Codeforces 501C - Misha and Forest (机智)

最新推荐文章于 2019-11-24 21:08:57 发布

原创最新推荐文章于 2019-11-24 21:08:57 发布 · 2.3k 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#ACM #codeforces

题意

给出每个点的度数和和这个点连接的点的异或和，输出图的每条边。

思路

看了squee_spoon的题解。

这题的突破口在于叶子。因为叶子的sum就是点的编号。

所以在读取输入的时候把num为1的点都加入到队列，然后取出来，之后就能得到一条边。以此类推

代码

#include <cstdio>
#include <stack>
#include <list>
#include <set>
#include <iostream>
#include <string>
#include <vector>
#include <queue>
#include <functional>
#include <cstring>
#include <algorithm>
#include <cctype>
#include <string>
#include <map>
#include <cmath>
using namespace std;
#define LL long long
#define ULL unsigned long long
#define SZ(x) (int)x.size()
#define Lowbit(x) ((x) & (-x))
#define MP(a, b) make_pair(a, b)
#define MS(arr, num) memset(arr, num, sizeof(arr))
#define PB push_back
#define F first
#define S second
#define ROP freopen("input.txt", "r", stdin);
#define MID(a, b) (a + ((b - a) >> 1))
#define LC rt << 1, l, mid
#define RC rt << 1|1, mid + 1, r
#define LRT rt << 1
#define RRT rt << 1|1
#define BitCount(x) __builtin_popcount(x)
#define BitCountll(x) __builtin_popcountll(x)
#define LeftPos(x) 32 - __builtin_clz(x) - 1
#define LeftPosll(x) 64 - __builtin_clzll(x) - 1
const double PI = acos(-1.0);
const int INF = 0x3f3f3f3f;
const double eps = 1e-8;
const int MAXN = 100000 + 10;
const int MOD = 1000007;
const int dir[][2] = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} };
typedef pair<int, int> pii;
typedef vector<int>::iterator viti;
typedef vector<pii>::iterator vitii;
struct POINT
{
    int num, sum;
}arr[MAXN];
queue<int> Q;
int main()
{
    //ROP;
    int n, i, j;
    scanf("%d", &n);
    int ans = 0;
    for (i = 0; i < n; i++)
    {
        scanf("%d%d", &arr[i].num, &arr[i].sum);
        ans += arr[i].num;
        if (arr[i].num == 1) Q.push(i);
    }
    printf("%d\n", (ans>>1));
    while (!Q.empty())
    {
        int leaf = Q.front(); Q.pop();
        if (arr[leaf].num == 0) continue;
        arr[leaf].num--;
        int u = arr[leaf].sum;
        arr[u].num--;
        arr[u].sum ^= leaf;
        if (arr[u].num == 1) Q.push(u);
        printf("%d %d\n", leaf, u);
    }
    return 0;
}