A - Concerts

思路：dp[i][j]表示前i天看了j场演唱会。转移方程：dp[i][j]=dp[i-1][j]+dp[由h数组决定的前一场的最晚日期][j-1]。

#include<bits/stdc++.h>
using namespace std;
int h[300];
char plan[305],schedule[100005];
int k,n;
int dp[100005][305];
const int mod=1e9+7;
int main()
{while(~scanf("%d%d",&k,&n)){for(int i='A';i<='Z';++i)scanf("%d",h+i);scanf("%s%s",plan,schedule);dp[1][1]=(plan[0]==schedule[0]);for(int i=2;i<=n;++i)dp[i][1]=(plan[0]==schedule[i-1])?dp[i-1][1]+1:dp[i-1][1];for(int i=2;i<=k;++i)dp[1][i]=0;for(int j=2;j<=k;++j){for(int i=2;i<=n;++i){dp[i][j]=dp[i-1][j];if(plan[j-1]==schedule[i-1]&&i-h[plan[j-2]]>=2)dp[i][j]=(1LL*dp[i][j]+dp[i-1-h[plan[j-2]]][j-1])%mod;}}printf("%d\n",dp[n][k]);}
}

D - Harry Potter and The Vector Spell

思路：并查集

#include<bits/stdc++.h>
using namespace std;
const int maxn=1e5+5;
vector<int> a[maxn];
int n,m;
int fa[maxn];
int find(int x){return x==fa[x]?x:fa[x]=find(fa[x]);}
int main()
{while(~scanf("%d%d",&m,&n)){for(int i=1;i<=m;++i){fa[i]=i;int k,tmp;scanf("%d",&k);for(int j=1;j<=k;++j){scanf("%d",&tmp);a[tmp].push_back(i);}}int ans=0;for(int i=1;i<=n;++i){int u=a[i][0],v=a[i][1];int x=find(u),y=find(v);if(x!=y){ans++;fa[x]=y;}}printf("%d\n",ans);}
}

G - Robots

水题

K - Escape room

水题

转载于:https://www.cnblogs.com/maoruimas/p/9741027.html

2016-2017 ACM-ICPC Southeastern European Regional Programming Contest (SEERC 2016)相关推荐

2020-2021 ICPC Southeastern European Regional Programming Contest (SEERC 2020)
2020-2021 ICPC Southeastern European Regional Programming Contest (SEERC 2020) B. Reverse Game 题目描述: ...
Gym 101964 2018-2019 ACM-ICPC Southeastern European Regional Programming Contest (SEERC 2018)
传送门 A.Numbers B.Broken Watch 先考虑最简单的情况,就是a,b,c都相等的情况,这个时候答案显然只会跟n有关系,在n个线段里面选3个的情况就是C(n,3),其中有一部分不合法 ...
2018-2019 ICPC Northwestern European Regional Programming Contest (NWERC 2018)
2018-2019 ICPC Northwestern European Regional Programming Contest (NWERC 2018) 题号题目知识点难度 A Access ...
2016-2017 ACM-ICPC Southwestern European Regional Programming Contest (SWERC 2016)
A. Within Arm's Reach 留坑. B. Bribing Eve 枚举经过$1$号点的所有直线,统计直线右侧的点数,旋转卡壳即可. 时间复杂度$O(n\log n)$. #includ ...
2018-2019 ICPC Northwestern European Regional Programming Contest (NWERC 2018)记录
A. Access Points 待更 B. Brexit Negotiations 题意:给一个 DAG,最小化[每个点点权与其在拓扑序内的编号之和的最大值],输出这个最小值. 解法:反向建图,然后 ...
The 2020 ICPC Asia Shenyang Regional Programming Contest I题 Rise of Shadows（数论）
题目链接The 2020 ICPC Asia Shenyang Regional Programming Contest 题目大意: 一天内有H小时,每小时M分钟,时针分针以恒定速率旋转. 现在若时针 ...
The 2020 ICPC Asia Yinchuan Regional Programming Contest
The 2020 ICPC Asia Yinchuan Regional Programming Contest A 开三个vector数组存储x,y,z轴上的点,unique+erase去重 #in ...
The 2022 ICPC Asia Hangzhou Regional Programming Contest
比赛连接:Dashboard - The 2022 ICPC Asia Hangzhou Regional Programming Contest - Codeforces A. Modulo Rui ...
2017 ACM ICPC Asia Shenyang Regional Contest 题解（10 / 13）【每日亿题2 / 16】
整理的算法模板合集: ACM模板点我看算法全家桶系列!!! 实际上是一个全新的精炼模板整合计划目录 A.(2017 ICPC shenyang I)Little Boxes B.(2017 ICP ...

2016-2017 ACM-ICPC Southeastern European Regional Programming Contest (SEERC 2016)

A - Concerts

D - Harry Potter and The Vector Spell

G - Robots

K - Escape room

2016-2017 ACM-ICPC Southeastern European Regional Programming Contest (SEERC 2016)相关推荐

最新文章

热门文章