长度不小于k的公共子串个数
题目链接：https://cn.vjudge.net/contest/318888#problem/J
思路：
1 #include <stdio.h> 2 #include <iostream> 3 #include <algorithm> 4 #include <string.h> 5 #include <stdlib.h> 6 #include <math.h> 7 #include <queue> 8 #include <set> 9 10 #define INF 0x3f3f3f3f 11 #define pii pair<int,int> 12 #define LL long long 13 using namespace std; 14 typedef unsigned long long ull; 15 const int MAXN = 2e5 + 10; 16 int t; 17 int wa[MAXN], wb[MAXN], wv[MAXN], ws_[MAXN]; 18 int sta[MAXN]; 19 int cnt[MAXN]; 20 void Suffix(int *r, int *sa, int n, int m) 21 { 22 int i, j, k, *x = wa, *y = wb, *t; 23 for(i = 0; i < m; ++i) ws_[i] = 0; 24 for(i = 0; i < n; ++i) ws_[x[i] = r[i]]++; 25 for(i = 1; i < m; ++i) ws_[i] += ws_[i - 1]; 26 for(i = n - 1; i >= 0; --i) sa[--ws_[x[i]]] = i; 27 for(j = 1, k = 1; k < n; j *= 2, m = k) 28 { 29 for(k = 0, i = n - j; i < n; ++i) y[k++] = i; 30 for(i = 0; i < n; ++i) if(sa[i] >= j) y[k++] = sa[i] - j; 31 for(i = 0; i < n; ++i) wv[i] = x[y[i]]; 32 for(i = 0; i < m; ++i) ws_[i] = 0; 33 for(i = 0; i < n; ++i) ws_[wv[i]]++; 34 for(i = 1; i < m; ++i) ws_[i] += ws_[i - 1]; 35 for(i = n - 1; i >= 0; --i) sa[--ws_[wv[i]]] = y[i]; 36 t = x; 37 x = y; 38 y = t; 39 for(x[sa[0]] = 0, i = k = 1; i < n; ++i) 40 x[sa[i]] = (y[sa[i - 1]] == y[sa[i]] && y[sa[i - 1] + j] == y[sa[i] + j]) ? k - 1 : k++; 41 } 42 } 43 int Rank[MAXN], height[MAXN], sa[MAXN], r[MAXN]; 44 char s[MAXN],ss[MAXN]; 45 int indexx[MAXN],vis[200]; 46 void calheight(int *r,int *sa,int n) 47 { 48 int i,j,k=0; 49 for(i=1; i<=n; i++)Rank[sa[i]]=i; 50 for(i=0; i<n; height[Rank[i++]]=k) 51 for(k?k--:0,j=sa[Rank[i]-1]; r[i+k]==r[j+k]; k++); 52 } 53 54 55 int main(){ 56 while (~scanf("%d",&t) && t){ 57 memset(r,0, sizeof(r)); 58 memset(sa,0, sizeof(sa)); 59 memset(Rank,0, sizeof(Rank)); 60 scanf("%s%s",s,ss); 61 int len1 = strlen(s); 62 int len2 = strlen(ss); 63 s[len1] = '@'; 64 for (int i=0;i<len2;i++){ 65 s[len1+1+i] = ss[i]; 66 } 67 s[len1+len2+1] = '\0'; 68 for (int i=0;i<len1+len2+1;i++){ 69 r[i] = s[i]; 70 } 71 r[len1+len2+2] = 0; 72 Suffix(r,sa,len1+len2+2,300); 73 calheight(r,sa,len1+len2+1); 74 LL sum = 0,ans = 0; //sum代表栈里面对答案的贡献值，cnt数组记录的是一个组内的后缀数 75 int top = 0; 76 for (int i=1;i<len1+len2+2;i++){ // 处理A和它前面B的公共前缀 77 if (height[i]<t){ 78 top = 0; 79 sum = 0; 80 continue; 81 } 82 int num = 0; 83 while (top && height[i]<sta[top]){ 84 sum -= (LL)(sta[top]-t+1)*cnt[top]; //在这儿比height[i]大的都要变成height[i]，所以先减去先前加上去的 85 sum += (LL)(height[i]-t+1)*cnt[top]; //乘cnt[top]的原因是可能有多个合并成一个了，比如4,5,3，那么4和5都只能以3来计算 86 num += cnt[top]; //那么此时这三个可以合并在一起，cnt个数就是3，和就是3*3=9 87 top--; 88 } 89 sta[++top] = height[i]; 90 if (sa[i-1]>len1){ 91 sum += (LL)(height[i]-t+1); 92 cnt[top] = num+1; 93 } 94 else 95 cnt[top] = num; 96 if (sa[i]<len1) 97 ans += sum; 98 } 99 sum = 0,top = 0; 100 101 for (int i=1;i<=len1+len2+2;i++){ //处理B和它前面的A的公共前缀 102 if (height[i]<t){ 103 top = 0; 104 sum = 0; 105 continue; 106 } 107 int num = 0; 108 while (top && height[i]<sta[top]){ 109 sum -= (LL)(sta[top]-t+1)*cnt[top]; 110 sum += (LL)(height[i]-t+1)*cnt[top]; 111 num += cnt[top]; 112 top--; 113 } 114 sta[++top] = height[i]; 115 if (sa[i-1]<len1){ 116 sum += (LL)(height[i]-t+1); 117 cnt[top] = num+1; 118 } 119 else 120 cnt[top] = num; 121 if (sa[i]>len1) 122 ans += sum; 123 } 124 printf("%lld\n",ans); 125 } 126 return 0; 127 }
