hdoj 3660 Alice and Bob's Trip

類型：樹形DP

題目：http://acm.hdu.edu.cn/showproblem.php?pid=3660

來源：2010 Asia Regional Harbin

思路：令狀態dp[i]表示以i爲根時到葉子節點的最優值，當i爲bob時，取最大值，當i爲alice時，取最小值。用dist[i]記錄從0節點到當前節點的路徑的長度。

當搜索到某條邊(u, v)時，根據題意知始終有：dist[u] + dp[v] + p[i].w >= l && dist[u] + dp[v] + p[i].w <= r成立。根據該條件判斷是否選擇當前邊。

// hdoj 3660 Alice and Bob's Trip
// tle ac 796MS 10012K
#include <iostream>
#include <string>
#include <algorithm>
#include <cmath>
#include <cstdio>
#include <cstring>
using namespace std;

#define FOR(i,a,b) for(i = (a); i < (b); ++i)
#define FORE(i,a,b) for(i = (a); i <= (b); ++i)
#define CLR(a,b) memset(a,b,sizeof(a))

const int MAXN = 500010;
const int INF = 0x7f7f7f7f;

int dp[MAXN], head[MAXN], dist[MAXN];
int cnt, n, r, l, minsum;
struct node {
    int v, nxt, w;
}p[MAXN];

void dfs(int u, int sign) {
    for(int i = head[u]; i != -1; i = p[i].nxt) {
        int v = p[i].v;
        dist[v] = dist[u] + p[i].w;
        dfs(p[i].v, !sign);
        //!!!
        if(dist[u] + dp[v] + p[i].w >= l && dist[u] + dp[v] + p[i].w <= r) {
            if(dp[u] == 0) {
                dp[u] = dp[v] + p[i].w;
                continue;
            }
            if(sign == 1) {
                if(dp[u] < dp[v] + p[i].w)
                    dp[u] = dp[v] + p[i].w;
            }
            else if(dp[u] > dp[v] + p[i].w)
                dp[u] = dp[v] + p[i].w;
        }
    }
}

int getint() {
    char c=getchar();
    int t=0;
    while (c<'0' || c>'9') {
        c=getchar();
    }
    while (c>='0' && c<='9') {
        t=t*10+c-'0';
        c=getchar();
    }
    return t;
}

int main() {
    int i;
    int u, v, w;

    while(scanf("%d %d %d", &n, &l, &r) != EOF) {
        CLR(head, -1);
        cnt = 0;
        CLR(dp, 0);
        CLR(dist, 0);
        FOR(i, 0, n - 1) {
            u = getint(), v = getint(), w = getint();
            p[cnt].v = v;
            p[cnt].w = w;
            p[cnt].nxt = head[u];
            head[u] = cnt++;
        }
        dfs(0, 1);
        if(dp[0] == 0)
            printf("Oh, my god!\n");
        else
            printf("%d\n", dp[0]);
    }
    return 0;
}