L Eventual … Journey

https://codeforc.es/gym/102056 

思路：步数只有1，2，3，判断一下10相连情况； 

#include <iostream>
#include <cstring>
#include <queue>
#include <vector>
#include <algorithm>
#include <cstdio>
#include <map>
#include<bits/stdc++.h>
using namespace std;
#define sfi(i) scanf("%d",&i)
#define sfl(i) scanf("%lld",&i)
#define sfs(i) scanf("%s",(i))
#define pri(i) printf("%dn",i)
#define sff(i) scanf("%lf",&i)
#define ll long long
#define ull unsigned long long
#define mem(x,y) memset(x,y,sizeof(x))
#define INF 0x3f3f3f3f
#define eps 1e-10
#define PI acos(-1.0)
#define lowbit(x) ((x)&(-x))
#define zero(x) (((x)>0?(x):-(x))<eps)
#define fl() printf("flagn")
#define MOD(x) ((x%mod)+mod)%mod
#define endl 'n'
#define pb push_back
#define FAST_IO ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)
/*
//---------------------------------------------------------
#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/hash_policy.hpp>
#include <ext/pb_ds/priority_queue.hpp>
using namespace __gnu_pbds;
*/
//gp_hash_table<string,int>mp2;
//__gnu_pbds::priority_queue<int>q;//因为放置和std重复，故需要带上命名空间
//__gnu_pbds::priority_queue<int,greater<int>,pairing_heap_tag> pq;//最快
//----------------------------------------------------------
/*
//----------------------------------------------------------
const int BufferSize = 1 << 16;
char buffer[BufferSize], *_head, *_tail;
inline char Getchar() {
if (_head == _tail) {
int l = fread(buffer, 1, BufferSize, stdin);
_tail = (_head = buffer) + l;
}
return *_head++;
}
inline int read() {
int x = 0, f = 1;char c = Getchar();
for (;!isdigit(c);c = Getchar()) if (c == '-') f = -1;
for (;isdigit(c);c = Getchar()) x = x * 10 + c - '0';
return x * f;
}
//----------------------------------------------------------
*/
const int maxn=1e5+9;
const int maxm=4e5+9;
const int mod=998244353;
bool a[maxn];
map<int,int>vis;
int main()
{
//FAST_IO;
//freopen("input.txt","r",stdin);
int n,m;
sfi(n);
sfi(m);
int yi=0;
int li=0;
for(int i=1;i<=n;i++)
{
sfi(a[i]);
if(a[i]) yi++;
else li++;
}
int k1=0,k2=0;
for(int i=1;i<=m;i++)
{
int u,v;
sfi(u);
sfi(v);
if(a[u]!=a[v])
{
if(vis[u]==0)
{
if(a[u]==1)k1++;
else k2++;
}
if(vis[v]==0)
{
if(a[v]==1) k1++;
else k2++;
}
vis[u]++,vis[v]++;
}
}
//cout<<yi<<li<<endl;
for(int i=1;i<=n;i++)
{
int tmp=vis[i];
if(i!=1) printf(" ");
if(a[i])
{
if(tmp>0)
{
if(li>tmp)
printf("%d",yi-1+tmp+(li-tmp)*2);
else printf("%d",yi-1+tmp);
}
else
{
//if(li>k2)
printf("%d",yi-1+(li-k2)*3+2*k2);
//else printf("%d",yi-1);
}
}
else
{
if(tmp>0)
{
if(yi>tmp)
printf("%d",li-1+tmp+(yi-tmp)*2);
else printf("%d",li-1+tmp);
}
else
{
//if(yi>k1)
printf("%d",li-1+(yi-k1)*3+2*(k1));
// else printf("%d",li-1);
}
}
}
puts("");
return 0;
}

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