这题首先每道菜如果够吃，就让吃它的人放到最后只吃这道菜，那对其他菜来说，就少了一些吃它的人，就又能有菜富裕，这样不断进行，如果每道菜都能在某个阶段富裕，那就alive，然后按照存入的顺序倒序输出，否则dead
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| Submission | Time | Verdict |
|---|---|---|
| 84867136 | Jun/24/2020 11:06 | Accepted |
→ Problem tags
constructive algorithms
data structures
greedy
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→ Contest materials
E. DeadLee
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
Lee bought some food for dinner time, but Lee's friends eat dinner in a deadly way. Lee is so scared, he doesn't want to die, at least not before seeing Online IOI 2020...
There are n
different types of food and m Lee's best friends. Lee has wi plates of the i-th type of food and each friend has two different favorite types of food: the i-th friend's favorite types of food are xi and yi (xi≠yi
).
Lee will start calling his friends one by one. Whoever is called will go to the kitchen and will try to eat one plate of each of his favorite food types. Each of the friends will go to the kitchen exactly once.
The only problem is the following: if a friend will eat at least one plate of food (in total) then he will be harmless. But if there is nothing left for him to eat (neither xi
nor yi), he will eat Lee instead ×_×
.
Lee can choose the order of friends to call, so he'd like to determine if he can survive dinner or not. Also, he'd like to know the order itself.
Input
The first line contains two integers n
and m (2≤n≤105; 1≤m≤2⋅105
) — the number of different food types and the number of Lee's friends.
The second line contains n
integers w1,w2,…,wn (0≤wi≤106
) — the number of plates of each food type.
The i
-th line of the next m lines contains two integers xi and yi (1≤xi,yi≤n; xi≠yi) — the favorite types of food of the i
-th friend.
Output
If Lee can survive the dinner then print ALIVE (case insensitive), otherwise print DEAD (case insensitive).
Also, if he can survive the dinner, print the order Lee should call friends. If there are multiple valid orders, print any of them.
Examples
Input
Copy
3 3 1 2 1 1 2 2 3 1 3
Output
Copy
ALIVE 3 2 1
Input
Copy
3 2 1 1 0 1 2 1 3
Output
Copy
ALIVE 2 1
Input
Copy
4 4 1 2 0 1 1 3 1 2 2 3 2 4
Output
Copy
ALIVE 1 3 2 4
Input
Copy
5 5 1 1 1 2 1 3 4 1 2 2 3 4 5 4 5
Output
Copy
ALIVE 5 4 1 3 2
Input
Copy
4 10 2 4 1 4 3 2 4 2 4 1 3 1 4 1 1 3 3 2 2 1 3 1 2 4
Output
Copy
DEAD
Note
In the first example, any of the following orders of friends are correct : [1,3,2]
, [3,1,2], [2,3,1], [3,2,1]
.
In the second example, Lee should call the second friend first (the friend will eat a plate of food 1
) and then call the first friend (the friend will eat a plate of food 2). If he calls the first friend sooner than the second one, then the first friend will eat one plate of food 1 and food 2 and there will be no food left for the second friend to eat.
#include<stdio.h>
#include<algorithm>
#include<iostream>
#include<string>
#include<queue>
#include<string.h>
#include<map>
#define ll long long
//#define ll int
#include <iostream>
#include <math.h>
using namespace std;
#define maxn 300005
#include<vector>
struct node
{
int id,x;
};
vector<node>e[maxn];
queue<int>q;
int d[maxn],w[maxn],ans[maxn],vis[maxn];
int main()
{
int n,m;
scanf("%d %d",&n,&m);
for(int i=1;i<=n;i++)
scanf("%d",&w[i]);
for(int i=1;i<=m;i++)
{
int x,y;
scanf("%d %d",&x,&y);
e[x].push_back({i,y});
e[y].push_back({i,x});
d[x]++;d[y]++;
}
for(int i=1;i<=n;i++)
{
if(d[i]<=w[i])
{
q.push(i);
}
}
int f=m;
while(!q.empty())
{
int i=q.front();
q.pop();
//printf("i=%d\n",i);
for(auto j:e[i])
{
if(vis[j.id]==1)
continue;
vis[j.id]=1;
d[j.x]--;
//printf("j.id=%d j.x=%d\n",j.id,j.x);
if(d[j.x]<=w[j.x]) q.push(j.x);
ans[f--]=j.id;
}
}
if(f)
{
printf("DEAD\n");
return 0;
}
printf("ALIVE\n");
for(int i=1;i<=m;i++)
{
printf("%d",ans[i]);
printf("%c",i==n?'\n':' ');
}
}