Gym 101149L Right Build

最新推荐文章于 2025-11-14 11:47:24 发布

转载最新推荐文章于 2025-11-14 11:47:24 发布 · 121 阅读

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原文链接：http://www.cnblogs.com/shenben/p/10420864.html

本文探讨了在MMORPG《Path of Exile》中，如何利用最少的天赋点数快速获取特定的两项强力天赋。通过分析游戏内角色升级过程中的依赖关系，采用图论中的最短路径算法，寻找从起始点到达目标点a和b的最优解。

In a MMORPG «Path of Exile» characters grow by acquiring talents for special talent points received in a game process. Talents can depend on others. A talent can be acquired only if a character has at least one of the talents it depends on. Acquiring one talent costs one talent point. At the beginning of the game a character has a single starting talent.

Schoolboy Vasiliy decided to record a video manual how to level-up a character in a proper way, following which makes defeating other players and NPCs very easy. He numbered all $\text{[math]}$ talents as integers from 0 to n, so that the starting talent is numbered as 0. Vasiliy thinks that the only right build is acquiring two different talents a and b as quickly as possible because these talents are imbalanced and much stronger than any others. Vasiliy is lost in thought what minimal number of talent points is sufficient to acquire these talents from the start of the game.

Input

The first line contains four space-separated integers: n, m, a and b (2 ≤ n, m ≤ 2·10⁵, 1 ≤ a, b ≤ n, a ≠ b) — the total number of talents, excluding the starting talent, the number of dependencies between talents, and the numbers of talents that must be acquired.

Each of the next m lines contains two space-separated integers: x_j and y_j (0 ≤ x_j ≤ n, 1 ≤ y_j ≤ n, x_j ≠ y_j), which means the talent y_jdepends on the talent x_j. It's guaranteed that it's possible to acquire all $\text{[math]}$ talents given the infinite number of talent points.

Output

Output a single integer — the minimal number of talent points sufficient to acquire talents a and b.

Examples
input
Copy
6 8 4 6
0 1
0 2
1 2
1 5
2 3
2 4
3 6
5 6
output
Copy
4
input
Copy
4 6 3 4
0 1
0 2
1 3
2 4
3 4
4 3
output
Copy
3

【题意】

给出一个n个点m条边的有向连通图每次只有起点可以用终点才可以用，初始时0点可以用，问最少要用多少点才能用到a点和b点

【分析】

如果只是要用一个点a，那么求0->a的最短路即可，但是要用两个点就可能0->a和0->b都不是最短路，但是由于路径重叠很多点可以被用两次以节省用的点数，但是注意到两条路径只会是开始时重叠在一次，之后分开就再也不会相交，因为如果再次相交不如之前全部重叠更优，那么我们枚举这个叉点c，求出0->c,a->c,b->c的最短路来更新最优解即可

【代码】

#include<queue>
#include<vector>
#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
const int N=2e5+5;
typedef long long ll;
#define pir pair<int,int>
#define m(s) memset(s,0,sizeof s);
int n,m,a,b,dis[3][N];
struct node{int v,w,next;}e[N<<2];int tot,head[N];bool vis[N];
struct data{int x,y,z;}ed[N];
inline void add(int x,int y,int z){
	e[++tot].v=y;e[tot].w=z;e[tot].next=head[x];head[x]=tot;
}
inline void Init(){
	scanf("%d%d%d%d",&n,&m,&a,&b);
	for(int i=1;i<=m;i++) scanf("%d%d",&ed[i].x,&ed[i].y),add(ed[i].x+1,ed[i].y+1,1);
	memset(dis[0],0x3f,sizeof dis[0]);
}
#define mp make_pair
inline void dijkstra(int S,int t){
	priority_queue<pir,vector<pir>,greater<pir> >q;
	q.push(mp(dis[t][S]=0,S));
	while(!q.empty()){
		pir p=q.top();q.pop();
		int x=p.second;
		if(vis[x]) continue;
		vis[x]=1;
		for(int i=head[x];i;i=e[i].next){
			int v=e[i].v;
			if(!vis[v]&&dis[t][v]>dis[t][x]+e[i].w){
				q.push(mp(dis[t][v]=dis[t][x]+e[i].w,v));
			}
		}
	}
}
inline void out(int x){
	for(int i=1;i<=n+1;i++) printf("%12d",dis[x][i]);puts("");
}
inline void Solve(){
	dijkstra(1,0);
	tot=0;m(head);
	for(int i=1;i<=m;i++) add(ed[i].y+1,ed[i].x+1,1);
	m(vis);memset(dis[1],0x3f,sizeof dis[1]);
	dijkstra(++a,1);
	m(vis);memset(dis[2],0x3f,sizeof dis[2]);
	dijkstra(++b,2);
	int ans=2e9;
	for(int i=1;i<=n+1;i++) ans=(int)min((ll)ans,(ll)dis[0][i]+dis[1][i]+dis[2][i]);
	printf("%d",ans);
}
int main(){
	Init();
	Solve();
	return 0;
}

转载于:https://www.cnblogs.com/shenben/p/10420864.html