ;
     dfs(, x);
     return sum[] <= m;
 }

 int main()
 {
     int i, x, y, z, mid;
     while()
     {
         cnt = ;
         n = read();
         m = read();
         if(!n && !m) break;
         memset(a, , sizeof(a));
         memset(head, -, sizeof(head));
         for(i = ; i < n; i++)
         {
             a[x = read()]++;
             a[y = read()]++;
             z = read();
             add(x, y, z);
             add(y, x, z);
         }
         x = , y = m, ans = -;
         while(x <= y)
         {
             mid = (x + y) >> ;
             if(check(mid)) ans = mid, y = mid - ;
             else x = mid + ;
         }
         printf("%d\n", ans);
     }
     return ;
 }