[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.

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

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 $1$.

Sample Input 2

Sample Output 2

inf

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

Sample Input 3

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]);

	}

}