問(wèn)題 C: A Path Plan

時(shí)間限制:?1 Sec??內(nèi)存限制:?128 MB
提交:?55??解決:?26
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題目描述

WNJXYK hates Destinys so that he does not want to meet him at any time. Luckily, their classrooms and dormitories are at different places. The only chance for them to meet each other is on their way to classrooms from dormitories.?
To simple this question, we can assume that the map of school is a normal rectangular 2D net. WNJXYK’s dormitory located at (0,y_1) and his classroom located at (x_1,0). Destinys’s dormitory located at (0,y_2) and his classroom is located at (x_2,0). On their way to classrooms, they can do two kinds of movement : (x,y)→(x,y-1) and (x,y)→(x+1,y).?
WNJXYK does not want to meet Destinys so that he thinks that it is not safe to let his path to classroom from dormitory has any intersect point with Destinys ‘s. And then he wonders how many different paths for WNJXYK and Destinys arriving their classrooms from dormitories safely.

輸入

The input starts with one line contains exactly one positive integer T which is the number of test cases.
Each test case contains one line with four positive integers x1,x2,y1,y2 which has been explained above.

輸出

For each test case, output one line containing “y” where y is the number of different paths modulo 10^9+7.

樣例輸入

3 1 2 1 2 2 3 2 4 4 9 3 13

樣例輸出

3 60 16886100

提示

T≤1000
x1<x2,y1<y2
0 < x1,x2,y1,y2≤100000
For Test Case 1, there are following three different ways.

題意:兩個(gè)人分別從(0,y1)-(x1,0)和(0,y2)-(x2,0),求他們路徑不相交的方案數(shù)。

題解:比賽的時(shí)候用組合數(shù)學(xué)想了好久也沒(méi)想到,最后沒(méi)想到有這么個(gè)定理,還是太渣了,定理如下

對(duì)于一張無(wú)邊權(quán)的DAG圖，給定n個(gè)起點(diǎn)和對(duì)應(yīng)的n個(gè)終點(diǎn)，這n條不相交路徑的方案數(shù)為

det()?（該矩陣的行列式）

其中e(a,b)為圖點(diǎn)a到點(diǎn)b的方案數(shù),如此答案只需要求解這個(gè)行列式即可。

2.行列式計(jì)算

https://blog.csdn.net/zhoufenqin/article/details/7779707

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ps:求階乘的逆元的小trick

inv[maxn] = qk_mod(fac[maxn],mod - 2);

for(int i = maxn - 1;i >= 0;i --) inv[i] = inv[i + 1] * (i + 1) % mod;

#include<bits/stdc++.h> #define rep(i,s,t) for(int i=s;i<=t;i++) #define SI(i) scanf("%lld",&i) #define PI(i) printf("%lld\n",i) using namespace std; typedef long long ll; const int mod=1e9+7; const int MAX=1e6+7; const int INF=0x3f3f3f3f; const double eps=1e-8; int dir[9][2]={0,1,0,-1,1,0,-1,0, -1,-1,-1,1,1,-1,1,1}; template<class T>bool gmax(T &a,T b){return a<b?a=b,1:0;} template<class T>bool gmin(T &a,T b){return a>b?a=b,1:0;} template<class T>void gmod(T &a,T b){a=((a+b)%mod+mod)%mod;} typedef pair<ll,ll> PII;ll qpow(ll a,ll b) {a%=mod;ll ans=1;while(b){if(b&1)ans=(ans*a)%mod;a=(a*a)%mod;b/=2;}return ans; }ll fac[MAX]; ll C(ll a,ll b) {return fac[a]*qpow(fac[a-b],mod-2)%mod*qpow(fac[b],mod-2)%mod; }int main() {fac[0]=1;for(int i=1;i<=200005;i++)fac[i]=fac[i-1]*i%mod;int T;scanf("%d",&T);while(T--){ll x1,x2,y1,y2;scanf("%lld%lld%lld%lld",&x1,&x2,&y1,&y2);ll ans=C(x1+y1,x1)*C(x2+y2,x2)%mod-C(x1+y2,x1)*C(x2+y1,x2)%mod;gmod(ans,(ll)mod);printf("%lld\n",ans);} }

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