260D Black and White Tree 優先佇列
The board has got a painted tree graph, consisting of n nodes. Let us remind you that a non-directed graph is called a tree if it is connected and doesn't contain any cycles.
Each node of the graph is painted black or white in such a manner that there aren't two nodes of the same color, connected by an edge. Each edge contains its value written on it as a non-negative integer.
A bad boy Vasya came up to the board and wrote number sv near each node v — the sum of values of all edges that are incident to this node. Then Vasya removed the edges and their values from the board.
Your task is to restore the original tree by the node colors and numbers sv.
Input
The first line of the input contains a single integer n
Output
Print the description of n - 1 edges of the tree graph. Each description is a group of three integers vi, ui, wi (1 ≤ vi, ui ≤ n, vi ≠ ui, 0 ≤ wi ≤ 109), where vi and ui — are the numbers of the nodes that are connected by the i-th edge, and wi is its value. Note that the following condition must fulfill cvi ≠ cui.
It is guaranteed that for any input data there exists at least one graph that meets these data. If there are multiple solutions, print any of them. You are allowed to print the edges in any order. As you print the numbers, separate them with spaces.
Examples
Input
3 1 3 1 2 0 5
Output
3 1 3 3 2 2
Input
6 1 0 0 3 1 8 0 2 0 3 0 0
Output
2 3 3 5 3 3 4 3 2 1 6 0 2 1 0
題解： 我們先把兩種顏色的點放在兩個優先佇列中，每次選出價值最大的，然後在把價值大的點，重新放進佇列，此時，該點的價值變小了，還要注意的是，若兩個點價值相等，並且已經出現空隊列了，那就把原先的點放進去，並且價值為0，為什麼呢？
因為可能會出現 比如 白色 還有2個點 價值為0 而黑色的點都pop掉了 所以 保證每個佇列都要有元素，剛開始寫的時候沒考慮，錯在了 第9個 納悶為什麼沒有報RT
#include<iostream>
#include<cstdio>
#include<queue>
using namespace std;
const int N=1e5+10;
struct node{
int id,val;
node(){}
node(int id,int val):id(id),val(val){}
bool operator < (const node &x)const {
return val < x.val;
}
};
struct edge{
int u,v,d;
}e[N];
int n;
int main()
{
while(~scanf("%d", &n))
{
priority_queue<node> p1,p2;
node now1,now2;
int op,val;
for(int i=1;i<=n;i++)
{
scanf("%d%d",&op,&val);
if(op==0)
p1.push(node(i,val));
else
p2.push(node(i,val));
}
int len=0;
for(int i=1;i<n;i++)
{
now1=p1.top();p1.pop();
now2=p2.top();p2.pop();
e[++len].u=now1.id;
e[len].v=now2.id;
e[len].d=min(now1.val,now2.val);
if(now1.val>now2.val)
p1.push(node(now1.id,now1.val-now2.val));
else if(now1.val<now2.val)
p2.push(node(now2.id,now2.val-now1.val));
else
{
if(p1.empty())
p1.push(node(now1.id,0));
else
p2.push(node(now2.id,0));
}
}
for(int i=1;i<n;i++)
printf("%d %d %d\n",e[i].u,e[i].v,e[i].d);
}
return 0;
}