HDU 4118 Holiday's Accommodation（树形DP）

来源：互联网发布：淘宝视频下载器破解版编辑：程序博客网时间：2024/05/22 08:57

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4118

题意：一棵树上每个结点住着一个人，让所有的人交换结点，求最终这些人所走过的边的权值之和

思路：对于每一条边，我们都需要使它尽可能多的被走过，那么最好的走过这条边的策略就是这边的权值乘上以该边为分界的两子树中较少的点数再×2，注意long long 和递归爆栈

#include <cstdio>#include <cstring>#include <cstdlib>#include <iostream>#include <algorithm>#include <vector>#include <queue>#include <string>#include <set>#pragma comment (linker, "/STACK:1024000000,1024000000")using namespace std;const int maxn = 100010;int n, cnt;long long sum;int head[maxn];struct edge{    int to, nxt, w;} e[maxn * 2];void init(){    cnt = sum = 0;    memset(head, -1, sizeof(head));}void add(int u, int v, int w){    e[cnt].to = v;    e[cnt].w = w;    e[cnt].nxt = head[u];    head[u] = cnt++;}int dfs(int u, int fa, int w){    int siz = 1;    for(int i = head[u]; ~i; i = e[i].nxt)    {        int v = e[i].to;        if(v == fa) continue;        siz += dfs(v, u, e[i].w);    }    int res = min(siz, n - siz);    sum += (long long) res * w * 2;    return siz;}int main(){    int t;    cin >> t;    for(int ca = 1; ca <= t; ca++)    {        init();        printf("Case #%d: ", ca);        scanf("%d", &n);        for(int i = 1; i <= n - 1; i++)        {            int u, v, w;            scanf("%d%d%d", &u, &v, &w);            add(u, v, w);            add(v, u, w);        }        dfs(1, -1, 0);        cout << sum << endl;    }    return 0;}

0 0