poj3258 二分 最小值最大化问题】的更多相关文章

River Hopscotch Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 10842   Accepted: 4654 Description Every year the cows hold an event featuring a peculiar version of hopscotch that involves carefully jumping from rock to rock in a river.…
题意: 给定一个有向图,每条边都有一个权值,每次你可以选择一个结点v和一个整数d,把所有以v为终点的边的权值减小d,把所有以v为起点的边的权值增加d,最后要让所有边权的最小值非负且尽量大 两个特判 1.图中存在负环 则 No Solution (构成差分约束系统的图后bk的最小值为w(u,v)- 1:所以check(1) 2.不存在最短路 则可以任意解  就是使x最大 看是否形成负环 还不形成负环 则说明 可以任意解 然后就是套最小值最大化的二分模板 #include <iostream> #…
题意: 就是给出n对坐标,每对只能选一个,以选出来的点为圆心,半径自定义,画圆,而这些圆不能覆盖,求半径最小的圆的最大值 解析: 看到最x值最x化,那二分变为判定性问题,然后...然后我就没想到... 好的吧...2-sat 还能在不是一对的两个值之间建边...求出连通分量后  看一对中的两个坐标是否在一个连通分量中 在一个说明半径太小 要加大  反之..反之. 因为涉及小数 要用eps #include <iostream> #include <cstdio> #include…
#include <iostream> #include <cstdio> #include <sstream> #include <cstring> #include <map> #include <set> #include <vector> #include <stack> #include <queue> #include <algorithm> #include <cma…
https://vjudge.net/problem/POJ-3258 二分最小值,判断需要删去的点的个数,如果大于给定,则直接return 0,则说明该数需要再小. 最后注意,起点是0终点是l,起点可以不加进数组,终点必须加进去!! #include<iostream> #include<cstdio> #include<queue> #include<cstring> #include<algorithm> #include<cmath…
River Hopscotch Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 15273   Accepted: 6465 Description Every year the cows hold an event featuring a peculiar version of hopscotch that involves carefully jumping from rock to rock in a river.…
Polycarp has a lot of work to do. Recently he has learned a new time management rule: "if a task takes five minutes or less, do it immediately". Polycarp likes the new rule, however he is not sure that five minutes is the optimal value. He suppo…
POJ - 3258 River Hopscotch Time Limit: 2000MS   Memory Limit: 65536KB   64bit IO Format: %I64d & %I64u [Submit]   [Go Back]   [Status] Description Every year the cows hold an event featuring a peculiar version of hopscotch that involves carefully jum…
emmm...去吃早饭了... rujia讲的很好.. 最小值最大化问题,,,二分枚举答案   设x1.x2为同一个集合中的元素,y1.y2为另一个集合中的元素,如果x1与y1之差小于mid,那么如果选了x1就必须选y2,反过来,选了y1就必须选x2.这样就是2-SAT模型了.只需找出使得这个2-SAT有解的最大mid即可. #include <iostream> #include <cstdio> #include <sstream> #include <cst…
River Hopscotch Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 6697   Accepted: 2893 Description Every year the cows hold an event featuring a peculiar version of hopscotch that involves carefully jumping from rock to rock in a river. T…