UVA - 12657 Boxes in a Line(双向链表)

来源：互联网发布：linux more 下一页编辑：程序博客网时间：2024/05/22 02:32

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int maxn = 100000+5;int n,m;int Left[maxn],Right[maxn];void link(int L ,int R){    Right[L] = R,Left[R] = L;}int main(){    int kase = 1;    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 x,y,check = 0,op;        while(m--){            scanf("%d",&op);            if(op == 4) check = !check;            else{                scanf("%d%d",&x,&y);                if(op == 3 && Right[y] == x) swap(x,y);                if(op != 3 && check) 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(check && n%2 == 0) ans = (long long) n * (n+1)/2 -ans;        printf("Case %d: %lld\n",kase++,ans);    }    return 0;}

题目是这样的，给出n个盒子,m条指令(1<=n,m<=100000)，

1 x y 把盒子x放在y的左边

2 x y 把盒子x放在y的右边

3 x y 把x,y调换

4 把整个序列反转

题解，利用双向链表的性质，有左链表和右链表。

主要是两个结点相互连接

void link(int L,int R){

right[L] = R, left[R] = L;

}

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