其他

在Hdu上是下面幾題：Hdu 2041 超級樓梯下面是解題報告（因為有的題有點水，和貼發）Hdu 2044 一隻小蜜蜂...   

/* 先排序，然後用序號來代替 */#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;struct node{ int value; int pos; }a[100];bool cmp(node a,node b){ return a.value<b.value;} int main(){ int b[100];

#include<cstdio>#include<cstring>#include<queue>#include<vector>#include<algorithm>using namespace std;const int maxn = 100;struct TwoSAT { int n; vector<int> G[maxn*2]; bool mark[maxn*2]; int S[maxn*2], c;

csdn真是神坑，剛寫的解題竟然被吞了，尼瑪，害我重寫。本題的突破口在每個投票人都要有超過一半的建議被採納。所以對於k<3的，每個建議都得採納，k>=3的，只有一個建議不能採納。都得採納的，我用了一個must數組記錄，然後dfs找出所有開始就定下來的建議的值；只有一個不能採納，即每兩個建議間都至少有一個被採納。最後輸出要判斷“？”，我的方法是對每個建議檢查是否無論它采不採納都有解。開始忘了dfs找所有應該must的建議，2y#include<cstdio>#include

#include<cstdio>#include<algorithm>#include<vector>#include<cstring>#include<stack>using namespace std;const int maxn = 1000 + 10;//圖中節點編號從0開始，scc從1開始vector<int> G[maxn];int pre[maxn], lowlink[maxn], sccno[maxn],

題目很簡單，資料不大，直接深搜。#include<stdio.h>#include<string.h>#define MAXN 105int map[MAXN][MAXN];int n,m;void dfs(int i,int j){ if(!map[i][j]) return; map[i][j]=0; dfs(i+1,j); dfs(i-1,j); dfs(i-1,j-1); dfs(i-1,j+1); dfs(i+1,j-1

文章目錄 Sample inputOutput for sample input Problem B: The Largest CliqueGiven a directed graph G, consider the following transformation. First, create a new graphT(G) to have the same vertex set as G. Create a directed edge

/*判斷有無歐拉迴路 還是老樣子： 先判斷圖是否連通，這道題是無向圖，然後判斷每個點的度，最後列印之*/#include<cstdio>#include<cstring>using namespace std;const int maxn = 1005;int G[maxn][maxn];int degree[maxn];int f[maxn];int n;int find(int x){ if(x != f[x]) { f[x] =

D. Cow Programtime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputFarmer John has just given the cows a program to play with! The program contains two integer variables,x andy, and performs the

// 題目連結 http://acm.fzu.edu.cn/contest/problem.php?cid=126&sortid=4#include<stdio.h>#include<string.h>struct node{ int x; int y; }a[110];int dp[110][1100];int max(int x,int y){ if(x>y) return x; return y; }int min(int

  這道題初看輸入有點煩，仔細一看就開個二位元組，直接遍曆數組求解就ok了。但是輸出還有括弧，在這裡跪了幾次。然後值得注意的是，節點不一定是字母，頁可以使其他的字元。最後注意邊界就ok啦！下面是代碼：#include<stdio.h>#include<string.h>#define MAXN 210char map[MAXN][MAXN];int n;void dfs(int x,int y){ printf("%c(",map[x][y]); if(x==

容易想到需要思考能否把對面全部殺死，如果能，那麼。。。不能，則一定不會去殺防守的。貪心的思路大致有了方向，然後,不能的話，要怎麼做呢？策略是用我這邊最大的比對手最小的。證明如下，假設我手上的任意一張牌x1和最大的牌x2（x1 < x2）和對面的任意一張牌y2和最小的牌（y1 < y2）,滿足x2 > y2（否則一個都殺不死），如果y1 < y2 < x1 < x2,則用最大的去殺最小的沒有問題，因為x中任意一張牌都比y大，如果y1 < x1 <

MatrixTime Limit: 6000MS Memory Limit: 65536KTotal Submissions: 4150 Accepted: 1007DescriptionGiven a N × N matrix A, whose element in the i-th row andj-th columnAij is an number that equals i2 + 100000 ×i +j2 - 100000 × j + i × j, you are to find

B. Hometasktime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputFurik loves math lessons very much, so he doesn't attend them, unlike Rubik. But now Furik wants to get a good mark for math. For that

Jessica's Reading ProblemTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 5967 Accepted: 1794DescriptionJessica's a very lovely girl wooed by lots of boys. Recently she has a problem. The final exam is coming, yet she has spent little time

文章目錄 Sample InputPossible Output for Sample Input Problem E: WeddingUp to thirty couples will attend a wedding feast, at which they will be seated on either side of a long table. The bride and groom sit at one end,

刷刷水題，入個門先：百練oj：2754 八皇后西電oj：1010 素數環Poj：1011 Sticks百練oj：2754 八皇后  不過這道題還用存下每個可能解。  下面是代碼：  #include<iostream>#include<cmath>using namespace std;const int maxn = 100;const int N = 8;int A[maxn];int Ans[maxn][10];int n,b,tot;void

Radar InstallationTime Limit: 1000MS Memory Limit: 10000KTotal Submissions: 43100 Accepted: 9543DescriptionAssume the coasting is an infinite straight line. Land is in one side of coasting, sea in the other. Each small island is a point locating in

/* 有一塊草坪，長為l，寬為w， 在它的水平中心線上有n個位置可以安裝噴水裝置， 各個位置上的噴水裝置的覆蓋範圍為以它們自己的半徑ri為圓。求出最少需要的噴水裝置個數。 將他映射到一邊，變成區間覆蓋問題 */#include<stdio.h>#include<string.h>#include<math.h>#include<algorithm>using namespace std;#define N 10005 struct

DryingTime Limit: 2000MS Memory Limit: 65536KTotal Submissions: 7104 Accepted: 1823DescriptionIt is very hard to wash and especially to dry clothes in winter. But Jane is a very smart girl. She is not afraid of this boring process. Jane has decided