#include#include

亚洲免费在线-亚洲免费在线播放-亚洲免费在线观看-亚洲免费在线观看视频-亚洲免费在线看-亚洲免费在线视频

【cf489】D. Unbearable Controversy of Being

系統(tǒng) 1905 0

http://codeforces.com/contest/489/problem/D

很顯然,我們只需要找對于每個點(diǎn)能到達(dá)的深度為3的點(diǎn)的路徑的數(shù)量,那么對于一個深度為3的點(diǎn),如果有a種方式到達(dá),那么有方案數(shù)(a-1+1)*(a-1)/2

可是我用dfs找路徑就tle了QAQ

于是orz別人的代碼,,,,是暴力。。。。。。。。。。。。。。。。。。。。。。。。直接兩重循環(huán)orz

      #include <cstdio>

#include <cstring>

#include <cmath>

#include <string>

#include <iostream>

#include <algorithm>

#include <queue>

#include <set>

#include <map>

using namespace std;

typedef long long ll;

#define rep(i, n) for(int i=0; i<(n); ++i)

#define for1(i,a,n) for(int i=(a);i<=(n);++i)

#define for2(i,a,n) for(int i=(a);i<(n);++i)

#define for3(i,a,n) for(int i=(a);i>=(n);--i)

#define for4(i,a,n) for(int i=(a);i>(n);--i)

#define CC(i,a) memset(i,a,sizeof(i))

#define read(a) a=getint()

#define print(a) printf("%d", a)

#define dbg(x) cout << (#x) << " = " << (x) << endl

#define error(x) (!(x)?puts("error"):0)

#define rdm(x, i) for(int i=ihead[x]; i; i=e[i].next)

inline const int getint() { int r=0, k=1; char c=getchar(); for(; c<'0'||c>'9'; c=getchar()) if(c=='-') k=-1; for(; c>='0'&&c<='9'; c=getchar()) r=r*10+c-'0'; return k*r; }

const int N=3005;

struct dat { int to, next; }e[N*10];

int cnt, vis[N], c[N], n, m, ihead[N];

void add(int u, int v) { e[++cnt].next=ihead[u]; ihead[u]=cnt; e[cnt].to=v; }

void bfs(int x, int dep) {

	rdm(x, i) {

		int y=e[i].to;

		rdm(y, j) {

			int z=e[j].to;

			if(x==z) continue;

			++c[z];

		}

	}

}

ll ans;

int main() {

	read(n); read(m);

	for1(i, 1, m) { int u=getint(), v=getint(); add(u, v); }

	for1(i, 1, n) {

		for1(j, 1, n) vis[j]=0, c[j]=0;

		bfs(i, 1);

		//for1(j, 1, n) cout << c[j] << ' '; cout << endl;

		for1(j, 1, n) if(c[j]>=2) {

			--c[j];

			ans+=(ll)(c[j]+1)*c[j]/2;

		}

	}

	printf("%I64d\n", ans);

	return 0;

}


    

?


?

?

Tomash keeps wandering off and getting lost while he is walking along the streets of Berland. It's no surprise! In his home town, for any pair of intersections there is exactly one way to walk from one intersection to the other one. The capital of Berland is very different!

Tomash has noticed that even simple cases of ambiguity confuse him. So, when he sees a group of four distinct intersections?a,?b,?c?and?d, such that there are two paths from?a?to?c?— one through?b?and the other one through?d, he calls the group a "damn rhombus". Note that pairs?(a,?b),?(b,?c),?(a,?d),?(d,?c)?should be directly connected by the roads. Schematically, a damn rhombus is shown on the figure below:

【cf489】D. Unbearable Controversy of Being(暴力)

Other roads between any of the intersections don't make the rhombus any more appealing to Tomash, so the four intersections remain a "damn rhombus" for him.

Given that the capital of Berland has?n?intersections and?m?roads and all roads are unidirectional and are known in advance, find the number of "damn rhombi" in the city.

When rhombi are compared, the order of intersections?b?and?d?doesn't matter.

Input

The first line of the input contains a pair of integers?n,?m?(1?≤?n?≤?3000,?0?≤?m?≤?30000) — the number of intersections and roads, respectively. Next?m?lines list the roads, one per line. Each of the roads is given by a pair of integers?ai,?bi?(1?≤?ai,?bi?≤?n;ai?≠?bi) — the number of the intersection it goes out from and the number of the intersection it leads to. Between a pair of intersections there is at most one road in each of the two directions.

It is not guaranteed that you can get from any intersection to any other one.

Output

Print the required number of "damn rhombi".

Sample test(s)
input
          5 4
          
1 2
2 3
1 4
4 3
output
          1
        
input
          4 12
          
1 2
1 3
1 4
2 1
2 3
2 4
3 1
3 2
3 4
4 1
4 2
4 3
output
          12
        

【cf489】D. Unbearable Controversy of Being(暴力)


更多文章、技術(shù)交流、商務(wù)合作、聯(lián)系博主

微信掃碼或搜索:z360901061

微信掃一掃加我為好友

QQ號聯(lián)系: 360901061

您的支持是博主寫作最大的動力,如果您喜歡我的文章,感覺我的文章對您有幫助,請用微信掃描下面二維碼支持博主2元、5元、10元、20元等您想捐的金額吧,狠狠點(diǎn)擊下面給點(diǎn)支持吧,站長非常感激您!手機(jī)微信長按不能支付解決辦法:請將微信支付二維碼保存到相冊,切換到微信,然后點(diǎn)擊微信右上角掃一掃功能,選擇支付二維碼完成支付。

【本文對您有幫助就好】

您的支持是博主寫作最大的動力,如果您喜歡我的文章,感覺我的文章對您有幫助,請用微信掃描上面二維碼支持博主2元、5元、10元、自定義金額等您想捐的金額吧,站長會非常 感謝您的哦!!!

發(fā)表我的評論
最新評論 總共0條評論
主站蜘蛛池模板: 欧美在线精品一区二区三区 | 自拍 亚洲 欧美 | 成人三级毛片 | 久久频这里精品99香蕉久 | 国产日韩欧美在线 | 天天看天天射天天碰 | 羞羞视频免费观看网站 | 亚洲依人 | 欧美精品成人免费视频 | 综合久久久久久 | 国内精品免费久久影院 | 不卡影视| 久久毛片免费看 | 亚洲婷婷在线视频 | 妖精视频免费在线观看 | 亚洲视频一区二区三区四区 | 青青青爽在线视频观看大全 | 中文字幕高清免费不卡视频 | 久久亚洲精品中文字幕亚瑟 | 亚洲成人欧美 | 中文字幕一区二区三区亚洲精品 | 午夜深夜福利网址 | 亚洲狠狠 | 久久国产成人精品麻豆 | 看全大色黄大色黄大片一级爽 | 性生活国产 | 在线综合+亚洲+欧美中文字幕 | xxxx免费国产在线视频 | 午夜影院免费 | 欧洲色网站 | 国产精品久久久久久一区二区三区 | 看全色黄大色大片免费久黄久 | 午夜视频网站 | 一级女人18毛片免费 | 日韩欧美一级毛片在线 | 一区二区三区四区日韩 | 大乳欲妇三级一区二区三区 | 免费乱理伦片在线观看老妇 | 亚洲我射 | 精品久久久久久中文字幕2017 | 国内国产精品天干天干 |