【HDOJ】1011 Starship Troopers
第一道树形DP。很容易理解。
#include <cstdio>
#include <cstring>
#include <cstdlib> #define MAXN 105 typedef struct {
int n, p;
} room_t; room_t rooms[MAXN];
int adj[MAXN][MAXN];
int dp[MAXN][MAXN];
bool visit[MAXN];
int n, m; int max(int a, int b) {
return a>b ? a:b;
} void dfs(int r) {
int i, j, k, num, v; visit[r] = true;
num = (rooms[r].n+) / ;
for (i=num; i<=m; ++i)
dp[r][i] = rooms[r].p; for (i=; i<=adj[r][]; ++i) {
v = adj[r][i];
if (visit[v])
continue;
dfs(v);
for (j=m; j>=num; --j) {
for (k=; k+j<=m; ++k) {
if (dp[v][k]) {
dp[r][j+k] = max(dp[r][j+k], dp[r][j]+dp[v][k]);
}
}
}
}
} int main() {
int i, j, k; #ifndef ONLINE_JUDGE
freopen("data.in", "r", stdin);
#endif while (scanf("%d %d", &n, &m) != EOF) {
if (n==- && m==-)
break;
memset(adj, , sizeof(adj));
memset(visit, false, sizeof(visit));
memset(dp, , sizeof(dp));
for (i=; i<=n; ++i) {
scanf("%d %d", &rooms[i].n, &rooms[i].p);
}
for (i=; i<n; ++i) {
scanf("%d %d", &j, &k);
++adj[j][];
++adj[k][];
adj[j][adj[j][]] = k;
adj[k][adj[k][]] = j;
}
if (m == ) {
printf("0\n");
continue;
}
dfs();
printf("%d\n", dp[][m]);
} return ;
}
【HDOJ】1011 Starship Troopers的更多相关文章
- HDU 1011 Starship Troopers【树形DP/有依赖的01背包】
You, the leader of Starship Troopers, are sent to destroy a base of the bugs. The base is built unde ...
- hdu 1011 Starship Troopers(树形背包)
Starship Troopers Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Other ...
- HD 1011 Starship Troopers(树上的背包)
Starship Troopers Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Other ...
- [HDU 1011] Starship Troopers
Starship Troopers Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Other ...
- 杭电OJ——1011 Starship Troopers(dfs + 树形dp)
Starship Troopers Problem Description You, the leader of Starship Troopers, are sent to destroy a ba ...
- hdu 1011 Starship Troopers(树形DP入门)
Starship Troopers Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Other ...
- hdu 1011 Starship Troopers 树形背包dp
Starship Troopers Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Other ...
- hdu 1011 Starship Troopers 经典的树形DP ****
Starship Troopers Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Other ...
- 【51NOD-0】1011 最大公约数GCD
[算法]欧几里德算法 #include<cstdio> int gcd(int a,int b) {?a:gcd(b,a%b);} int main() { int a,b; scanf( ...
随机推荐
- Android 开发经验
学习社区 eoe移动开发者社区 (link) 链接:http://www.eoeandroid.com/ 环境配置 Cocos2d-x 3.x 全平台新手开发配置教程 链接:http://www.co ...
- codeforces 148D之概率DP
http://codeforces.com/problemset/problem/148/D D. Bag of mice time limit per test 2 seconds memory l ...
- [Javascript] Advanced Reduce: Additional Reducer Arguments
Sometimes we need to turn arrays into new values in ways that can't be done purely by passing an acc ...
- STL之hash_set和hash_map
Contents 1 hash_set和hash_map的创建与遍历 2 hash_set和hash_map的查找 3 建议 一句话hash_set和hash_map:它们皆由Hashtable(St ...
- NYOJ-569最大公约数之和
题目链接:http://acm.nyist.net/JudgeOnline/problem.php?pid=569 此题目可以用筛选法的思想来做,但是用到一个欧拉函数 gcd(1,12)=1,gcd( ...
- spring源码分析构建
命令如下: ant ant install-maven ant jar package E:\download\spring-framework-3.1.3.RELEASE\build-spring- ...
- PHP常用代码大全
1.连接MYSQL数据库代码 <?php $connec=mysql_connect("localhost","root","root" ...
- new date() 函数在浏览器中的兼容问题!!
引言: 同一种语言javascript,在不同的浏览器中,存在语言兼容性问题,本质上是由于不同的浏览器是支持的语言标准和实现上各有差异.本文将基于new Date来创建Date对象来分析这个问题. v ...
- 你好,C++(25)函数调用和它背后的故事5.1.2 函数调用机制
5.1.2 函数调用机制 在前面的学习中,我们多次提到了“调用函数”的概念.所谓调用函数,就是将程序的执行控制权从调用者(某个函数)交给被调用的函数,同时通过参数向被调用的函数传递数据,然后程序进入 ...
- jQuery get/post区别及contentType取值
1.GET访问 浏览器 认为 是等幂的 就是 一个相同的URL 只有一个结果[相同是指 整个URL字符串完全匹配]所以 第二次访问的时候 如果 URL字符串没变化浏览器是直接拿出了第一次访问的结果,表 ...