摘要： http://acm.timus.ru/problem.aspx?space=1&num=1770假设有可行解根据所给最短路的情况进行分类 枚举两点最短路假设最短路是dist(i,j) 相连情况为 k(i,j)(-1表示没有直接相连，0表示相连了值为0,1表示相连了值为1）1, dist(i,j)>1||dist(j,i)>1 --> k(i,j)==k(j,i)==-1.2, dist(i,j)==1&&dist(j,i)==1 -->k(i,j)==k(j,i)==-1.3, dist(i,j)==0&&dist(j,i)== 阅读全文

摘要： http://acm.timus.ru/problem.aspx?space=1&num=1065分两种情况1，M==0 的情况 这时枚举任意三个点(不共线) 算长度 取最短2，M!=0 的情况 这时枚举任意点到其它每个点的距离 比如说从i到j 距离 如果所有的M个点都在从i到j向量的右面 则距离可以直接求出 否则暂时为无穷大 然后用 floyd求最短路 最后求最小的环形路代码：#include<iostream>#include<cstdio>#include<cstring>#include<string>#include<ma 阅读全文

摘要： http://acm.timus.ru/problem.aspx?space=1&num=1487A可以打赢了B并不代表A打得比B好 stronger！=better注意这句话：Denis claims that the “Katraps” team plays better than the “Kolomotiv” team, namely, that “Katraps” is not weaker than any team which is stronger than “Komolotiv”.所以 如果A和B有共同祖先 No 否则 YES自己写的程序时间复杂度很高 水过代码：#i 阅读全文

摘要： http://acm.hdu.edu.cn/showproblem.php?pid=2141二分代码：#include<iostream>#include<cstdio>#include<cstring>#include<string>#include<map>#include<vector>#include<stack>#include<set>#include<map>#include<queue>#include<deque>#include<algo 阅读全文

摘要： http://acm.sdut.edu.cn/sdutoj/problem.php?action=showproblem&problemid=1483代码：#include<iostream>#include<cstdio>#include<algorithm>#include<cstring>#include<map>#include<cmath>#define LL long longusing namespace std;const int N=100005;LL a[N],l1[N],l2[N],r1[N] 阅读全文

摘要： http://acm.timus.ru/problem.aspx?space=1&num=1447最优比率生成树(最小生成树+二分）黑书 p303代码：#include<iostream>#include<cstdio>#include<cstring>#include<string>#include<map>#include<vector>#include<stack>#include<set>#include<map>#include<queue>#include 阅读全文

摘要： http://acm.timus.ru/problem.aspx?space=1&num=1699从一个点到另一个点正好有一条路 所以原图是一棵树先求出每一对点的最近公共祖先lca然后一遍dfs 求出每个点到它的lca的转弯数 最后还要判定两个点在lca处是否还要一个转弯代码：#include<iostream>#include<cstdio>#include<cstring>#include<string>#include<map>#include<vector>#include<stack>#inc 阅读全文

摘要： http://acm.timus.ru/problem.aspx?space=1&num=1544bfs代码：#include<iostream>#include<cstdio>#include<cstring>#include<string>#include<map>#include<vector>#include<stack>#include<set>#include<map>#include<queue>#include<deque>#include 阅读全文

摘要： http://poj.org/problem?id=3017dp+单调队列算法的理论时间复杂度应该还是接近 O(n^2) 但为什么过得还挺快呢 应该是后台数据的问题吧代码：#include<iostream>#include<cstdio>#include<cstring>#include<string>#include<map>#include<vector>#include<stack>#include<set>#include<map>#include<queue>#i 阅读全文

摘要： http://poj.org/problem?id=3378要想求以第k个数为5个数中最后一个数的组合数量就要求从1到k-1这些数中比第k个小且是4个数中最后一个数的这些组合数量 之和依次类推求和用树状数组来维护 最后结果可能超long long 我用了大整数代码：#include<iostream>#include<cstdio>#include<cstring>#include<string>#include<map>#include<vector>#include<stack>#include<se 阅读全文