Description
Ratish is a young man who always dreams of being a hero. One day his friend Luffy was caught by Pirate Arlong. Ratish set off at once to Arlong‘s island. When he got there, he found the secret place where his friend was kept, but he could not go straight in. He saw a large door in front of him and two locks in the door. Beside the large door, he found a strange rock, on which there were some odd words. The sentences were encrypted. But that was easy for Ratish, an amateur cryptographer. After decrypting all the sentences, Ratish knew the following facts:
Behind the large door, there is a nesting prison, which consists of M floors. Each floor except the deepest one has a door leading to the next floor, and there are two locks in each of these doors. Ratish can pass through a door if he opens either of the two locks in it. There are 2N different types of locks in all. The same type of locks may appear in different doors, and a door may have two locks of the same type. There is only one key that can unlock one type of lock, so there are 2N keys for all the 2N types of locks. These 2N keys were divided into N pairs, and once one key in a pair is used, the other key will disappear and never show up again.
Later, Ratish found N pairs of keys under the rock and a piece of paper recording exactly what kinds of locks are in the M doors. But Ratish doesn‘t know which floor Luffy is held, so he has to open as many doors as possible. Can you help him to choose N keys to open the maximum number of doors?
Input
There are several test cases. Every test case starts with a line containing two positive integers N (1 <= N <= 210) and M (1 <= M <= 211) separated by a space, the first integer represents the number of types of keys and the second integer represents the number of doors. The 2N keys are numbered 0, 1, 2, ..., 2N - 1. Each of the following N lines contains two different integers, which are the numbers of two keys in a pair. After that, each of the following M lines contains two integers, which are the numbers of two keys corresponding to the two locks in a door. You should note that the doors are given in the same order that Ratish will meet. A test case with N = M = 0 ends the input, and should not be processed.
Output
For each test case, output one line containing an integer, which is the maximum number of doors Ratish can open.
Sample Input
3 6 0 3 1 2 4 5 0 1 0 2 4 1 4 2 3 5 2 2 0 0
Sample Output
4 今天学习了 2-sat问题 这是一个现实中经常遇到的问题可以转化为图论去解决
有m扇门，每扇门都有两把锁，只要打开其中一把就可以打开这扇门，
打开了第i扇门才能继续开第i+1扇门。
现在有2n个钥匙，每两个绑在一起，使用了一个就不能使用另一个。
问最多能开几扇门。
这个就是 2-sat问题的模板题
以下是这类问题的基本的解题思路
- 构造有向图，这个有向图是一个表示了选择了一个点，一定要选择它能够到达的点的图。
- 对这个图求强连通图。
- 假如有两个点处在同一个强连通图内，这两个点是必须只能选择一个的一组内的点，就不可以。
- 否则可以
a->b+2*n 表示 选a不选b
门的建边就要 a+2*n->b
然后判断 a和a+2*n 在不在同一个强连通分量内
回到题目来 这题的话网上普遍二分解决 因为这样快啊
1 #include <iostream>
2 #include <cstdio>
3 #include <cstring>
4 #include <vector>
5 #include <string>
6 #include <algorithm>
7 #include <queue>
8 #include <stack>
9
10 using namespace std;
11 const int maxn = 1e5 + 10;
12 const int mod = 1e9 + 7 ;
13 const int INF = 0x7ffffff;
14 struct node {
15 int v, next;
16 } edge[maxn];
17 int head[maxn], dfn[maxn], low[maxn];
18 int s[maxn], belong[maxn], instack[maxn];
19 int tot, cnt, top, flag, n, m;
20 void init() {
21 tot = cnt = top = flag = 0;
22 memset(s, 0, sizeof(s));
23 memset(head, -1, sizeof(head));
24 memset(dfn, 0, sizeof(dfn));
25 memset(instack, 0, sizeof(instack));
26 }
27 void add(int u, int v ) {
28 edge[tot].v = v;
29 edge[tot].next = head[u];
30 head[u] = tot++;
31 }
32 void tarjin(int v) {
33 dfn[v] = low[v] = ++flag;
34 instack[v] = 1;
35 s[top++] = v;
36 for (int i = head[v] ; ~i ; i = edge[i].next ) {
37 int j = edge[i].v;
38 if (!dfn[j]) {
39 tarjin(j);
40 low[v] = min(low[v], low[j]);
41 } else if (instack[j]) low[v] = min(low[v], dfn[j]);
42 }
43 if (dfn[v] == low[v]) {
44 cnt++;
45 int t;
46 do {
47 t = s[--top];
48 instack[t] = 0;
49 belong[t] = cnt;
50 } while(t != v) ;
51 }
52 }
53 struct p {
54 int u, v;
55 } qu[maxn];
56 int check(int mid) {
57 init();
58 for (int i = 0 ; i < n ; i++) {
59 add(qu[i].v, qu[i].u + 2 * n);
60 add(qu[i].u, qu[i].v + 2 * n);
61 }
62 for (int i = 0 ; i < mid ; i++ ) {
63 add(qu[i + n].u + 2 * n, qu[i+n].v);
64 add(qu[i + n].v + 2 * n, qu[i+n].u);
65 }
66 for (int i = 0 ; i < 4 * n ; i++)
67 if (!dfn[i]) tarjin(i);
68 for (int i = 0 ; i < 2 * n ; i++)
69 if (belong[i] == belong[i + 2 * n]) return 0;
70 return 1;
71 }
72 int main() {
73 while(scanf("%d%d", &n, &m) != EOF) {
74 if (n == 0 && m == 0 ) break;
75 for (int i = 0 ; i < n ; i++)
76 scanf("%d%d", &qu[i].u, &qu[i].v);
77 for (int i = 0 ; i < m ; i++)
78 scanf("%d%d", &qu[i + n].u, &qu[i + n].v);
79 int low = 0, high = m, mid;
80 while(low <= high) {
81 mid = (low + high) / 2;
82 if (check(mid)) low = mid + 1;
83 else high = mid - 1;
84 }
85 printf("%d\n", high);
86 }
87 return 0;
88 }
原文地址：https://www.cnblogs.com/qldabiaoge/p/9092918.html