2012: [Ceoi2010]Pin

2012: [Ceoi2010]Pin

Time Limit: 8 Sec  Memory Limit: 259 MB
Submit: 266  Solved: 132
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Description

给出N(2<=N<=50000)个长度为4的字符串，问有且仅有D(1<=D<=4)处不相同的字符串有几对。

Input

第1行： N，D 以下N行每行一个字符串

Output

一个数：有多少对有且仅有处不相同的字符串。

Sample Input

4 2
0000
a010
0202
a0e2

Sample Output

3

HINT

Source

鸣谢sachs

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搜索 && 容斥果然我太弱，还是要信仰Mektpoy先hash + dfs搜出g[i]:至少有i个位置相同的对数然后容斥出f数组

#include<iostream>#include<cstdio>#include<queue>#include<vector>#include<bitset>#include<algorithm>#include<cstring>#include<map>#include<stack>#include<set>#include<cmath>#include<ext/pb_ds/priority_queue.hpp>using namespace std;typedef long long LL;const int maxn = 5E4 + 50;const LL hash = 131;int n,d;LL ans,h[5],ha[maxn][5],g[5],f[5],C[5][5],sum[maxn];char ch[maxn][5];bool bo[5];void dfs(int x,int tot){if (x == 5) {if (tot) {map <LL,int> m; ans = 0;for (int i = 1; i <= n; i++) {LL now = 0;for (int j = 1; j <= 4; j++)if (bo[j]) now += ha[i][j];if (m[now] > 1) ans -= sum[m[now]-1];++m[now]; ans += sum[m[now]-1];}g[tot] += ans;}return;} bo[x] = 0; dfs(x + 1,tot);bo[x] = 1; dfs(x + 1,tot + 1);}int main(){#ifdef DMCfreopen("DMC.txt","r",stdin);#endifcin >> n >> d;for (LL i = 1; i < maxn; i++) sum[i] = sum[i-1] + i;for (int i = 1; i <= n; i++) scanf("%s",1+ch[i]);h[0] = 1; for (int i = 1; i <= 4; i++) h[i] = h[i-1]*hash;for (int i = 1; i <= n; i++)for (int j = 1; j <= 4; j++)ha[i][j] = 1LL*ch[i][j]*h[j];dfs(1,0);C[0][0] = 1;for (int i = 1; i <= 4; i++) {C[i][0] = 1;for (int j = 1; j <= i; j++) C[i][j] = C[i-1][j] + C[i-1][j-1];}f[4] = g[4];f[3] = g[3] - C[4][3]*f[4];f[2] = g[2] - C[3][2]*f[3] - C[4][2]*f[4];f[1] = g[1] - C[2][1]*f[2] - C[3][1]*f[3] - C[4][1]*f[4];f[0] = sum[n-1] - C[1][0]*f[1] - C[2][0]*f[2] - C[3][0]*f[3] - C[4][0]*f[4];cout << f[4 - d];return 0;}