这个搜索算法不是很有效率 一时还想不出来什么好方法  放假了再说

＃i nclude <stdio.h>
＃i nclude <string.h>

int over;
int p, sud[10][10];
int stack[82][2];
int r[10][10], c[10][10], s[10][10];

int f(int i, int j)
{
	i = (i-1)/3;
	j = (j-1)/3+1;
	return i*3+j;
}

void output()
{
	int i, j;
	
	for(i = 1; i <= 9; i++)
	{
		for(j = 1; j <= 9; j++)
			printf("%d",sud[i][j]);
		printf("\n");
	}
}

void solve(int n)
{
	int i, j, k;
	int F, tmp[9];

	i = stack[n][0], j = stack[n][1];
	F = 0;
	for(k = 1; k <= 9; k++)
	{
		if(r[i][k]==0&
……

一个比较常见的 dp吧

code by 20053565

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

int N, G;
double R;
int valid[151][151];
long dp[151][151];
struct node
{
	double x;
	double y;
}p[151];

struct Node
{
	double x;
	double y;
}c[151];

int check(int i, int j)
{
	int q;
	double A, B, C, D, E;

	for(q = 0; q < G; q++)
	{
		A = p[j].x-p[i].x; B = p[i].y-p[j].y; C = p[j].y*p[i].x-p[i].y*p[j].x;
		D = (A*c[q].y+B*c[q].x+C);
		E = A*A*c[q].x-B*C-A*B*c[q].y;
		if(D*D<=R*R*(A*A+B*B))
		{
			if((c[q].x-p[i].x
……

官方解析
Problem E: Edge Pairing
A simple undirected graph with an even number of edges can always have its edges grouped into incident pairs. Below is a proof by construction.
Let r be some vertex in the graph. We grow a depth-first search tree T rooted at r. 
Since the graph is connected, T actually spans every vertex. 
We then construct the pairing bottom up,  starting from the leaves. 
We mark all edges connecting a leave and its parent as unpaired. 

……

RE了数N次 才过的, 但是在本机上始终未能运行成功,无奈之下交之,AC.  RP again!!

code by 20053565

source Colombia 2006

这个题是我迄今最长的代码......

离散课本 第四章函数 f(i,j) = (i+j)*(i+j+1)/2+j;
令u = f(i,j);
令A = [（sqrt(1+8*u)-1）/2](取整数部分)
则j = u-A*(A+1)/2;i = A-j;

用到了高精度的开方 还用到了凸多边形的面积公式 之前以为这个公式只有在所有顶点按顺时针排序后才适用.

可是没排序也能的出结果,只是符号的差异,不知是巧合还是......

听说java里的Biginteger用于高精度很爽,下学期就要学了,期待ing.

(手动开方部分angelpin指点)

code by 20053565

苦想之后还是TLE......

经fzk大牛指点才800+MS飘过

大牛说:"按（x,y)将所有顶点排序，扫描一边，看看每条竖着的边的两个顶点排序后是否相邻
再按（y,x)排序，再扫描一边";

感谢所有帮助过我的大牛们......

Code By 20053565