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;}请前辈们帮忙看一下！
与其用泪水悔恨今天，不如用汗水拼搏今天。
