You have n boxes in a line on the table numbered 1 . . . n from left to right. Your task is to simulate 4
kinds of commands:
• 1 X Y : move box X to the left to Y (ignore this if X is already the left of Y )
• 2 X Y : move box X to the right to Y (ignore this if X is already the right of Y )
• 3 X Y : swap box X and Y
• 4: reverse the whole line.
Commands are guaranteed to be valid, i.e. X will be not equal to Y .
For example, if n = 6, after executing 1 1 4, the line becomes 2 3 1 4 5 6. Then after executing
2 3 5, the line becomes 2 1 4 5 3 6. Then after executing 3 1 6, the line becomes 2 6 4 5 3 1.
Then after executing 4, then line becomes 1 3 5 4 6 2
Input
There will be at most 10 test cases. Each test case begins with a line containing 2 integers n, m
(1 ≤ n, m ≤ 100, 000). Each of the following m lines contain a command.
Output
For each test case, print the sum of numbers at odd-indexed positions. Positions are numbered 1 to n
from left to right.
Sample Input
6 4
1 1 4
2 3 5
3 1 6
4
6 3
1 1 4
2 3 5
3 1 6
100000 1
4
Sample Output
Case 1: 12
Case 2: 9
Case 3: 2500050000
题解：
给你一个从1~n的数，然后给你操作方案
• 1 X Y : move box X to the left to Y (ignore this if X is already the left of Y )
• 2 X Y : move box X to the right to Y (ignore this if X is already the right of Y )
• 3 X Y : swap box X and Y
• 4: reverse the whole line.
用双向链表，用着刘汝佳的模版，却无限wa
我的wa代码：
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
using namespace std;
const int INF=0x3f3f3f3f;
const double PI=acos(-1.0);
#define mem(x,y) memset(x,y,sizeof(x))
#define SI(x) scanf("%d",&x)
#define SL(x) scanf("%lld",&x)
#define PI(x) printf("%d",x)
#define PL(x) printf("%lld",x)
#define P_ printf(" ");
typedef long long LL;
const int MAXN=1e5+100;
int lef[MAXN],righ[MAXN];
void link(int l,int r){
lef[r]=l;
righ[l]=r;
}
int main(){
int n,m;
int s,x,y,kase=0;
while(~scanf("%d%d",&n,&m)){
for(int i=1;i<=n;i++){
lef[i]=i-1;
righ[i]=(i+1)%(n+1);
}
lef[0]=n;righ[0]=1;
int flot=0;
while(m--){
scanf("%d",&s);
if(s==4){
flot=!flot;
continue;
}
scanf("%d%d",&x,&y);
int lx=lef[x],rx=righ[x],ly=lef[y],ry=righ[y];
if(s!=3&&flot)s=3-s;
/****************************************/
if(s == 3 && righ[y] == x) swap(x, y);
if(s == 1 && x == lef[y]) continue;
if(s == 2 && x == righ[y]) continue;
/****************************************/
if(s==1){
link(lx,rx);link(ly,x);link(x,y);
}
else if(s==2){
link(lx,rx);link(y,x);link(x,ry);
}
else if(s==3){
if(righ[x] == y) { link(lx, y); link(y, x); link(x, ry); }
else {
link(lx,y);link(y,rx);link(ly,x);link(x,ry);
}
}
}
int cur=0;
LL ans=0;
for(int i=1;i<=n;i++){
cur=righ[cur];
if(i&1)ans+=cur;
}
if(flot&&n%2==0)ans=(LL)n*(n+1)/2-ans;
printf("Case %d: %lld\n",++kase,ans);
}
return 0;
}
刘汝佳的ac代码：
#include<cstdio>
#include<algorithm>
using namespace std;
const int maxn = 100000 + 5;
int n, left[maxn], right[maxn];
inline void link(int L, int R) {
right[L] = R; left[R] = L;
}
int main() {
freopen("1.txt","r",stdin);
freopen("2.txt","w",stdout);
int m, kase = 0;
while(scanf("%d%d", &n, &m) == 2) {
for(int i = 1; i <= n; i++) {
left[i] = i-1;
right[i] = (i+1) % (n+1);
}
right[0] = 1; left[0] = n;
int op, X, Y, inv = 0;
while(m--) {
scanf("%d", &op);
if(op == 4) inv = !inv;
else {
scanf("%d%d", &X, &Y);
if(op == 3 && right[Y] == X) swap(X, Y);
if(op != 3 && inv) op = 3 - op;
if(op == 1 && X == left[Y]) continue;
if(op == 2 && X == right[Y]) continue;
int LX = left[X], RX = right[X], LY = left[Y], RY = right[Y];
if(op == 1) {
link(LX, RX); link(LY, X); link(X, Y);
}
else if(op == 2) {
link(LX, RX); link(Y, X); link(X, RY);
}
else if(op == 3) {
if(right[X] == Y) { link(LX, Y); link(Y, X); link(X, RY); }
else { link(LX, Y); link(Y, RX); link(LY, X); link(X, RY); }
}
}
}
int b = 0;
long long ans = 0;
for(int i = 1; i <= n; i++) {
b = right[b];
if(i % 2 == 1) ans += b;
}
if(inv && n % 2 == 0) ans = (long long)n*(n+1)/2 - ans;
printf("Case %d: %lld\n", ++kase, ans);
}
return 0;
}
uva-12657 - Boxes in a Line（双向链表）
原文：http://www.cnblogs.com/handsomecui/p/5083213.html