Magic Maze dfs + dp

http://swjtuoj.cn/problem/2387/

设dp[cur]表示以cur这个节点为起点的时候，能走的最大贡献。

题目保证没环，也就是没回路。

感觉和树dp差不多了。

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <assert.h>
#define IOS ios::sync_with_stdio(false)
using namespace std;
#define inf (0x3f3f3f3f)
typedef long long int LL;


#include <iostream>
#include <sstream>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <bitset>
const int maxn = 2000 + 20;
struct Node {
    int u, v, w;
    int tonext;
}e[maxn * maxn];
int first[maxn], num;
void add(int u, int v, int w) {
    ++num;
    e[num].u = u, e[num].v = v, e[num].w = w;
    e[num].tonext = first[u];
    first[u] = num;
}
int vis[maxn], dp[maxn], DFN;
int dfs(int cur) {
    if (vis[cur] == DFN) return dp[cur];
    vis[cur] = DFN;
    int res = 0;
    for (int i = first[cur]; i; i = e[i].tonext) {
        int v = e[i].v;
        res = max(res, dfs(v) + e[i].w);
    }
    return dp[cur] = res;
}
void work() {
    memset(first, 0, sizeof first);
    int n, m;
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= m; ++i) {
        int u, v, w;
        scanf("%d%d%d", &u, &v, &w);
        u++;
        v++;
        add(u, v, w);
    }
    ++DFN;
    int ans = 0;
    memset(dp, 0, sizeof dp);
    for (int i = 1; i <= n; ++i) {
        if (vis[i] == DFN) continue;
        dp[i] = dfs(i);
    }
    for (int i = 1; i <= n; ++i) ans = max(ans, dp[i]);
    printf("%d\n", ans);
}

int main() {
#ifdef local
    freopen("data.txt", "r", stdin);
//    freopen("data.txt", "w", stdout);
#endif
    int t;
    scanf("%d", &t);
    while (t--) work();
    return 0;
}