HDU

树形dp

结构体数组存两点之间关系

dp [ i ] 表示以 i 结点为结尾时 最大的盈利值，进行更新后遍历dp 数组找最大的 answer

#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <string>#include <cmath>#include <set>#include <map>#include <stack>#include <queue>#include <ctype.h>#include <vector>#include <algorithm>#include <sstream>#define PI acos(-1.0)#define in freopen("in.txt", "r", stdin)#define out freopen("out.txt", "w", stdout)using namespace std;typedef long long ll;const int maxn = 2e5 + 7, INF = 0x3f3f3f3f, mod = 1e9 + 7;int T, n, pos, a[maxn], dp[maxn];struct node {    int x, y, cost;}g[maxn];void init() {    pos = 0;    for(int i = 1; i <= n; ++i)        scanf("%d", &a[i]);    int u, v, co;    for(int i = 1; i < n; ++i) {        scanf("%d%d%d", &u, &v, &co);        g[pos].x = u, g[pos].y = v, g[pos].cost = co;pos++;        g[pos].x = v, g[pos].y = u, g[pos].cost = co;pos++;    }}int main() {    scanf("%d", &T);    while(T--) {        scanf("%d", &n);        init();        memset(dp, 0, sizeof dp);        for(int i = 0; i < pos; ++i) {            dp[g[i].y] = max(dp[g[i].y], dp[g[i].x] - a[g[i].x] - g[i].cost + a[g[i].y]);        }        int ans = 0;        for(int i = 0; i < pos; ++i)            ans = max(ans, dp[i]);        cout << ans << endl;    }    return 0;}