ACM之路

Qin Shi Huang's National Road SystemTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 5381    Accepted Submission(s): 1884Problem Desc

Network of SchoolsTime Limit: 1000MS Memory Limit: 10000KTotal Submissions: 13509 Accepted: 5392DescriptionA number of schools are connected to a computer netwo

#include #include #include #include #include #include #include #include #include #include #include #include using namespace std; #define rep(i,a,n) for (int i=a;i<n;i++)#define per(i

void tarjan (int u, int x) { DFN[u] = low[u] = ++ord; instack[u] = 1; st[top++] = u; for (int i = head[u]; ~i; i = edge[i].next) { int v = edge[i].to;

#include #include #include #include #include #include #include #include #include #include #include #include using namespace std; //hdu4378 #define rep(i,a,n) for (int i=a;i<n;i++)#

#pragma comment(linker, "/STACK:102400000,102400000")c ++ 提交。

Warm upTime Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)Total Submission(s): 4971    Accepted Submission(s): 1105Problem Description　　N planets are c

/**********************************************二分图匹配（Hopcroft-Carp的算法）。初始化：g[][]邻接矩阵调用：res=MaxMatch(); Nx,Ny要初始化！！！时间复杂大为 O(V^0.5 E)适用于数据较大的二分匹配 ***********************************************

The Accomodation of StudentsTime Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3690    Accepted Submission(s): 1730Problem Description

MazeTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65768/65768 K (Java/Others)Total Submission(s): 2059    Accepted Submission(s): 829Special JudgeProblem DescriptionWhen wake

#include //hdu 4408#include #include #include #include #include #include #include #include #include #include #include using namespace std;#define rep(i,a,n) for (int i=a;i<n;i++)#defin

The Unique MSTTime Limit: 1000MS Memory Limit: 10000KTotal Submissions: 24427 Accepted: 8691DescriptionGiven a connected undirected graph, tell if its minimum s

#include #include #include #include #include #include #include #include #include #include #include using namespace std;#define rep(i,a,n) for (int i=a;i<n;i++)#define per(i,a,n) for (int

#include #include #include #include #include #include #include #include #include #include #include #include using namespace std;#define rep(i,a,n) for (int i=a;i<n;i++)#define per(i,a,n

uva10766#include #include #include #include #include #include #include #include #include #include #include #include using namespace std;#define rep(i,a,n) for (int i=a;i<n;i++)#define

Ice_cream’s world IITime Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3605    Accepted Submission(s): 870Problem DescriptionAfter

Scout YYF ITime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 6162 Accepted: 1755DescriptionYYF is a couragous scout. Now he is on a dangerous mission wh

Rain on your ParadeTime Limit: 6000/3000 MS (Java/Others)    Memory Limit: 655350/165535 K (Java/Others)Total Submission(s): 3235    Accepted Submission(s): 1031Problem DescriptionYou’

int EK(int s,int t){ queue q; int ans = 0; while(true) { memset(a,0,sizeof(a)); a[s] = INF; q.push(s); while(!q.empty()) { int u = q.front(); q.pop(); for(int v=0;v<=t;v++)

#include #include #include #include #include #include #include #include #include #include #include #include using namespace std; #define rep(i,a,n) for (int i=a;i<n;i++)#define per(i

MPI MaelstromTime Limit: 1000MS Memory Limit: 10000KTotal Submissions: 6623 Accepted: 4102DescriptionBIT has recently taken delivery of their new supercomputer,

Cyclic NacklaceTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4475    Accepted Submission(s): 2040Problem DescriptionCC always

1370 - Bi-shoe and Phi-shoePDF (English)StatisticsForumTime Limit: 2 second(s)Memory Limit: 32 MBBamboo Pole-vault is a massively popular sport in Xzhil

#include #include #include #include #include #include #include #include #include #include #include #include using namespace std; const int maxn = 1e7 + 10;bool t[maxn];int pri[maxn/

#define prln(x) cout

#include // hdu5497#include #include #include using namespace std;#define lowbit(x) x&(-x)const int maxn = 1e5+10;int a[maxn];int t[2][maxn]; int n,m; typedef long long ll;int sum(in

for (int i = 0; i <= n; ++i) { C[i][0] = 1 % P; for (int j = 1; j <= i; ++j) { C[i][j] = (C[i - 1][j] + C[i - 1][j - 1]) % P; } }

上界改容量，下界先流走。 U -> V之间添加一条 cap - down的边S->U 为U多出来的入的下界- 出的下界V -> T 为V多出来的出的下界-入的下界Max_Flow MF;int n,m;const int maxn = 210;const int maxm = 40010;int a[maxn];int id[maxm];int up[maxm]

int sumFlow;const int MAXN = 1010;const int MAXM = 1000200;const int INF = 1000000000;struct Edge{ int u, v, cap, cost; int next;}edge[MAXM<<2];int NE;int head[MAXN], dist[MAXN], pp[MA

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Redundant PathsTime Limit: 3000/1000MS (Java/Others)     Memory Limit: 65535/65535KB (Java/Others)Submit StatusIn order to get from one of the F(1≤F≤5,000) grazing fields (whic

DiningTime Limit: 2000MS Memory Limit: 65536KTotal Submissions: 11698 Accepted: 5376DescriptionCows are such finicky eaters. Each cow has a preference for certa

Warm up 2Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)Total Submission(s): 2031    Accepted Submission(s): 926Problem Description　　Some 1×2 domino

bool dfs(int u){ for(int i=1;i<=n;i++) { if(!used[i] && g[u][i]) { used[i] = true; if(Left[i] == -1 || dfs(Left[i])) { Left[i] = u; return true; } } } return false;}in

WormholesTime Limit: 2000MS Memory Limit: 65536KTotal Submissions: 36771 Accepted: 13465DescriptionWhile exploring his many farms, Farmer John has discovered a

Jamie's Contact GroupsTime Limit: 7000MS Memory Limit: 65536KTotal Submissions: 6967 Accepted: 2292DescriptionJamie is a very popular girl and has quite a lot o

#include #include #include #include using namespace std;/* KM算法 * 复杂度O(nx*nx*ny) * 求最大权匹配 * 若求最小权匹配，可将权值取相反数，结果取相反数 * 点的编号从0开始 */const int N = 310;const int INF = 0x3f3f3f3f;int n

单次lognint phi(int x){ int ret = x; for (int i = 2; i * i <= x; i++) { if (x % i == 0) { ret = ret / i * (i - 1); for (; x % i == 0; x /= i);

LayoutTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 8651 Accepted: 4165DescriptionLike everyone else, cows like to stand close to their friends when

还是畅通工程Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 33080 Accepted Submission(s): 14922Problem Description某省调查乡村交通状况，得到的统计表中列出了任意两村庄