Educational Codeforces Round 40 (Rated for Div. 2)

这场没打啊

A. Diagonal Walking

签到

B. String Typing

签到

C. Matrix Walk

题意

给一个x*y的矩阵A，对于每个 , Ai, j = y(i - 1) + j，矩阵中的格子一步可以走到相邻的格子，现给出路径序列，问y的大小，以及路径是否合理

分析

分析可知相邻两个点差值的绝对值不为1的话，则为y

只要注意一下，

1 2 3

4 5 6， 3->4 / 4->3 这两种不可一步互相到达即可

D. Fight Against Traffic

题意

给一个n个点，m条边的无向图，给定起点s，终点t，问现增加一条边使得s到t的最短路不变的方案数（n，m<=1e3 ）

分析

最短路？（X）

分别从s和t跑一次bfs，记录下所有点到s和t的距离，枚举所有没有给的边，check一下即可

check：如果加的这条边使得s到t的最短路变小，则一定经过这条边，只需枚举一下边的端点即可

#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<queue>
#include<map>
#include<vector>
#include<set>
#include<ctime>
#include<bits/stdc++.h>
#define LL long long
#define db double
#define EPS 1e-15
#define inf 1e16
#define pa pair<int,int>using namespace std;const int maxn = 1000+5;int ds[maxn], dt[maxn];
int n,m,s,t;
vector<int>g[maxn];
bool vis[maxn];
int mp[maxn][maxn];void bfs(int ss)
{memset(vis,0,sizeof(vis));vis[ss]=1;ds[ss]=0;queue<int>q;while(!q.empty())q.pop();q.push(ss);while(!q.empty()){int u=q.front();q.pop();for(int i=0;i<int(g[u].size());i++){if(!vis[g[u][i]]){vis[g[u][i]]=1;ds[g[u][i]]=ds[u]+1;q.push(g[u][i]);}}}
}
void bfsa(int tt)
{memset(vis,0,sizeof(vis));vis[tt]=1;dt[tt]=0;queue<int>q;while(!q.empty())q.pop();q.push(tt);while(!q.empty()){int u=q.front();q.pop();for(int i=0;i<int(g[u].size());i++){if(!vis[g[u][i]]){vis[g[u][i]]=1;dt[g[u][i]]=dt[u]+1;q.push(g[u][i]);}}}
}int main()
{int u, v;scanf("%d%d%d%d", &n, &m, &s, &t);for(int i=1;i<=m;i++){scanf("%d%d", &u, &v);g[u].push_back(v);g[v].push_back(u);mp[u][v]=mp[v][u]=1;}bfs(s);bfsa(t);int cnt=0;for(int i=1;i<=n;i++){for(int j=i+1;j<=n;j++){if(!mp[i][j]){if((ds[i]+dt[j]+1)>=ds[t] && (dt[i]+ds[j]+1)>=ds[t])cnt++;}}}printf("%d\n" ,cnt);return 0;
}

View Code

E. Water Taps

题意

分析

F. Runner's Problem

题意

分析

G. Castle Defense

题意

分析

转载于:https://www.cnblogs.com/Superwalker/p/8641400.html

Educational Codeforces Round 40 (Rated for Div. 2)相关推荐

Educational Codeforces Round 90 (Rated for Div. 2)（A, B, C, D, E）
Educational Codeforces Round 90 (Rated for Div. 2) Donut Shops 思路分三种情况: a==c/ba == c / ba==c/b这个时候两 ...
Educational Codeforces Round 89 (Rated for Div. 2)（A, B, C, D）
Educational Codeforces Round 89 (Rated for Div. 2) A. Shovels and Swords 思路题意非常简单,就是得到最多的物品嘛,我们假定a, ...
Educational Codeforces Round 100 (Rated for Div. 2)
文章目录 Educational Codeforces Round 100 (Rated for Div. 2) A. Dungeon B. Find The Array C. Busy Robot ...
Educational Codeforces Round 112 (Rated for Div. 2)-A. PizzaForces-题解
目录 Educational Codeforces Round 112 (Rated for Div. 2)-A. PizzaForces Problem Description Input Outp ...
Educational Codeforces Round 114 (Rated for Div. 2) （A ~ F）全题解
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 Educational Codeforces Round 114 (Rated for Div. 2) ...
Educational Codeforces Round 106 (Rated for Div. 2)（A ~ E）题解（每日训练 Day.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 Educational Codeforces Round 106 (Rated for Div. ...
Educational Codeforces Round 37 (Rated for Div. 2) 1
Educational Codeforces Round 37 (Rated for Div. 2) A.Water The Garden 题意:Max想给花园浇水.花园可被视为长度为n的花园床,花园 ...
Educational Codeforces Round 114 (Rated for Div. 2) D. The Strongest Build 暴力 + bfs
传送门文章目录题意: 思路: 题意: 你有nnn个装备槽,每个槽里面有cic_ici个力量加成,对于每个槽只能选一个力量加成,现在给你mmm个力量组合[b1,b2,...,bn][b_1,b_2 ...
Educational Codeforces Round 72 (Rated for Div. 2) D. Coloring Edges dfs树/拓扑找环
传送门文章目录题意: 思路: 题意: 给你一张图,你需要给这个图的边染色,保证如果有环那么这个环内边的颜色不全相同,输出染色方案和用的颜色个数. n,m≤5e3n,m\le5e3n,m≤5e3 思 ...

Educational Codeforces Round 40 (Rated for Div. 2)

Educational Codeforces Round 40 (Rated for Div. 2)相关推荐

最新文章

热门文章