bzoj3875 [Ahoi2014]骑士游戏

传送门：http://www.lydsy.com/JudgeOnline/problem.php?id=3875

【题解】

容易列出dp，但是是有后效性，我们用spfa来去除这种后效性即可，如果一个点可更新，把前面所有点加进来即可。

# include <queue>
# include <stdio.h>
# include <string.h>
# include <iostream>
# include <algorithm>
// # include <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
const int M = 2e6 + 10, N = 5e5 + 10;
const int mod = 1e9+7;

# define RG register
# define ST static

int n;
ll S[N];
int head[N], nxt[M], to[M], tot=0;
inline void add(int u, int v) {
    ++tot; nxt[tot] = head[u];
    head[u] = tot; to[tot] = v;
}

int head2[N];
inline void add2(int u, int v) {
    ++tot; nxt[tot] = head2[u];
    head2[u] = tot; to[tot] = v;
}

queue<int> q;
ll d[M];
bool vis[M];
inline void spfa() {
    while(!q.empty()) q.pop();
    for (int i=1; i<=n; ++i) q.push(i), vis[i] = 1;
    while(!q.empty()) {
        int top = q.front(); q.pop(); vis[top] = 0;
        ll a = S[top];
        for (int i=head[top]; i; i=nxt[i]) a += d[to[i]];
        if(a >= d[top]) continue;
        d[top] = a;
        for (int i=head2[top]; i; i=nxt[i])
            if(!vis[to[i]]) q.push(to[i]), vis[to[i]] = 1;
    }
}

int main() {
    cin >> n;
    for (int i=1, t, tt; i<=n; ++i) {
        scanf("%lld%lld%d", &S[i], &d[i], &t);
        while(t--) {
            scanf("%d", &tt);
            add(i, tt);
            add2(tt, i);
        }
    }
    spfa();
    cout << d[1];
    return 0;
}