USACO: Riding the Fences

Riding the Fences

Farmer John owns a large number of fences that must be repaired annually. He traverses the fences by riding a horse along each and every one of them (and nowhere else) and fixing the broken parts.

Farmer John is as lazy as the next farmer and hates to ride the same fence twice. Your program must read in a description of a network of fences and tell Farmer John a path to traverse each fence length exactly once, if possible. Farmer J can, if he wishes, start and finish at any fence intersection.

Every fence connects two fence intersections, which are numbered inclusively from 1 through 500 (though some farms have far fewer than 500 intersections). Any number of fences (>=1) can meet at a fence intersection. It is always possible to ride from any fence to any other fence (i.e., all fences are "connected").

Your program must output the path of intersections that, if interpreted as a base 500 number, would have the smallest magnitude.

There will always be at least one solution for each set of input data supplied to your program for testing.

PROGRAM NAME: fence

INPUT FORMAT

Line 1:	The number of fences, F (1 <= F <= 1024)
Line 2..F+1:	A pair of integers (1 <= i,j <= 500) that tell which pair of intersections this fence connects.

SAMPLE INPUT (file fence.in)

9
1 2
2 3
3 4
4 2
4 5
2 5
5 6
5 7
4 6

OUTPUT FORMAT

The output consists of F+1 lines, each containing a single integer. Print the number of the starting intersection on the first line, the next intersection's number on the next line, and so on, until the final intersection on the last line. There might be many possible answers to any given input set, but only one is ordered correctly.

SAMPLE OUTPUT (file fence.out)

1
2
3
4
2
5
4
6
5
7

 

解体思路：

第一次写欧拉路，为了提速，用了邻接矩阵+邻接表

写这道题时我受到的启发是：邻接矩阵删除边比较方便，邻接表建堆查找下一条最小编号边比较方便，二者不可兼得，配合使用比较好

删除边的时候在邻接矩阵中操作，而邻接表中的相应边没有改动，而是留待下一次弹出最小编号的边的时候与邻接矩阵比较，看看这条边是否还是存在的，就是这几行代码：

while (degr[curPs] && edge[curPs][cont[curPs][0]] == 0){ cont[curPs][0] = cont[curPs][-- degr[curPs]]; siftDw(curPs); }

 最终代码如下：

/* ID: geterns1 PROG: fence LANG: C++ */ #include <iostream> #include <fstream> #include <cstdio> #include <cmath> #include <cstdlib> #include <cstring> #include <algorithm> using namespace std; const int size = 500 + 1; const int pnum = 1024 + 1; //原始数据 int realSize, realPnum; int edge[size][size]; //记录边的条数（一对结点允许有多条边） int cont[size][size], degr[size]; //堆记录的连接点和出入度 inline void qSwap(int &a, int &b){ a ^= b; b ^= a; a ^= b; } inline int fath(int idx){ return (idx - 1) / 2; } inline int left(int idx){ return 2 * idx + 1; } inline int rght(int idx){ return 2 * (idx + 1); } void siftDw(int idx){ int curIdx = 0, tarIdx = 0; while (true){ if (left(curIdx) < degr[idx] && cont[idx][left(curIdx)] < cont[idx][tarIdx]) tarIdx = left(curIdx); if (rght(curIdx) < degr[idx] && cont[idx][rght(curIdx)] < cont[idx][tarIdx]) tarIdx = rght(curIdx); if (tarIdx ^ curIdx){ qSwap(cont[idx][curIdx], cont[idx][tarIdx]); curIdx = tarIdx; } else return; } } void siftUp(int idx){ int tarIdx = degr[idx] - 1; while (tarIdx){ if (cont[idx][fath(tarIdx)] > cont[idx][tarIdx]){ qSwap(cont[idx][fath(tarIdx)], cont[idx][tarIdx]); tarIdx = fath(tarIdx); } else return; } } //辅助数组 int stack[pnum], top; int curPs; int ePath[pnum], cnt; //假设已知图是连通的 bool cir, pat; void check(){ int oddCnt = 0; cir = true; pat = false; for (int i = 1; i <= realSize; i ++){ if (degr[i] & 1){ cir = false; oddCnt ++; } } if (oddCnt == 0 || oddCnt == 2) pat = true; } // void eulerTours(){ top = cnt = 0; for (int i = 1; i <= realSize; i ++){ if (degr[i] && (cir || (degr[i] & 1))){ curPs = i; break; } } while (true){ while (degr[curPs] && edge[curPs][cont[curPs][0]] == 0){ cont[curPs][0] = cont[curPs][-- degr[curPs]]; siftDw(curPs); } if (degr[curPs]){ edge[curPs][cont[curPs][0]] --; edge[cont[curPs][0]][curPs] --; stack[top ++] = curPs; curPs = cont[curPs][0]; } else{ ePath[cnt ++] = curPs; if (top == 0) return; curPs = stack[-- top]; } } } int main(){ freopen("fence.in", "r", stdin); freopen("fence.out", "w", stdout); int x, y; scanf("%d/n", &realPnum); for (int i = 0; i < realPnum; i ++){ scanf("%d %d/n", &x, &y); edge[x][y] ++; edge[y][x] ++; cont[x][degr[x] ++] = y; cont[y][degr[y] ++] = x; siftUp(x); siftUp(y); if (x > realSize) realSize = x; if (y > realSize) realSize = y; } check(); eulerTours(); for (int i = cnt - 1; i > -1; i --) printf("%d/n", ePath[i]); fclose(stdin); fclose(stdout); return 0; }