其他

題目：http://acm.hdu.edu.cn/showproblem.php?pid=3549Flow ProblemTime Limit: 5000/5000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)Total Submission(s): 5111    Accepted Submission(s): 2385Problem DescriptionNetwork flow is a well-known

先來看一個代碼：#include <iostream>using namespace std;class base{ public: virtual void fun() { cout<<"base"<<endl; }};class son:public base{ public: void fun() {

#include <iostream>using namespace std;class base{ public: virtual void fun() { cout<<"base"<<endl; }};class son:public base{ public: void fun() { cout<<"son"<&

Sequence operationTime Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4603    Accepted Submission(s): 1338Problem Descriptionlxhgww got a sequence contains n characters which are all '0's or

MatrixTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1008    Accepted Submission(s): 560Problem DescriptionYifenfei very like play a number game in the n*n Matrix. A positive integer number is

1：cin.clear()  清楚錯誤狀態，c++ primer中cin.clear(istream::failbit)  錯了~這是吧流置於錯誤狀態應cin.clear(istream::goodbit) 注意：此時錯誤的輸入，如字元，仍在緩衝區             中。可以看看另一篇cin注意點。2:  類的定義應放在標頭檔中，函式宣告後即可使用（在後面定義），但是只聲明類（儘管後面定義類，但使用類在定義之前），類處於不完全狀態，此時可定義類的指標和引用，      

很綜合的一道題額~考二分 不過 沒用線上段樹上~~#include <iostream>#include <cstring>#include <cstdio>#include <algorithm>#define lson pos<<1#define rson pos<<1|1using namespace std;const int MAXN=50005;int blocks[MAXN*2];struct node{

預設情況下從大到小輸出#include <iostream>#include <queue>using namespace std;int main(){ priority_queue<int> q; //等價priority_queue<int,vector<int>,less<int> >  q; q.push(2); q.push(6); q.push(1); int i;

第一道額~~理解了好久~思路基本都是網上的~#include <iostream>#include <cstring>#include <cstdio>#include <algorithm>#include <map>#define lson pos<<1#define rson pos<<1|1using namespace std;const int MAXN=1000005;struct line{

矩形周長和~#include <iostream>#include <cstring>#include <cstdio>#include <algorithm>#define lson pos<<1#define rson pos<<1|1using namespace std;const int MAXN=30000;struct line{ int l,r,h,f; line(int a=0,int b=0,

Problem DescriptionOne day, ZHJ came across a warrior novel named Kunlun in library. Soon, ZHJ fall into it.He tells everybody how deeply he loves Kunlun no matter whom he meets.He found it attractive so he decided to read the novel in the next N

雖然也是矩形面積並~但讓我更深刻了理解了~~本來想的是將大矩形正常打標記，將小矩形反向打標記就是下面的是-1上面是1~~發現範例過不去~~發現思想錯了~~~只能分割成小矩形~~#include <iostream>#include <cstring>#include <cstdio>#include <algorithm>#define lson pos<<1#define rson pos<<1|1using

這個題有一點要考慮到，就是某人從a站坐到b站，他在b站就下車了，他在車上乘坐的區間為a到b-1。#include <iostream>#include <cstring>#include <cstdio>using namespace std;const int MAXN=1000005;struct node{ int l,r; int sum; int lazy; int mid() { return (l+r)

以前在我們學校的OJ上做過一道類似的題目，一開始想類比操作過程的但是實現不了，建立哈夫曼樹又不會，後來跟同學學習了優先隊列，所以就無節操的水過了啊、、、、下面是代碼#include<iostream>#include<algorithm>#include<queue>using namespace std;int main(){ long long n,i,t,tt,ans; priority_queue<long long

Can you answer these queries?Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65768/65768 K (Java/Others)Total Submission(s): 6849    Accepted Submission(s): 1566Problem DescriptionA lot of battleships of evil are arranged in a line before

這是POJ討論上的題目解釋，寫的很好摘過來的，感謝提供者、、、對於每一個月來說，是盈利如果則盈利S，如果虧空則虧d。每五個月進行一次統計，共統計八次（1-5月一次，2-6月一次.......）統計的結果是這八次都是虧空。問題：判斷全年是否能盈利，如果能則求出最大的盈利。如果不能盈利則輸出Deficit簡單的貪心，分類討論就行了啊、、、、#include <stdio.h>int main(){ int i, in, out, count, t; while(~scanf(

第一道額~~#include <iostream>#include <cstring>#include <cstdio>using namespace std;const int MAXN=100005;struct node{ int l,r; int lsum,rsum,msum; int mid() { return (l+r)>>1; }};node tree[MAXN*4];inline

~~#include <iostream>#include <cstring>#include <cstdio>#define lson pos<<1#define rson pos<<1|1using namespace std;const int MAXN=100005;struct node{ int l,r; int msum; int lsum,rsum; int mid() {

A new Graph GameTime Limit: 8000/4000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1010    Accepted Submission(s): 454Problem DescriptionAn undirected graph is a graph in which the nodes are connected by

分析後就是求樹的直徑兩次bfs即可證明： 樹的直徑是指樹的最長簡單路。求法: 兩遍BFS :先任選一個起點BFS找到最長路的終點，再從終點進行BFS，則第二次BFS找到的最長路即為樹的直徑；              原理: 設起點為u,第一次BFS找到的終點v一定是樹的直徑的一個端點              證明: 1) 如果u 是直徑上的點，則v顯然是直徑的終點(因為如果v不是的話，則必定存在另一個點w使得u到w的距離更長，則於BFS找到了v矛盾)