Never Wait for Weights
In a laboratory, an assistant, Nathan Wada, is measuring weight differences between sample pieces pair by pair. He is using a balance because it can more precisely measure the weight difference between two samples than a spring scale when the samples have nearly
the same weight.
He is occasionally asked the weight differences between pairs of samples. He can or cannot answer based on measurement results already obtained.
Since he is accumulating a massive amount of measurement data, it is now not easy for him to promptly tell the weight differences. Nathan asks you to develop a program that records measurement results and automatically tells the weight differences.
Input
The input consists of multiple datasets. The first line of a dataset contains two integers N and M. N denotes
the number of sample pieces (2
N100,
000). Each sample is assigned a unique number from 1 to N as
an identifier. The rest of the dataset consists of M lines (1M100,
000), each of which corresponds to either a measurement result or an inquiry. They are given in chronological order.
A measurement result has the format,
! a b w
which represents the sample piece numbered b is heavier than one numbered a by w micrograms (a 
b).
That is, w =wb - wa,
where wa and wb are
the weights of a and b,
respectively. Here, w is a non-negative integer not exceeding 1,000,000.
You may assume that all measurements are exact and consistent.
An inquiry has the format,
? a b
which asks the weight difference between the sample pieces numbered a and b (a b).
The last dataset is followed by a line consisting of two zeros separated by a space.
Output
For each inquiry, ? a b,
print the weight difference in micrograms between the sample pieces numbered a and b, wb - wa,
followed by a newline if the weight difference can be computed based on the measurement results prior to the inquiry. The difference can be zero, or negative as well as positive. You can assume that its absolute value is at most 1,000,000. If the difference
cannot be computed based on the measurement results prior to the inquiry, print `UNKNOWN' followed by a newline.
Sample Input
2 2 ! 1 2 1 ? 1 2 2 2 ! 1 2 1 ? 2 1 4 7 ! 1 2 100 ? 2 3 ! 2 3 100 ? 2 3 ? 1 3 ! 4 3 150 ? 4 1 0 0
Sample Output
1 -1 UNKNOWN 100 200 -50
并查集求集合内任意两点距离，find函数里面稍微调整下，dis[a][b]=s[a]-s[b]，任意两点的距离等于这两点到根的距离之差。
代码：
#include<iostream>
#include<vector>
#include<string>
#include<queue>
#include<cstdio>
#include<cstring>
#define maxn 100005
#define INF 0xfffffff
using namespace std;
int n,m;
int f[maxn],s[maxn];
char str[10];
void init()
{
for(int i=1;i<=n;i++) f[i]=i;
memset(s,0,sizeof(s));
}
int find(int x)
{
if(x==f[x]) return x;
int xx=find(f[x]);
s[x]+=s[f[x]];//更新祖先改变后的值
return f[x]=xx;
}
int main()
{
int a,b,w,p,q;
while(scanf("%d%d",&n,&m)==2)
{
if(!n&&!m) break;
init();
for(int i=1;i<=m;i++)
{
scanf("%s",str);
scanf("%d%d",&a,&b);
if(str[0]=='!')
{
scanf("%d",&w);
p=find(a),q=find(b);
f[p]=q;
s[p]=-s[a]+w+s[b];
}
else if(str[0]=='?')
{
if(find(a)==find(b))
printf("%d\n",s[a]-s[b]);
else
printf("UNKNOWN\n");
}
}
}
}