a451907-优快云博客

原创 zoj1163 The Staircases （简单dp）

The StaircasesTime Limit: 2 Seconds Memory Limit: 65536 KB One curious child has a set of N little bricks. From these bricks he builds different staircases. Staircase consists of steps

2013-02-26 17:09:36 426

原创 poj1002 487-3279（电话号码重复次数）

487-3279Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 205583 Accepted: 35862DescriptionBusinesses like to have memorable telephone numbers. One way to

2013-02-23 09:02:25 754

原创 zoj1239 Hanoi Tower Troubles Again!

Hanoi Tower Troubles Again!Time Limit: 2 Seconds Memory Limit: 65536 KB People stopped moving discs from peg to peg after they know the number of steps needed to complete the entire tas

2013-02-21 15:53:55 808

原创 poj3487 The Stable Marriage Problem（最稳定的婚姻）

The Stable Marriage ProblemTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 1966 Accepted: 838DescriptionThe stable marriage problem consists of matching

2013-02-21 10:32:21 888

原创 uva10004 Bicoloring（交叉染色法）

题目：In 1976 the ``Four Color Map Theorem" was proven with the assistance of a computer. This theorem states that every map can be colored using only four colors, in such a way that no region is color

2013-02-20 09:05:08 1035

原创 poj1523（无向图求割点）

SPFTime Limit: 1000MS Memory Limit: 10000KTotal Submissions: 4085 Accepted: 1888DescriptionConsider the two networks shown below. Assuming that data moves around th

2013-02-19 10:21:23 1869

原创 poj1144-Network（求割点）

NetworkTime Limit: 1000MS Memory Limit: 10000KTotal Submissions: 7278 Accepted: 3419DescriptionA Telephone Line Company (TLC) is establishing a new telephone cable

2013-01-31 11:43:26 2817

原创 poj3114 Countries in War(强连通分量+最短路）

Countries in WarTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 1757 Accepted: 553DescriptionIn the year 2050, after different attempts of the UN to mai

2013-01-28 20:49:48 1186

原创 poj2553-The Bottom of a Graph（强连通分量）

The Bottom of a GraphTime Limit: 3000MS Memory Limit: 65536KTotal Submissions: 7098 Accepted: 2905DescriptionWe will use the following (standard) definitions from g

2013-01-26 19:17:00 1276

原创 hdu 2767 Proving Equivalences(强连通分量)

Proving EquivalencesTime Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1426 Accepted Submission(s): 535Problem DescriptionConsider t

2013-01-26 11:15:04 478

原创 poj1236-network of schools(强连通分量)

Network of SchoolsTime Limit: 1000MS Memory Limit: 10000KTotal Submissions: 8137 Accepted: 3221DescriptionA number of schools are connected to a computer network. A

2013-01-26 09:56:56 382

原创 CF—98A—Help Victoria the Wise

A. Help Victoria the Wisetime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputVasilisa the Wise from a far away kingdom g

2013-01-26 09:43:41 813

原创 poj2186Popular Cows

牛群里的牛相互崇拜，这种崇拜是单向可传递的，要求最受崇拜的牛有几个，最受崇拜的牛是受其余所有牛的崇拜用tarjan算法完全抄袭的模板，写出来了还是各种转不出来#include#include#include#includeusing namespace std;int n,m;vector vec[100005];int vis[100005];int

2013-01-25 10:10:52 279

原创 hdu1272小希的迷宫

通过已给的路径判断每个顶点之间是否都能互相到达并且没有回路#include#includeusing namespace std;int f[100005];int v[100005];int mm;int find(int x){ int r=x; while(f[r]!=r) r=f[r]; f[x]=r;

2013-01-24 16:51:13 240

原创最短路-find the safest road

#includedouble map[1005][1005];double a[1005];int v[1005];int main(){ int n,m; while(scanf("%d",&n)!=EOF) { for(int i=1; i { for(int j=1; j

2013-01-22 19:25:18 246

原创最短路-1011

首先建图，给出的2n+1行字符串一行横着一行竖着，建好图直接dij#include#includeusing namespace std;int map[450][450];int a[450],v[450];int n,m;int poit;void buildMap(){ int a; char b; for(int i=0;

2013-01-22 19:19:54 277

原创最短路-1010

Einbahnstrasse Time Limit : 2000/1000ms (Java/Other) Memory Limit : 32768/32768K (Java/Other)Total Submission(s) : 17 Accepted Submission(s) : 5Problem DescriptionEinbahnstra e (German fo

2013-01-22 19:15:36 334

原创最短路—1009

开始直接建图，超时，将图反过来建，一次Dij就可以了#include#includeusing namespace std;int n,m,s,w;int map[1005][1005];int a[1005],v[1005];int start[1005];void dij(){ int min,min_f,i,j; for(i=1; i

2013-01-22 19:13:42 307

原创 hdu1789

贪心算法，时间按从小到大排序，时间相同按分数从小到大排序 #include#include#includeusing namespace std;struct sa{ int x; //天数 int y; //分数} data[1000];int cmp(const sa &a,const sa &b){ if(a.x==b

2013-01-20 11:06:03 305

原创最短路—1008

计算汇率，用Floyd，最后检查对角线上的树是不是大于1#include#includestruct node{ char name[30];} c[33];double map[33][33];int main(){ int n,m,i,j,k,cc=0; while(scanf("%d",&n)&&n!=0) {

2013-01-20 11:04:46 267

原创最短路—1007

#includeint n,start,end;int map[205][205];int a[205],v[205];void dij(){ int i,j,min,min_f; for(i=1;i { a[i]=map[start][i]; v[i]=0; } v[start]=1;

2013-01-19 19:01:53 249

原创 2013寒假SD高校ACM周赛3—Steps

#includeint main(){ int n,a,b; scanf("%d",&n); while(n--) { scanf("%d%d",&a,&b); int s; s=b-a; int sum=0; int i=1; while(s

2013-01-19 18:18:18 231

原创最短路—1006

#includeint map[1005][1005];int path[1005][1005];int charge[1005];int n;int main(){ int i,j,k,start,end; while(scanf("%d",&n)&&n!=0) { for(i=1;i for(j=

2013-01-19 14:02:08 277

原创最短路—1005

#include#include#includeusing namespace std;vector road[1005];int n;int a[1005],map[1005][1005],v[1005];int s[1005];void dij(){ int i,j,min,min_f; for(i=1;i {

2013-01-19 14:01:41 331

原创最短路—1004

#include#define inf 100000000int map[201][201];int a[201],v[201];int n,s,e;void dij(){ int i,j,min,min_f; for(i=0; i { a[i]=map[s][i]; v[i]=0; }

2013-01-12 22:47:09 486

原创最短路—1003

指定开始和结束的位置，求最短路，要保存顶点#include#includestruct node{ char a[31];} bi[151];int poit;int map[151][151];int v[151],t[151];void dij(){ int i,j,min,min_f; for(i=0; i {

2013-01-12 22:45:26 220

原创最短路—1002

很多条路找到最短的#include#includeint map[101][101],n,a[101],v[101];void dij(){ int i,j,min,min_f=0; for(i=1;i { a[i]=map[1][i]; v[i]=0; } v[1]=1; for

2013-01-11 00:25:58 528

原创最短路—1001

#include#includeint map[101][101],n,a[101],v[101];void dij(){ int i,j,min,min_f=0; for(i=1;i { a[i]=map[1][i]; v[i]=0; } v[1]=1; for(i=1;i

2013-01-11 00:24:48 323

原创 poj3624

01背包#include#includeint dp[12881];int w[3403],d[3403];int main(){ int n,m,i,j; scanf("%d%d",&n,&m); for(i=0;i scanf("%d%d",&w[i],&d[i]); memset(dp,0,sizeof(dp)

2012-12-18 12:04:36 346

原创山东理工大学-2012秋季ACM集训队测试赛4-I-矩形回旋

有一个由 n*m(n是行数，m为列数) 的方格组成的矩形阵列。现在从左上角开始顺时针旋转走动，遇到矩形边框或者遇到已经过的方格就转向继续，直到走完所有方格。求最后结束点的坐标(横坐标位置从1到m，纵坐标从1到n)。矩阵回旋，要比较输入的两个数的大小，按小的来走，然后判断小的的奇偶性。#includeint nn,mm;int p(int a,int b){

2012-12-08 18:12:51 611

原创山东理工大学-2012秋季ACM集训队测试赛4-H-火星计数法

火星人的计数规则里只有a,b,c,d四个字母，计数规则从小到大是 a,b,c,d,aa,ab,ac,ad,ba，……。给出来由a,b,c,d四种字母组成的火星数字，算出该数字是第几个(从1开始)。 #include#include#includeint p(int x){ int sum=1,i; for(i=1;i sum=(sum*4)

2012-12-08 18:09:17 617

原创山东理工大学-2012秋季ACM集训队测试赛4-F-多项式求和

多项式描述如下：1 - 1/2 + 1/3 - 1/4 + 1/5 - 1/6 ……先请你求出多项式前n项的和。当大于300的时候输出值不变#include#includeint main(){ int t,i,n; double sum; scanf("%d",&t); while(t--) {

2012-12-08 18:07:25 508

原创 NEFU-4-2-Center Alignment

把输入的所有句子居中放置，不能完全居中的一左一右放置，四周用*围满。#include#includestruct node{ char a[1001]; int l;} p[1001];int main(){ int i,j,k,n=0,max=0,mm,f=0; while(gets(p[n].a)) {

2012-12-04 22:30:54 300

原创 NEFU-4-1-Chat Server's Outgoing Traffic

#include#include#includeusing namespace std;char a[101];int main(){ int i,sum=0,pm=0; while(gets(a)) { if(a[0]=='+') { pm++; }

2012-12-04 22:26:50 322

原创 NEFU-3-C-Registration system

#include#includestruct node{ char a[33]; int f;}name[100005];int main(){ int i,j,n; scanf("%d",&n); int sum=0; for(i=0;i { char s[33];

2012-11-30 21:33:14 339

原创 NEFU-3-A-Watermelon

判断给出的数能不能分成两个偶数#includeint main(){ int i,n; while(scanf("%d",&n)!=EOF) { if(n%2) printf("NO\n"); else { if(n>2)

2012-11-30 21:30:54 276

原创 NEFU-2-C-Tic-tac-toe

错了无数遍竟然是因为判断的是'A'与'0'……#includeint map[3][3];int main(){ int i,j; int f1=0,f2=0; int s1=0,s2=0,s3=0; for(i=1; i { for(j=1; j { scanf("%c

2012-11-30 21:28:00 469

原创 10038 - Jolly Jumpers

判断给出的长度为n的数组相邻两个数的差值是不是从1到n-1。#include#includeint a[3001],b[3001];int main(){ int i,n,flag=0; while(scanf("%d",&n)!=EOF) { flag=0; memset(b,0,sizeof(b));

2012-11-29 22:54:57 348

原创 10137 - The Trip

#include#includedouble a[1000],sum,s1,s2;char tmp[1000];int main(){ int i,n; while(scanf("%d",&n)&&n!=0) { sum=0.0; s1=0.0; s2=0.0; for(i=0

2012-11-26 23:10:16 296

原创 10054 - The Necklace

给出n个珠子，珠子两边有不同的颜色，把珠子串起来，要求相邻珠子的颜色一样#include#includeint n,d[55],map[55][55];void dfs(int s){ int i; for(i=1;i { if(map[s][i]) { map[i][s]--;

2012-11-23 21:29:04 241

空空如也

空空如也