\(C_{x+y}^y\)的公式,DP容斥删多余贡献. #include <cstdio> #include <iostream> #include <cstring> #include <algorithm> #include <cmath> #define R(a,b,c) for(register int a = (b); (a) <= (c); ++(a)) #define nR(a,b,c) for(register int a…

C. Gerald and Giant Chess time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Giant chess is quite common in Geraldion. We will not delve into the rules of the game, we'll just say that the ga…

LINK 题目大意 有一个wxh的网格,上面有n个黑点,问你从(1,1)走到(w,h)不经过任何黑点的方案数 思路 考虑容斥 先把所有黑点按照x值进行排序方便计算 \(dp_{i}\)表示从起点走到第i个黑点不经过任何的黑点的方案数 然后\(dp_{i}=C(x_i+y_i-2,x_i-1)-\sum_{j|x_j\leq x_i,y_j\leq y_i}dp_{j}\times C(j->i)\) 这样容斥为什么是正确的,\(dp_{j}\)考虑了所有经过j的情况,其他的包含j的点都不会考虑j…

题意:在h×w的棋盘中从左上角走到右下角,只能向右或向下走,有n个点不可以经过,一共有多少种方案. 解法:dp.先对点按横坐标排序(横坐标相等按纵坐标,也可以反过来)dp[i]表示不经过其他非法点走到第i个非法点的路径数,则有dp方程:dp[i] = c(point[i].x + point[i].y - 2, point[i].x - 1) - Σj = 0...i - 1 dp[j] * c(point[i].x - point[j].x + point[i].y - point[j].y,…

题目链接:http://codeforces.com/contest/560/problem/E 给你一个n*m的网格,有k个坏点,问你从(1,1)到(n,m)不经过坏点有多少条路径. 先把这些坏点排序一下. dp[i]表示从(1,1)到第i个坏点且不经过其他坏点的路径数目. dp[i] = Lucas(x[i], y[i]) - sum(dp[j]*Lucas(x[i]-x[j], y[i]-x[j])) , x[j] <= x[i] && y[j] <= y[i] //到i…

传送门 令f[i]f[i]f[i]表示对于第iii个棋子,从(1,1)(1,1)(1,1)出发到它不经过其它棋子的方案数. 于是我们假设(h,w)(h,w)(h,w)有一个棋子,求出它的fff值就可以了. 然后考虑容斥转移fff数组. 根据定义,我们求出从(1,1)(1,1)(1,1)出发到它的总方案数,再减去经过了其它棋子的方案数. 然后再考虑如何才会补充不漏. 发现从之前每一个fff转移过来就行了. fi=(xi+yi−2xi−1)−∑fj∗(xi−yi+xj−yjxi−xj)f_i=\bi…

Gerald and Giant Chess Problem's Link: http://codeforces.com/contest/559/problem/C Mean: 一个n*m的网格,让你从左上角走到右下角,有一些点不能经过,问你有多少种方法. analyse: BZOJ上的原题. 首先把坏点和终点以x坐标为第一键值,y坐标为第二键值排序 . 令fi表示从原点不经过任何坏点走到第i个点的个数,那么有DP方程: fi=Cxixi+yi−∑(xj<=xi,yj<=yi)C(xi−xj)…

C. Gerald and Giant Chess Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/559/problem/C Description Gerald got a very curious hexagon for his birthday. The boy found out that all the angles of the hexagon are equal to . The…

C. Gerald and Giant Chess time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Giant chess is quite common in Geraldion. We will not delve into the rules of the game, we'll just say that the ga…

[题解]CF559C C. Gerald and Giant Chess(容斥+格路问题) 55336399 Practice: Winlere 559C - 22 GNU C++11 Accepted 186 ms 1608 KB 2019-06-09 17:03:21 2019-06-09 17:03:21 一道小水题(为什么2400??我为什么之前被一道2200锤QAQ) 有个很显然的公式,在组合数学那本书上面也有. 从坐标\((0,0)\)到坐标\((x,y)\)总共有\({x+y}\c…