[题解] [JSOI2014] 骑士游戏

题面

题解

有一个很简单的 DP 式

[f[i] = min(k[i], s[i] + sum f[j]) ]

其中 (j) 是普攻 (i) 后产生的小怪编号
但是这样转移可能有环
我们考虑使用最短路转移, 对于一对 ((i, j)) 连边 (i o j)
然后初始化每个点都为法术攻击, 如果当前点被更新了, 就拿他去更新他的前驱

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
#include <vector>
#include <queue>
typedef long long ll; 
const int N = 2e5 + 5;
const int M = 1e6+ 5; 
using namespace std;

int n, head[N], headf[N], cnt, tot;
ll s[N], k[N], dis[N];
struct edge { int to, nxt; } e[M], fe[M]; 
bool vis[N];
vector<int> t[N], ft[N]; 
queue<int> q; 

template < typename T >
inline T read()
{
    T x = 0, w = 1; char c = getchar();
    while(c < '0' || c > '9') { if(c == '-') w = -1; c = getchar(); }
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * w; 
}

inline void adde(int u, int v)
{
    e[++cnt] = (edge) { v, head[u] }, head[u] = cnt;
    fe[++tot] = (edge) { u, headf[v] }, headf[v] = cnt; 
}

ll SPFA()
{
    for(int i = 1; i <= n; i++)
	dis[i] = k[i], q.push(i), vis[i] = 1;
    ll tmp; 
    while(!q.empty())
    {
	int u = q.front(); q.pop();
	vis[u] = 0, tmp = 0; 
/*	for(int v, i = head[u]; i; i = e[i].nxt)
	    v = e[i].to, tmp += dis[v];
	tmp += s[u];
	if(tmp < dis[u])
	{
	    dis[u] = tmp; 
	    for(int f, i = headf[u]; i; i = fe[i].nxt)
	    {
		f = fe[i].to;
		if(!vis[f]) q.push(f), vis[f] = 1; 
	    }
	}
*/
	for(int i = 0; i < t[u].size(); i++)
	    tmp += dis[t[u][i]];
	tmp += s[u];
	if(dis[u] <= tmp) continue;
	dis[u] = tmp;
	for(int i = 0; i < ft[u].size(); i++)
	    if(!vis[ft[u][i]]) vis[ft[u][i]] = 1, q.push(ft[u][i]); 
    }
    return dis[1]; 
}

int main()
{
    n = read <int> ();
    for(int num, i = 1; i <= n; i++)
    {
	s[i] = read <ll> (), k[i] = read <ll> (), num = read <int> (); 
	for(int v, j = 1; j <= num; j++)
	    v = read <int> (), adde(i, v), t[i].push_back(v), ft[v].push_back(i); 
    }
    printf("%lld
", SPFA()); 
    return 0; 
}