//网络流SAP模板,复杂度O(N^2*M) //使用前调用init(源点,汇点,图中点的个数),然后调用add_edge()加边 //调用getflow得出最大流 #define N 55 #define M 500500 #define INF 0x3fffff struct Max_Flow { struct node { int to,w,next; }edge[M]; int s,t; int nn; int cnt,pre[N]; int lv[N],gap[N]; void ini…
Description Every time it rains on Farmer John's fields, a pond forms over Bessie's favorite clover patch. This means that the clover is covered by water for awhile and takes quite a long time to regrow. Thus, Farmer John has built a set of drainage…
Going Home Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 15944 Accepted: 8167 Description On a grid map there are n little men and n houses. In each unit time, every little man can move one unit step, either horizontally, or vertical…
Drainage Ditches Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 8599 Accepted Submission(s): 4005 Problem Description Every time it rains on Farmer John's fields, a pond forms over Bessie's…
\(EK\) 算法模板 #include <iostream> #include <queue> #include<string.h> using namespace std; #define arraysize 201 int maxData = 0x7fffffff; int capacity[arraysize][arraysize]; //记录残留网络的容量 int flow[arraysize]; //标记从源点到当前节点实际还剩多少流量可用 int pre[…
http://poj.org/problem?id=1273 Drainage Ditches Time Limit: 1000MS Memory Limit: 10000K Total Submissions:87219 Accepted: 33916 Description Every time it rains on Farmer John's fields, a pond forms over Bessie's favorite clover patch. This means…
Drainage Ditches Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 13273 Accepted Submission(s): 6288 Problem Description Every time it rains on Farmer John's fields, a pond forms over Bessie's…
Stars Time Limit: 5000/2000 MS (Java/Others) Memory Limit: 32768/65536 K (Java/Others) Total Submission(s): 785 Accepted Submission(s): 335 Problem Description Yifenfei is a romantic guy and he likes to count the stars in the sky. To make the p…
#include<cstring> #include<cstdio> #define FOR(i,f_start,f_end) for(int i=f_startl;i<=f_end;i++) #define MS(arr,arr_value) memset(arr,arr_value,sizeof(arr)) ; ; int size; int n; const int inf=0x3f3f3f3f; using namespace std; int head[maxn];…
#include <iostream> #include <cstdio> #include <cstring> #include <queue> #define mem(a,b) memset(a,b,sizeof(a)) using namespace std; , INF = 0x7fffffff; int d[maxn], head[maxn]; int n, m, s, t; struct edge{ int u, v, f, c, next; }…
//这个是邻接矩阵的#include<iostream> #include<queue> #include<string.h> #include<stdio.h> #include<algorithm> using namespace std; ; const int inf=0x3f3f3f; int N; int depth[maxn]; int a[maxn][maxn]; bool bfs(int s,int e)//广搜求深度 { qu…
接着Django模板-基础知识继续写模板相关知识. if标签 {% if %} 标签接受 and , or 或者 not 关键字来对多个变量做判断 ,或者对变量取反( not ). 但是不允许在同一个标签中同时使用 and 和 or,下边是不合法的: {% if athlete_list and coach_list or cheerleader_list %} 多次使用同一个逻辑操作符是没有问题的,但是我们不能把不同的操作符组合起来. {% if athlete_list or coach_l…
#6004. 「网络流 24 题」圆桌聚餐 内存限制:256 MiB时间限制:5000 ms标准输入输出 题目类型:传统评测方式:Special Judge 上传者: 匿名 提交提交记录统计讨论测试数据 题目描述 假设有来自 n nn 个不同单位的代表参加一次国际会议.每个单位的代表数分别为 ri r_iri.会议餐厅共有 m mm 张餐桌,每张餐桌可容纳 ci c_ici 个代表就餐.为了使代表们充分交流,希望从同一个单位来的代表不在同一个餐桌就餐. 试设计一个算法,给出满足要…
地址:http://poj.org/problem?id=1279 题目: Art Gallery Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 7329 Accepted: 2938 Description The art galleries of the new and very futuristic building of the Center for Balkan Cooperation have the f…
