标签:air mes type pair mmm 代码 其他 fine class
Sereja and the Arrangement of Numbers
题解:
ummm。
在一副图中,如果全部点的度数是偶数/只有2个点是奇数,则能一笔画。
考虑图的点数k为奇数的时候,那么每个点的度数都是偶数点,所以就是可以一笔画,答案为 1 +k * (i - kll) / 2;
k为偶数的时候,所有的点是奇数点,我们保留2个点是奇数点,将其他的点改为偶数点,就可以一笔画了。 1 +k * (i - kll) / 2 + k/2 - 1.
代码:
#include<bits/stdc++.h> using namespace std; #define Fopen freopen("_in.txt","r",stdin); freopen("_out.txt","w",stdout); #define LL long long #define ULL unsigned LL #define fi first #define se second #define pb push_back #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 #define lch(x) tr[x].son[0] #define rch(x) tr[x].son[1] #define max3(a,b,c) max(a,max(b,c)) #define min3(a,b,c) min(a,min(b,c)) typedef pair<int,int> pll; const int inf = 0x3f3f3f3f; const int _inf = 0xc0c0c0c0; const LL INF = 0x3f3f3f3f3f3f3f3f; const LL _INF = 0xc0c0c0c0c0c0c0c0; const LL mod = (int)1e9+7; const int N = 1e5 + 100; int n, m; int a[N], b[N]; LL dp[N]; int main(){ scanf("%d%d", &n, &m); for(int i = 1; i <= m; ++i){ scanf("%d%d", &a[i], &b[i]); if(i&1) dp[i] = 1 + i * (i - 1ll) / 2; else dp[i] = 1 + i * (i-1ll)/2 + i/2 - 1; } int k = m; while(dp[k] > n) --k; sort(b+1, b+1+m, greater<int>()); LL ans = 0; for(int i = 1; i <= k; ++i) ans += b[i]; cout << ans << endl; return 0; }
CodeForces 367 C Sereja and the Arrangement of Numbers 欧拉回路
标签:air mes type pair mmm 代码 其他 fine class
原文地址:https://www.cnblogs.com/MingSD/p/10868609.html