[ABC280F] Pay or Receive

Problem Statement

There are $N$ towns numbered $1,\ldots,N$ and $M$ roads numbered $1,\ldots,M$.

Road $i$ connects towns $A_i$ and $B_i$. When you use a road, your score changes as follows:

• when you move from town $A_i$ to town $B_i$ using road $i$, your score increases by $C_i$; when you move from town $B_i$ to town $A_i$ using road $i$, your score decreases by $C_i$.

Your score may become negative.

Answer the following $Q$ questions.

• If you start traveling from town $X_i$ with initial score $0$, find the maximum possible score when you are at town $Y_i$.
  
  Here, if you cannot get from town $X_i$ to town $Y_i$, print nan instead; if you can have as large a score as you want when you are at town $Y_i$, print inf instead.

Constraints

• $2\leq N \leq 10^5$
• $0\leq M \leq 10^5$
• $1\leq Q \leq 10^5$
• $1\leq A_i,B_i,X_i,Y_i \leq N$
• $0\leq C_i \leq 10^9$
• All values in the input are integers.

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Input

The input is given from Standard Input in the following format:

$N$ $M$ $Q$
$A_1$ $B_1$ $C_1$
$\vdots$
$A_M$ $B_M$ $C_M$
$X_1$ $Y_1$
$\vdots$
$X_Q$ $Y_Q$

Output

Print $Q$ lines as specified in the Problem Statement.

The $i$-th line should contain the answer to the $i$-th question.

---

Sample Input 1

5 5 3
1 2 1
1 2 2
3 4 1
4 5 1
3 5 2
5 3
1 2
3 1

Sample Output 1

-2
inf
nan

For the first question, if you use road $5$ to move from town $5$ to town $3$, you can have a score $-2$ when you are at town $3$.

Since you cannot make the score larger, the answer is $-2$.

For the second question, you can have as large a score as you want when you are at town $2$ if you travel as follows:
repeatedly "use road $2$ to move from town $1$ to town $2$ and then use road $1$ to move from town $2$ to town $1$" as many times as you want,
and finally use road $2$ to move from town $1$ to town $2$.

For the third question, you cannot get from town $3$ to town $1$.

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Sample Input 2

2 1 1
1 1 1
1 1

Sample Output 2

inf

The endpoints of a road may be the same, and so may the endpoints given in a question.

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Sample Input 3

9 7 5
3 1 4
1 5 9
2 6 5
3 5 8
9 7 9
3 2 3
8 4 6
2 6
4 3
3 8
3 2
7 9

Sample Output 3

inf
nan
nan
inf
-9

nan 明显就是不同连通块的情况，而当且仅当一个连通块中存在的环都是0环，他这个连通块的点的答案才不是 inf。

但是怎么判断一个连通块是否存在非0环呢？其实可以从某一个点开始搜索，如果到达点 \(x\) 存在两条长度不相等的路径，那么就一定存在非0环。否则就无环或者只有0环。

那么现在已经确定了起始点到某个点的距离了，设起始点为 \(a\) 到点 \(x\) 距离为 \(dis_x\),点 \(x\) 到 点 \(y\) 的距离易得为 \(dis_y-dis_x\)。这是因为没有0环，所有 \(x\) 到 \(y\) 的路径都是同样距离，，当中存在一条路径为 \(x\rightarrow a\rightarrow y\)。

#include<bits/stdc++.h>
typedef long long LL;
const int N=1e5+5;
struct edge{
	int v,nxt,w;
}e[N<<1];
int n,m,q,u,v,w,fa[N],hd[N],e_num,vs[N];
LL dp[N];
void add_edge(int u,int v,int w)
{
	e[++e_num]=(edge){v,hd[u],w};
	hd[u]=e_num;
}
int find(int x)
{
	if(fa[x]==x)
		return x;
	return fa[x]=find(fa[x]);
}
void dfs(int x,LL w)
{
	if(dp[x]==dp[0])
		dp[x]=w;
	else
	{
		if(dp[x]!=w)
			vs[find(x)]=1;
		return;
	}
	for(int i=hd[x];i;i=e[i].nxt)
		dfs(e[i].v,w+e[i].w);
}
int main()
{
	memset(dp,-0x7f,sizeof(dp));
	scanf("%d%d%d",&n,&m,&q);
	for(int i=1;i<=n;i++)
		fa[i]=i;
	for(int i=1;i<=m;i++)
	{
		scanf("%d%d%d",&u,&v,&w);
		add_edge(u,v,w);
		add_edge(v,u,-w);
		fa[find(u)]=find(v);
	}
	for(int i=1;i<=n;i++)
		if(fa[i]==i)
			dfs(i,0);
	while(q--)
	{
		scanf("%d%d",&u,&v);
		if(find(u)!=find(v))
			printf("nan\n");
		else if(vs[find(u)])
			printf("inf\n");
		else
			printf("%lld\n",dp[v]-dp[u]);
	}
}