【日常学习】【模拟双向链表】【疑问】Uva12657 - Boxes in a Line题解

标签：日常学习   疑问   链表   

这道题目我做的不对。事实上，我按书上的标程抄的，几乎一模一样，我认为他没有什么错误，可我就是不知道为什么我在代码仓库下的刘汝佳写的程序就AC，我写的就WA。跳了一下午，两程序样例输出完全一样（奇怪的是和书上答案不一样）一个字一个字的比对，就是找不出哪里不一样。我觉得极少不一样的地方应该没有影响，哪位大神愿意给看看？

这是一道双向链表，同样没有用指针，而是用两个数组模拟，道理和上面的那道非指针单向链表题目一样

刘汝佳的代码

// UVa12657 Boxes in a Line
// Rujia Liu
#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;
}

我的代码

//Boxes in a line-doubly linked list
#include<cstdio>
#include<algorithm>
using namespace std;

int left[100010],right[100010];

void linked(int l,int r){
	right[l]=r;left[r]=l;                                        
} 

int main(){
	freopen("1.txt","r",stdin);
	freopen("2.txt","w",stdout);
	int n,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);//good deal
		}
		right[0]=1;left[0]=n;//notice!!!
		int op,x,y,inv;
		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);//all are changed!!!
				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){
					linked(lx,rx);linked(ly,x);linked(x,y);
				}
				else if (op==2){
					linked(lx,rx);linked(y,x);linked(x,ry);//here is wrongTUT
				}
				else if (op==3){
					if (right[x]==y){
						linked(lx,y);linked(y,x);linked(x,ry);
					}
					else {
						linked(ly,x);linked(x,ry);linked(lx,y);linked(y,rx);
					}
				}
			}
		}
		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;
}

【日常学习】【模拟双向链表】【疑问】Uva12657 - Boxes in a Line题解

标签：日常学习   疑问   链表   

原文地址：http://blog.csdn.net/ametake/article/details/44018179