ACM 省赛题 求算法

[ice_cream]

(1) Magic Sequence

Ivankevin loves sequences.

A sequence of n + m elements is called a MAGIC SEQUENCE it.
1. Each elements ai (1 <= i <= n + m ) is +1 or -1.
2. There are total n “+1”s and m “-1”s.
3. For every j ( 1 <= j <= n + m), we have the sum of ai (1 <= i <= j ) is nonnegative.

Now given n and m, Ivankevin wants to know the number of magic sequences consists of n “+1”s and m “-1”s.

Input
The first line is an integer T, the number of test cases.
Each case is a line of two integers n and m, 1 <= n,m <= 100.
Output
For each case ,print an integer x, which is the number of magic sequences modulo 100000007.

Sample Input
2
2 1
3 3
Sample Output
2
5


(2) The largest Arrow

Three points A(x1 ,y1), B(x2,y2), and C(x3,y3) form an arrow if and only if the following conditions are fulfilled:
x1 = x3 < x2
y1 < y2 < y3

The length of the arrow (x2 - x1) must not shorter than the width of the arrow y3 - y1 .
Figure 3.1 shows an example of an arrow. We define the size of an arrow as its length plus its width . Given N points in the plane, what’s the largest possible arrow can be formed from three of the given points ?



Input:
The first line of the input contains a positive integer T specifying the number of test cases to follow.
The fist line each test case contains a positive integer N ( 1 <= N <= 2000 ). Then follows N lines ,each describes a point with coordinates (xi, yi). ( 0 <= xi, yi <= 107)
It’s assumed that no two points are at the same location.

Output:
For each test case you should output a single line containing “Case X : Y” ( quotes for clarity) where X is the number of the test case (starting at 1). Y is the size of the largest arrow in the plane. If there’s not any arrow in the plan, Y should be equal to -1.

Sample Input:
1
5
2 2
2 5
6 4
2 4
2 10
Sample Output
Case 1: 7

[绿色夹克衫]

不是删除点，只是删除线段，并且是按照x排序后。比如存在x = 0，(1, 10000)这样的线段，那么x = 1，(2,6)这样的线段不就没用了么？因为任何一个可以组成箭头的组合，同x = 0的点组成的箭头，Size都要大于同x=1组成的箭头。同理在x相等的情况下，只需要取y的最大和最小值就可以了。

如果存在x = 0，(1, 10)这样的线段，x = 1，(2, 11)，那么等价于 x = 0 (1, 10), x = 1 (10,11)。
如果存在x = 0，(2, 5)这样的线段，x = 1，(1, 10)，那么等价于 x = 0 (2, 5), x = 1 (5,10)。 x = 1 (1,2) 这3条线。


[Quote=引用 41 楼 的回复:]

引用 15 楼 的回复:

先把点按照x排序，从中找出可以作为左边起点的点对pStart[i] pEnd[i]。这样就获得了k条线段，再对这k条线段做处理，让所有x' > x的线段不能遮住坐标为x的线段（删除遮……

这个存在一种可能性，比如 (1,10000),(1,5),(1,1),(6,2).如果按照你所说的要删除了(1,5)(1,1)组成的线段的化，就没有箭头可以生成了。
我……
[/Quote]

[bingo_ms]

感觉第二题算法O（n^2）的可以吗？

[ice_cream]

通不过...

[Quote=引用 31 楼 的回复:]

感觉第二题算法O（n^2）的可以吗？
[/Quote]

[ice_cream]

不知道你是怎么做的， 能说下吗， 学习下..

[Quote=引用 31 楼 的回复:]

感觉第二题算法O（n^2）的可以吗？
[/Quote]

[shangguangshuang]

想当年参加MCM，不过那是好多年前的事儿了，顶楼主

[gotoac]

省赛? 哪个省的? 浙江吗?

[gotoac]

在哪个oj上的题啊?
[Quote=引用 43 楼 的回复:]

省赛? 哪个省的? 浙江吗?
[/Quote]

[bingo_ms]

。。就是很简单的枚举过程。。
不过常数的系数可能只有0.5左右。。2000的数据不知道要运行多久。
要求是1S吗？
[Quote=引用 34 楼 的回复:]
不知道你是怎么做的， 能说下吗， 学习下..


引用 31 楼 的回复:

感觉第二题算法O（n^2）的可以吗？
[/Quote]

[hklove]

这个不知道啊~~

[思想的浪潮]

果断连题目都未看懂

[Sawyer]

用C写一个


#include "stdio.h"
#include "stdlib.h"
int arrow(int **array,int sum);
void show(int **array,int sum);
int find(int hight,int **array,int x,int sum);
int main(){
int total=0,cur=0;
int *final;
scanf("%d",&total);
final=(int*)malloc(total*sizeof(int));
while(cur<total){
int sum=0,i=0,j=0;
scanf("%d",&sum);
int **array;
array=(int**)malloc(sum*sizeof(int*));
while(i<sum){
array[i]=(int*)malloc(2*sizeof(int));
i++;
}
i=0;
while(i<sum){
scanf("%d %d",&array[i][j],&array[i][j+1]);
i++;
}
//show(array,sum);
int l=arrow(array,sum);
final[cur]=l;
cur++;
}
cur=0;
while(cur<total){
printf("%d",final[cur]);
cur++;
}
return 0;
}
int arrow(int **array,int sum){
int i=0,j=0,min_x=0,max_x=0,min_y=0,max_y=0;
int x=0,hight=0,size=0,maxsize=0,y=0;
while(i<sum){
x=array[i][0];
y=array[i][1];
while(i<sum){
if(array[i][0]==x && (array[i][1]-y)!=0){
hight=abs(array[i][1]-y);
size=find(hight,array,x,sum);
if(maxsize<size)
maxsize=size;
}
i++;
}
i++;
}
return maxsize;
}
void show(int **array,int sum){
int i=0,j=0;
while (i<sum) {
printf("%d %d\n",array[i][j],array[i][j+1]);
i++;
}
}
int find(int hight,int **array,int x,int sum){
int i=0;
while(i<sum){
if(array[i][0]>x && abs(array[i][0]-x)>hight)
return abs(array[i][0]-x+hight);
i++;
}
}

[Payden]

果断很纠结啊

[lwouyang]

第二题的题意应该是说形成箭头要满足3个条件吧 ？！
1、x1 = x3 < x2
2、y1 < y2 < y3
3、width(y3 - y1) <= length(x2 - x1)，也就是箭头尖的程度是有要求的。
LZ 的原文中 第三个条件 是文字叙述 不是数学式子，而且还跟前面两个隔开了一行，但从文意上来看应该是连续的。不知是不是我看错了！ ^_^
因此（如果我没有看错题意的话），借用 litaoye 的表达方式，
像 x1 = 0，(1, 10000) x2 = 10，(2,6) 无法形成箭头，
像 x1 = 0，(1, 10) x2 = 10，(2,6) 则可以，
因此为预防这种意外，不能轻易删除那些较短的线段吧？

[ssd189]

F(n, m) = F(n-1, m) + F(n, m-1)
这个公式是很有问题的. 第一题的.

[FelixLiu_5]

我也是来学习的，暂时只能围观

[yishuangpangxie]

厉害，高手一点，少走半年

[ssd189]

第一题的:

#include <stdlib.h>
#include <stdio.h>

//n个+1, m个-1的MagicSequence的种类, 且前面数组之和为s(s>=0)
int kindsOfMagicSequenceWithFormer1(int n, int m, int s) {
	int a, b;
	
	if(m + n == 0)
		return 1;
	
	a = (n > 0)?kindsOfMagicSequenceWithFormer1(n - 1, m, s + 1):0;
	b = (m > 0 && s - 1 >= 0)?kindsOfMagicSequenceWithFormer1(n, m - 1, s - 1):0;
	
	return a + b;
}

//n个+1, m个-1的MagicSequence的种类
int kindsOfMagicSequence(int n, int m) {
	if(n < m) return 0;
	return kindsOfMagicSequenceWithFormer1(n, m, 0);
}

void doTest() {
	int n, m, k;
	
	scanf("%d %d", &n, &m);
	k = kindsOfMagicSequence(n, m);
	printf("Output: %d\n", k);
}

int main() {
	int numOfCase, i;
	
	scanf("%d", &numOfCase);
	for(i = 0; i < numOfCase; ++i)
		doTest();

	return 0;	
}

[ssd189]

第二题的, 但是是错的.

#include <stdlib.h>
#include <stdio.h>

typedef struct {
	int x;
	int y;
}Point;

Point newPoint(int x, int y) {
	Point p;
	
	p.x = x;
	p.y = y;
	return p;
}

int pointLess(Point p1, Point p2) {
	if(p1.y < p2.y)
		return 1;
	else if(p1.y > p2.y)
		return 0;
	else if(p1.x <= p2.x)
		return 1;
	else
		return 0;
}

//已经验证的有
//pA.y < pB.y < pC.y
//pA.x < pB.x
int sufficeArrow(Point pA, Point pB, Point pC) {
	if(pA.x != pC.x)
		return 0;
	else if((pB.x - pA.x) < (pC.y - pA.y))
		return 0;
	else {
		return 1;
	}
}

int kindsOfArrowWithSortedPointsAndPointA_PointBFixed(
	Point *points, int npoints, int pBIndex) {
	int i, sum;
	Point pA, pB;
	
	pA = points[0];
	pB = points[pBIndex];
	sum = 0;
	for(i = pBIndex + 1; i < npoints; ++i) {
		Point pC;
		
		pC = points[i];
		if(pC.y == pB.y)
			continue;
		if(sufficeArrow(pA, pB, pC))
			++sum;
	}
	
	return sum;
}

//points[0]是pA
int kindsOfArrowWithSortedPointsAndPointAFixed(Point *points, int npoints) {
	int i, sum;
	Point pA;
	
	pA = points[0];
	sum = 0;
	for(i = 1; npoints - (i + 1) >= 1; ++i) {
		Point pB;
		
		pB = points[i];
		if(pB.x <= pA.x || pB.y == pA.y)
			continue;
		else
			sum += kindsOfArrowWithSortedPointsAndPointA_PointBFixed(points, npoints, i);
	}
}

//points以纵坐标升序排列
int kindsOfArrowWithSortedPoints(Point *points, int npoints) {
	int i, sum;
	
	sum = 0;
	for(i = 0; npoints - i >= 3; ++i) {
		sum += kindsOfArrowWithSortedPointsAndPointAFixed(points + i, npoints - i);
	}
}

int kindsOfArrow(Point *points, int npoints) {
	int i;
	
	for(i = 1; i < npoints; ++i) {
		int j; 
		Point key;
		
		key = points[i];
		for(j = i - 1; j >= 0; --j) {
			if(pointLess(key, points[j]))
				points[j + 1] = points[j];
			else break;
		}
		points[j + 1] = key;
	}
	
	return kindsOfArrowWithSortedPoints(points, npoints);
}

void doTest() {
	int i, npoints, k;
	Point *points;
	
	npoints;
	scanf("%d", &npoints);
	points = malloc(sizeof(Point) * npoints);
	for(i = 0; i < npoints; ++i) {
		int x, y;
		
		scanf("%d %d", &x, &y);
		points[i] = newPoint(x, y);
	}
	
	k = kindsOfArrow(points, npoints);
	printf("Output: %d\n", k);
}

int main() {
	int N, i;
	
	scanf("%d", &N);
	for(i = 0; i < N; ++i)
		doTest();

	return 0;	
}

[lyyyj123]

高手如云啊 强

[半山th]

我在线提交了，楼上各位的代码，都不能AC啊！谁有能AC的代码啊？