codeforces Round #269(div2) D解题报告
2014-09-30 09:52
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D. MUH and Cube Walls
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
Polar bears Menshykov and Uslada from the zoo of St. Petersburg and elephant Horace from the zoo of Kiev got hold of lots of wooden cubes somewhere. They started making cube towers by placing the cubes one on top of the other. They defined multiple towers standing
in a line as a wall. A wall can consist of towers of different heights.
Horace was the first to finish making his wall. He called his wall an elephant. The wall consists of w towers.
The bears also finished making their wall but they didn't give it a name. Their wall consists of n towers.
Horace looked at the bears' tower and wondered: in how many parts of the wall can he "see an elephant"? He can "see an elephant" on a segment of w contiguous
towers if the heights of the towers on the segment match as a sequence the heights of the towers in Horace's wall. In order to see as many elephants as possible, Horace can raise and lower his wall. He even can lower the wall below the ground level (see the
pictures to the samples for clarification).
Your task is to count the number of segments where Horace can "see an elephant".
Input
The first line contains two integers n and w (1 ≤ n, w ≤ 2·105)
— the number of towers in the bears' and the elephant's walls correspondingly. The second line contains n integers ai (1 ≤ ai ≤ 109)
— the heights of the towers in the bears' wall. The third line contains w integers bi (1 ≤ bi ≤ 109)
— the heights of the towers in the elephant's wall.
Output
Print the number of segments in the bears' wall where Horace can "see an elephant".
Sample test(s)
input
output
Note
The picture to the left shows Horace's wall from the sample, the picture to the right shows the bears' wall. The segments where Horace can "see an elephant" are in gray.
int n, m;
int tmp[200010], a[200010], b[200010], p[200010];
void init() {
scanf("%d%d", &n, &m);
for (int i = 1; i <= n; i++) {
scanf("%d", &tmp[i]);
a[i-1] = tmp[i]-tmp[i-1];
}
for (int i = 1; i <= m; i++) {
scanf("%d", &tmp[i]);
b[i-1] = tmp[i]-tmp[i-1];
}
}
void p_treat() {
int j = 0;
p[1]=0;
for (int i = 2; i <= m; i++) {
while (j>0 && b[j+1]!= b[i]) j=p[j];
if (b[j+1] == b[i]) j++;
p[i]=j;
}
}
void solve() {
if (m == 1) {
printf("%d", n);
return;
}
p_treat();
int j=0, ans=0;
for (int i = 1; i < n; i++) {
while (j>0 && b[j+1]!=a[i]) j=p[j];
if (b[j+1] == a[i]) j++;
if (j == m-1) {
ans++;
j=p[j];
}
}
printf("%d", ans);
}
int main() {
init();
solve();
}
官方解题报告中,还有另外两种算法,Z函数与后缀数组,这个我还不会= =,先留着
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
Polar bears Menshykov and Uslada from the zoo of St. Petersburg and elephant Horace from the zoo of Kiev got hold of lots of wooden cubes somewhere. They started making cube towers by placing the cubes one on top of the other. They defined multiple towers standing
in a line as a wall. A wall can consist of towers of different heights.
Horace was the first to finish making his wall. He called his wall an elephant. The wall consists of w towers.
The bears also finished making their wall but they didn't give it a name. Their wall consists of n towers.
Horace looked at the bears' tower and wondered: in how many parts of the wall can he "see an elephant"? He can "see an elephant" on a segment of w contiguous
towers if the heights of the towers on the segment match as a sequence the heights of the towers in Horace's wall. In order to see as many elephants as possible, Horace can raise and lower his wall. He even can lower the wall below the ground level (see the
pictures to the samples for clarification).
Your task is to count the number of segments where Horace can "see an elephant".
Input
The first line contains two integers n and w (1 ≤ n, w ≤ 2·105)
— the number of towers in the bears' and the elephant's walls correspondingly. The second line contains n integers ai (1 ≤ ai ≤ 109)
— the heights of the towers in the bears' wall. The third line contains w integers bi (1 ≤ bi ≤ 109)
— the heights of the towers in the elephant's wall.
Output
Print the number of segments in the bears' wall where Horace can "see an elephant".
Sample test(s)
input
13 5 2 4 5 5 4 3 2 2 2 3 3 2 1 3 4 4 3 2
output
2
Note
The picture to the left shows Horace's wall from the sample, the picture to the right shows the bears' wall. The segments where Horace can "see an elephant" are in gray.
题目大意:
给出2堵墙,A墙已经被命名。找出在B墙中,有多少处与A墙相似解法:
其实就是找出相似的轮廓线,轮廓线可以抽象成字符串,例如样例 的A墙(0, 1, 0 ,-1 ,-1)。第一想法想到了KMP算法。代码:
#include <cstdio>int n, m;
int tmp[200010], a[200010], b[200010], p[200010];
void init() {
scanf("%d%d", &n, &m);
for (int i = 1; i <= n; i++) {
scanf("%d", &tmp[i]);
a[i-1] = tmp[i]-tmp[i-1];
}
for (int i = 1; i <= m; i++) {
scanf("%d", &tmp[i]);
b[i-1] = tmp[i]-tmp[i-1];
}
}
void p_treat() {
int j = 0;
p[1]=0;
for (int i = 2; i <= m; i++) {
while (j>0 && b[j+1]!= b[i]) j=p[j];
if (b[j+1] == b[i]) j++;
p[i]=j;
}
}
void solve() {
if (m == 1) {
printf("%d", n);
return;
}
p_treat();
int j=0, ans=0;
for (int i = 1; i < n; i++) {
while (j>0 && b[j+1]!=a[i]) j=p[j];
if (b[j+1] == a[i]) j++;
if (j == m-1) {
ans++;
j=p[j];
}
}
printf("%d", ans);
}
int main() {
init();
solve();
}
官方解题报告:
地址:http://codeforces.com/blog/entry/13986官方解题报告中,还有另外两种算法,Z函数与后缀数组,这个我还不会= =,先留着
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