Codeforces915E Physical Education Lessons

题目链接:http://codeforces.com/contest/915/problem/E

题意:初始化区间1~N,全为1,N是1E9级别,Q个操作与询问。对于当前询问输出1~N还有多少个1。

操作1:把L-R这段区间全变0

操作2:吧L-R这段区间全变1

题解:把区间左右端点暴力存在set里, 对于当前l,r在set里二分下找到第一个在他右边的区间。

首先不管K=1还是K=2,都当成K=2,然后答案加上交集的长度,表示恢复了这段区间,然后删除set里当前的区间,如果L < l,表示删多了,把<L, l - 1>这段区间放进set里, R > r 也表示删除多了,把<r+1 , R>放进set里, 一直找到尾,如果发现当前set里的区间L,R与此时的l,r没有交集了,就break掉。 最后如果K=1的话,就把<l,r>放进set里,这里为了方便二分,把r当做第一优先级放在set<pari<int,int>>里。

#include <bits/stdc++.h>

using namespace std;
#define SZ(X) ((int)X.size())
#define mp make_pair
#define ALL(X) begin(X),end(x)
#define Pii pair<int,int>
using ll = long long ;
using ld = long double ;

set<Pii> S;
int main()
{
    std::ios::sync_with_stdio(false), cin.tie(0), cout.tie(0); cout << fixed;
    int n, q;
    cin >> n >> q;
    int res = n;
    while(q --) {
        int l, r, k;
        cin >> l >> r >> k;
        auto it = S.lower_bound(mp(l, 0));
        while(it != S.end()) {
            int L, R;
            tie(R, L) = *it;
            int len = min(R, r) - max(l, L) + 1;
            if(len <= 0)  break;
            res += len;
            S.erase(it++);

            if(L < l) S.insert(mp(l - 1, L));
            if(R > r) S.insert(mp(R, r + 1));
        }
        if(k == 1) {
            res -= (r - l + 1);
            S.insert(mp(r, l));
        }
        cout << res << "\n";
    }
    return 0;
}


### Codeforces 887E Problem Solution and Discussion The problem **887E - The Great Game** on Codeforces involves a strategic game between two players who take turns to perform operations under specific rules. To tackle this challenge effectively, understanding both dynamic programming (DP) techniques and bitwise manipulation is crucial. #### Dynamic Programming Approach One effective method to approach this problem utilizes DP with memoization. By defining `dp[i][j]` as the optimal result when starting from state `(i,j)` where `i` represents current position and `j` indicates some status flag related to previous moves: ```cpp #include <bits/stdc++.h> using namespace std; const int MAXN = ...; // Define based on constraints int dp[MAXN][2]; // Function to calculate minimum steps using top-down DP int minSteps(int pos, bool prevMoveType) { if (pos >= N) return 0; if (dp[pos][prevMoveType] != -1) return dp[pos][prevMoveType]; int res = INT_MAX; // Try all possible next positions and update 'res' for (...) { /* Logic here */ } dp[pos][prevMoveType] = res; return res; } ``` This code snippet outlines how one might structure a solution involving recursive calls combined with caching results through an array named `dp`. #### Bitwise Operations Insight Another critical aspect lies within efficiently handling large integers via bitwise operators instead of arithmetic ones whenever applicable. This optimization can significantly reduce computation time especially given tight limits often found in competitive coding challenges like those hosted by platforms such as Codeforces[^1]. For detailed discussions about similar problems or more insights into solving strategies specifically tailored towards contest preparation, visiting forums dedicated to algorithmic contests would be beneficial. Websites associated directly with Codeforces offer rich resources including editorials written after each round which provide comprehensive explanations alongside alternative approaches taken by successful contestants during live events. --related questions-- 1. What are common pitfalls encountered while implementing dynamic programming solutions? 2. How does bit manipulation improve performance in algorithms dealing with integer values? 3. Can you recommend any online communities focused on discussing competitive programming tactics? 4. Are there particular patterns that frequently appear across different levels of difficulty within Codeforces contests?
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