Codeforces Round #207 (Div. 2)

A. Group of Students

链接：http://codeforces.com/contest/357/problem/A

描述：一共m(2<=m<=100)个人，c1，c2，c3....cm(0<=ci<=100)代表分数为1，2，3...m的人数。求一个分界分数点，使得该分界点左边所有人数和右边所有人数（右边包括分界点人数）都在x和y之内（包括x，y).(1<=x<=y<=10000).

 思路：水题，直接模拟。

#include <iostream>#include <cstdio>int main(){int m;while (scanf("%d", &m) != EOF){int a[110] = {0}, sum = 0;for (int i = 1; i <= m; ++i){scanf("%d", a + i);sum += a[i];}int x, y;scanf("%d %d", &x, &y);int sum1 = 0, sum2;int flag = 0;for (int i = 1; i <= m; ++i){sum1 += a[i-1];sum2 = sum - sum1;if (sum1 >= x && sum1 <= y && sum2 >= x && sum2 <= y){flag = i;break;}}printf("%d\n", flag);}return 0;}

B. Flag Day

链接：http://codeforces.com/contest/357/problem/B

描述：在一场舞会上，有m(1<=m<=10^5)个dance，n(3<=n<=10^5)个dancer。每个dance要求必须3人参加，且三个人的衣服颜色各是白，红，蓝色。由于n可能小于3m，所以每个人有可能参加多个dance。但是每个dance最多只有一个dancer参加过以前的dance。任务就是输出这n个人的衣服颜色（1 for white, 2 for red, 3 for blue）使得条件满足。

思路：一开始想多了，想用dfs染色。但是后来注意到一个条件非常重要“ Your agency cannot allow that to happen, so each dance has at most one dancer who has danced in some previous dance.”这样一来就变成一个大水题了。只要记录以前被染色过的点，以后遇到一个三人组里有一个被染色的，就随便给另外两个染色就可以了。要是三人都没有被染色，就可以随便给这仨货染色。看题要仔细，抓住一个信息就变成大水题了。代码有点逆天，还好一次过。

#include <iostream>#include <cstdio>#include <cstring>#include <vector>#include <set>using namespace std;const int M = 100000 + 10;int color[M];//记录颜色，0表示未染色int main(){int n, m;int a, b, c;while (scanf("%d %d", &n, &m) != EOF){memset(color, 0, sizeof(color));for (int i = 1; i <= m; ++i){scanf("%d %d %d", &a, &b, &c);if (i == 1 || (!color[a] && !color[b] && !color[c])){color[a] = 1, color[b] = 2, color[c] = 3;continue;}if (color[a]){if (color[a] == 1)color[b] = 2, color[c] = 3;if (color[a] == 2)color[b] = 1, color[c] = 3;if (color[a] == 3)color[b] = 1, color[c] = 2;}else if (color[b]){if (color[b] == 1)color[a] = 2, color[c] = 3;if (color[b] == 2)color[a] = 1, color[c] = 3;if (color[b] == 3)color[a] = 1, color[c] = 2;}else if (color[c]){if (color[c] == 1)color[a] = 2, color[b] = 3;if (color[c] == 2)color[a] = 1, color[b] = 3;if (color[c] == 3)color[a] = 1, color[b] = 2;}}for (int i = 1; i < n; ++i)printf("%d ", color[i]);printf("%d\n", color[n]);}return 0;}

C. Knight Tournament

链接：http://codeforces.com/contest/357/problem/C

描述：骑士锦标赛。有n(2<=n<=3*10^5)个骑士，每人被唯一分配一个一到n之间的数字。有m(1<=m<=3*10^5)场比赛，每场比赛编号至少为li，至多为ri的一些人比赛，胜者为xi.(1<=li<ri<=n, li<=xi<=ri)，并且胜者留下，其他人离开不再参赛。第m场会有一个唯一的胜者。要求输出从1到n每个人被打败的编号，即战胜这个人的编号。最后一个胜者输出0.

思路：显然可以直接模拟出来，但是肯定会超时。换个数据结构就可以了，用查询和删除都是O(logn)的<set>。关于set的具体内容可以参考：http://www.cplusplus.com/reference/set/set/?kw=set

#include <cstdio>#include <iostream>#include <vector>#include <algorithm>#include <cstring>#include <set>using namespace std;set<int> s;typedef set<int>::iterator s_it;typedef vector<s_it>::iterator v_it;int a[300000+10];int main(){int n, m;while (scanf("%d %d", &n, &m) != EOF){s.clear();memset(a, 0, sizeof(a));for (int i = 1; i <= n; ++i)s.insert(i);int li, ri, xi;s_it it, beg, end;vector<s_it> t;v_it vit;while (m--){scanf("%d %d %d", &li, &ri, &xi);beg = s.lower_bound(li), end = s.upper_bound(ri);for (it = beg; it != end; ++it)if (*it != xi){a[*it] = xi;t.push_back(it);  //为了防止迭代器失效，先将需要删除的保存，后面一起删除}for (vit = t.begin(); vit != t.end(); ++vit)//将保存的点删除s.erase(*vit);t.clear();}for (int i = 1; i <= n; ++i)printf(i == n ? "%d\n" : "%d ", a[i]);}return 0;}

思路：