BZOJ4025 二分图

首先对时间分治，每个节点表示一个时间的区间$[l, r]$

然后对于每个节点记录一个可以撤销的并查集，维护图的信息即可

(这里的并查集不用路径压缩，只要按秩合并，这样子可以保证单次操作的时间复杂度是$O(logn)$的)

我去啊。。。把边分类这一段代码调的我QAQ了。。。代码能力太弱QAQQQ

 

  1 /**************************************************************
  2     Problem: 4025
  3     User: rausen
  4     Language: C++
  5     Result: Accepted
  6     Time:4344 ms
  7     Memory:6668 kb
  8 ****************************************************************/
  9  
 10 #include <cstdio>
 11 #include <algorithm>
 12  
 13 using namespace std;
 14 const int N = 1e5 + 5;
 15 const int M = 2e5 + 5;
 16  
 17 inline int read();
 18  
 19 struct edge {
 20     int x, y, st, ed;
 21  
 22     inline void get() {
 23         x = read(), y = read();
 24         st = read() + 1, ed = read();
 25     }
 26 } e[M];
 27  
 28 int n, m, time;
 29  
 30 namespace set {
 31     int fa[N], d[N], a[N];
 32     int s[N << 2], top = 0;
 33      
 34     int find(int x) {
 35         while (fa[x] != x) x = fa[x];
 36         return x;
 37     }
 38     int dep(int x) {
 39         static int res;
 40         for (res = 0; fa[x] != x; x = fa[x])
 41             res ^= a[x];
 42         return res;
 43     }
 44      
 45     void set_union(int x, int y, int _d) {
 46         if (d[x] > d[y]) swap(x, y);
 47         if (d[x] == d[y]) ++d[y], s[++top] = -y;
 48         fa[x] = y, a[x] = _d, s[++top] = x; 
 49     }
 50      
 51     void set_resume(int t) {
 52         for (; top != t; --top)
 53             if (s[top] < 0) --d[-s[top]];
 54             else fa[s[top]] = s[top], a[s[top]] = 0;
 55     }
 56 }
 57 using namespace set;
 58  
 59 void work(int l, int r, int m) {
 60     int mid = l + r >> 1, now = top;
 61     int i, j, fx, fy, _d;
 62     for (i = 1; i <= m; ++i)
 63         if (e[i].st <= l && r <= e[i].ed) {
 64             fx = find(e[i].x), fy = find(e[i].y), _d = !(dep(e[i].x) ^ dep(e[i].y));
 65             if (fx != fy) set_union(fx, fy, _d);
 66             else if (_d) {
 67                 while (l <= r)
 68                     puts("No"), ++l;
 69                 set_resume(now);
 70                 return;
 71             }
 72             swap(e[m--], e[i--]);
 73         };
 74     if (l == r) puts("Yes");
 75     else {
 76         for (i = 1, j = m; i <= j; ++i)
 77             if (e[i].st > mid) swap(e[j--], e[i--]);
 78         work(l, mid, j);
 79         for (i = 1, j = m; i <= j; ++i)
 80             if (e[i].ed <= mid) swap(e[j--], e[i--]);
 81         work(mid + 1, r, j);
 82     }
 83     set_resume(now);
 84 }
 85  
 86 int main() {
 87     int i;
 88     n = read(), m = read(), time = read();
 89     for (i = 1; i <= n; ++i)
 90         fa[i] = i, d[i] = 1, a[i] = 0;
 91     for (i = 1; i <= m; ++i) e[i].get();
 92     for (i = 1; i <= m; ++i)
 93         if (e[i].st > e[i].ed) swap(e[i--], e[m--]);
 94     work(1, time, m);
 95     return 0;
 96 }
 97  
 98 inline int read() {
 99     static int x, sgn;
100     static char ch;
101     x = 0, sgn = 1, ch = getchar();
102     while (ch < '0' || '9' < ch) {
103         if (ch == '-') sgn = -1;
104         ch = getchar();
105     }
106     while ('0' <= ch && ch <= '9') {
107         x = x * 10 + ch - '0';
108         ch = getchar();
109     }
110     return sgn * x;
111 }
View Code

 

posted on 2015-05-13 23:08  Xs酱~  阅读(1150)  评论(1)  编辑  收藏  举报