hdu 5348 MZL's endless loop 欧拉回路

MZL's endless loop

Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1502    Accepted Submission(s): 331
Special Judge

Problem Description

As we all kown, MZL hates the endless loop deeply, and he commands you to solve this problem to end the loop.
You are given an undirected graph with n vertexs and m edges. Please direct all the edges so that for every vertex in the graph the inequation |out degree − in degree|≤1 is satisified.
The graph you are given maybe contains self loops or multiple edges.

 

Input

The first line of the input is a single integer T, indicating the number of testcases.
For each test case, the first line contains two integers n and m.
And the next m lines, each line contains two integers ui and vi, which describe an edge of the graph.
T≤100, 1≤n≤105, 1≤m≤3∗105, ∑n≤2∗105, ∑m≤7∗105.

 

Output

For each test case, if there is no solution, print a single line with −1, otherwise output m lines,.
In ith line contains a integer 1 or 0, 1 for direct the ith edge to ui→vi, 0 for ui←vi.

 

Sample Input

23 31 22 33 17 61 21 31 41 51 61 7

 

Sample Output

111010101

 

Source

2015 Multi-University Training Contest 5

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5348

题目大意：将一个无向图构造出一个有向图，要求每个点的入度和出度之差绝对值不超过1。

思路，重点在于删边，边表删边相对于vector存储而言更方便一些，因为.next自带边号，每次把head[]，以i为起点的第一条边存储的位置(实际上是最后输入的那条边的边号)改为edge[i].next，即实现了“拆边”。

AC代码：

#include <iostream>#include <cstdio>#include <cmath>#include <cstring>#include <algorithm>#include <vector>using namespace std;#define MAXN 100010#define MAXM 300010int t, n, m;int targ[MAXN][2];int degree[MAXN], ans[MAXM << 1];int head[MAXN], edge_cnt;struct node {    int v, next;} edge[MAXM << 1];void init() {    memset(head, -1, sizeof head);    memset(degree, 0, sizeof degree);    memset(targ, 0, sizeof targ);    memset(ans, -1, sizeof ans);    edge_cnt = 0;}void add_edge(int u, int v) {    edge[edge_cnt].v = v;    edge[edge_cnt].next = head[u];    head[u] = edge_cnt++;}void DFS(int u, int col) {    for(int i = head[u]; ~i; i = edge[i].next) {        if(ans[i] != -1) {            head[u] = edge[i].next;//拆边            continue;        }        int v = edge[i].v;        if(u != v && targ[v][col ^ 1] > targ[v][col]) continue;        ans[i] = col;        ans[i ^ 1] = col ^ 1;        head[u] = edge[i].next;        targ[u][col]++;        targ[v][col ^ 1]++;        DFS(v, col);        break;    }}int main() {    scanf("%d", &t);    while(t--) {        scanf("%d%d", &n, &m);        init();        int a, b;        for(int i = 1; i <= m; i++) {            scanf("%d%d", &a, &b);            add_edge(a, b);            add_edge(b, a);            degree[a]++; degree[b]++;        }        for(int i = 1; i <= n; i++) {            while(targ[i][0] + targ[i][1] < degree[i]) {                if(targ[i][0] < targ[i][1]) DFS(i, 0);                else DFS(i, 1);            }        }        for(int i = 0; i < edge_cnt; i += 2) {            printf("%d\n", ans[i]);        }    }    return 0;}