Problem Description Conflicts are everywhere in the world, from the young to the elderly, from families to countries. Conflicts cause quarrels, fights or even wars. How wonderful the world will be if all conflicts can be eliminated. Edward contribute…

HDU 4115 Eliminate the Conflict pid=4115">题目链接 题意:Alice和Bob这对狗男女在玩剪刀石头布.已知Bob每轮要出什么,然后Bob给Alice一些限制,1表示i轮和j轮Alice必须出不一样的,0表示必须出一样的.假设Alice有一局输了就算输了,否则就是赢,问Alice能否赢 思路:2-sat问题,已经Bob出什么,Alice要么就出赢的要么就出平的,然后加上m个约束就是2-sat问题了 代码: #include <cstdio>…

Eliminate the Conflict Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1315    Accepted Submission(s): 563 Problem Description Conflicts are everywhere in the world, from the young to the elder…

Problem Description Conflicts are everywhere in the world, from the young to the elderly, from families to countries. Conflicts cause quarrels, fights or even wars. How wonderful the world will be if all conflicts can be eliminated.Edward contributes…

Problem Description Ali has taken the Computer Organization and Architecture course this term. He learned that there may be dependence between instructions, like WAR (write after read), WAW, RAW. If the distance between two instructions is less than…

2-SAT,拆成六个点. #include<cstdio> #include<cstring> #include<cmath> #include<stack> #include<queue> #include<algorithm> using namespace std; +; int T,N,M; stack<int>S; vector<int>G[maxn]; vector<int>FG[max…

Eliminate the Conflict Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1114 Accepted Submission(s): 468 Problem Description Conflicts are everywhere in the world, from the young to the elderly, fr…

题意: 石头剪刀布 分别为1.2.3,有n轮,给出了小A这n轮出什么,然后m行,每行三个数a b k,如果k为0 表示小B必须在第a轮和第b轮的策略一样,如果k为1 表示小B在第a轮和第b轮的策略不一样,如果又一轮小B输了,那整个就输了,求小B能否战胜小A 解析: 写出来矛盾的情况  建图就好啦 可能我建的麻烦了...不过..我喜欢 hhhhh #include <iostream> #include <cstdio> #include <sstream> #inclu…

题目转自hdu 1102,题目传送门 题目大意: 输入一个n*n的邻接矩阵,其中i行j列代表从i到j的路径的长度 然后又m条路已经帮你修好了,求最短要修多长的路才能使所有村庄连接 不难看出,这道题就是标准的最小生成树模板,多水啊 解题思路 虽然很水,但本人还是调了近1h才把代码调好...... 下面介绍一下解决最小生成树的两个方法: Prim 和 Kruskal 一,Prim(不懂的点这里) Prim的思想和dijkstra的想法很想(如果不知道dijkstra算法的请点这里) 那么Prim的复…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2612 思路:从两个起点出发,有多个终点,求从两个起点同时能到达的终点具有的最小时间,开两个数组分别保存两个起点到达每一个终点的用时,最后将两个 数组里的时间加起来求最小的一组,必须对应相加,因为终点必须同时到达. #include <iostream> #include <string> #include <cstdio> #include <cmath> #i…