标签：
HDU - 1711
Description Given two sequences of numbers : a[1], a[2], ...... , a[N], and b[1], b[2], ...... , b[M] (1 <= M <= 10000, 1 <= N <= 1000000). Your task is to find a number K which make a[K] = b[1], a[K + 1] = b[2], ...... , a[K + M - 1] = b[M].
If there are more than one K exist, output the smallest one.
Input The first line of input is a number T which indicate the number of cases. Each case contains three lines. The first line is two numbers N and M (1 <= M <= 10000, 1 <= N <= 1000000). The second line contains N integers which indicate
a[1], a[2], ...... , a[N]. The third line contains M integers which indicate b[1], b[2], ...... , b[M]. All integers are in the range of [-1000000, 1000000].
Output For each test case, you should output one line which only contain K described above. If no such K exists, output -1 instead.
Sample Input
Sample Output
第一类型代码：
/*
Problem : 1711 ( Number Sequence ) Judge Status : Accepted
RunId : 14382956 Language : G++ Author : 24862486
Code Render Status : Rendered By HDOJ G++ Code Render Version 0.01 Beta
*/
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>
using namespace std;
const int MAXN = 10000 + 5;
const int MAXM = 1000000 + 5;
int T, N, M, a[MAXM], b[MAXN], nexts[MAXN];
void Get_Next() {
int i = 0, j = -1;
nexts[0] = -1;
while(i < M - 1) {
if(j == -1 || b[j] == b[i]) {
j ++;
i ++;
if(b[j] == b[i]) {
nexts[i] = nexts[j];
} else {
nexts[i] = j;
}
} else {
j = nexts[j];
}
}
}
int Get_KMP() {
int i = 0, j = 0;
while(i < N && j < M) {
if(j == -1 || a[i] == b[j]) {
i ++;
j ++;
} else {
j = nexts[j];//失配匹配
}
}
if(j == M) {
return i - j + 1;
}
return -1;
}
int main() {
scanf("%d", &T);
while(T --) {
scanf("%d %d", &N, &M);
for(int i = 0; i < N; i ++) {
scanf("%d", &a[i]);
}
for(int i = 0; i < M; i ++) {
scanf("%d", &b[i]);
}
Get_Next();
int cnt = Get_KMP();
printf("%d\n", cnt);
}
return 0;
}
第二中类型：效率较低，但是代码简单，便于记忆以及理解 /*
Problem : 1711 ( Number Sequence ) Judge Status : Accepted
RunId : 14382956 Language : G++ Author : 24862486
Code Render Status : Rendered By HDOJ G++ Code Render Version 0.01 Beta
*/
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>
using namespace std;
const int MAXN = 10000 + 5;
const int MAXM = 1000000 + 5;
int T, N, M, a[MAXM], b[MAXN], nexts[MAXN];
void Get_Next() {
int i = 0, j = -1;
nexts[0] = -1;
while(i < M) {
if(j == -1 || b[j] == b[i]) {
nexts[++ i] = ++ j;
} else {
j = nexts[j];
}
}
}
int Get_KMP() {
int i = 0, j = 0;
while(i < N && j < M) {
if(j == -1 || a[i] == b[j]) {
i ++;
j ++;
} else {
j = nexts[j];//失配匹配
}
}
if(j == M) {
return i - j + 1;
}
return -1;
}
int main() {
scanf("%d", &T);
while(T --) {
scanf("%d %d", &N, &M);
for(int i = 0; i < N; i ++) {
scanf("%d", &a[i]);
}
for(int i = 0; i < M; i ++) {
scanf("%d", &b[i]);
}
Get_Next();
int cnt = Get_KMP();
printf("%d\n", cnt);
}
return 0;
}
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HDU - 1711 Number Sequence KMP字符串匹配
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原文地址：http://blog.csdn.net/qq_18661257/article/details/47323835
