POJ 1417 True Liars 并查集+背包

时间：2017-08-07 13:46:55 阅读：187 评论：0 收藏：0 [点我收藏+]

标签：测试数据 print future 一个 scan always down 统计 ict

解题思路：比较容易想到的是并查集，然后把第三组数据测试一下之后发现这并不是简单的并查集，而是需要合并之后然后判断的。并且鉴于题目要求输出数据，因此还要记录数据，可以说是非常有意思的题目。

首先，如果a b yes，那么a与b一定都是圣人或者恶人；反之，如果a b no，那么两者一个圣人，一个恶人。因此可以将所有元素分为若干个集合，每个集合中放两个小集合，表示两种不一样的人，当然并不知道到底哪个里面是圣人，恶人。

另一个比较棘手的问题就是集合区分的情况，可以使用带权并查集，val[i]记录每个元素与祖先的关系，0为同种人，1为不同种人。那么在合并的时候就需要计算：val[ra] = (val[b] - val[a] + tmp + 2) % 2; 其中ra，rb为a，b的祖先，tmp表示a，b关系是yes -- 0， no -- 1。另外就是并查集find路径压缩的时候要更新val值。

还有一个问题就是DP了，可以这样：dp[i][j]表示前i个集合恰好选j个人圣人的方案数目，pre[i][j]则用来记录在背包i - 1处选择的人数，dp[bgcnt][p1]不为1的话输出no。

状态转移方程：

 1 dp[0][0] = 1;
 2 for(int i = 1; i <= bgcnt; i++){
 3     for(int j = 0; j <= p1; j++){
 4         if(j >= bagcnt[i][0] && dp[i - 1][j - bagcnt[i][0]]){
 5             dp[i][j] += dp[i - 1][j - bagcnt[i][0]];
 6             pre[i][j] = j - bagcnt[i][0];
 7         }
 8         if(j >= bagcnt[i][1] && dp[i - 1][j - bagcnt[i][1]]){
 9             dp[i][j] += dp[i - 1][j - bagcnt[i][1]];
10             pre[i][j] = j - bagcnt[i][1];
11         }
12     }
13 }

整体步骤就比较好说了，大致就是：并查集合并--枚举每个人放入对应的集合，同时统计大小集合的人数--动态规划统计方案数目，同时记录从大集合选择的人数--逆序求每次选择的小集合并保存其所有元素--排序输出

整体过程比较复杂，写了好长时间，需要耐着性子。

另外给出一些测试数据，选择poj discuss块，感谢各位，如有不妥，敬请指出。

  1 6 4 4
  2 1 3 yes
  3 1 5 no
  4 6 6 yes
  5 5 7 no
  6 2 8 no
  7 2 4 yes
  8 
  9 6 6 2
 10 1 3 yes
 11 1 5 no
 12 6 6 yes
 13 5 7 no
 14 2 8 no
 15 2 4 yes
 16 
 17 6 5 3
 18 1 3 yes
 19 1 5 no
 20 6 6 yes
 21 5 7 no
 22 2 8 no
 23 2 4 yes
 24 0 0 1
 25 
 26 
 27 1  1 0 
 28 1 1 yes
 29 
 30 3 2 0
 31 1 2 no
 32 2 2 yes
 33 1 1 yes
 34 
 35 3 2 0
 36 1 1 yes
 37 2 2 yes
 38 1 1 yes
 39 
 40 7 4 2
 41 2 3 no
 42 3 4 no
 43 4 3 no
 44 5 5 yes
 45 5 3 no
 46 5 5 yes
 47 1 5 yes
 48 
 49 11 9 7
 50 6 12 no
 51 4 8 yes
 52 3 12 yes
 53 5 9 no
 54 6 11 yes
 55 6 5 no
 56 15 4 yes
 57 16 15 yes
 58 10 3 no
 59 6 16 no
 60 2 6 no
 61 6 2 4
 62 1 1 yes
 63 2 3 yes
 64 3 4 yes
 65 4 5 no
 66 5 6 yes
 67 6 6 yes
 68 3 3 2
 69 1 2 no
 70 3 4 yes
 71 3 5 no
 72 
 73 0 0 0
 74 
 75 输出：
 76 no
 77 1
 78 2
 79 3
 80 4
 81 6
 82 7
 83 end
 84 no
 85 end
 86 1
 87 end
 88 no
 89 1
 90 2
 91 end
 92 1
 93 2
 94 4
 95 5
 96 end
 97 no
 98 5
 99 6
100 end
101 no

View Code

代码：

  1 const int maxm = 700;
  2 int bagcnt[maxm][2];
  3 vector<int> bagele[maxm][2];
  4 int bgcnt;
  5 int dp[maxm][maxm], pre[maxm][maxm];
  6 int fa[maxm], val[maxm], fb[maxm];
  7 int vis[maxm];
  8 int n, p1, p2, xn;
  9 vector<int> anss;
 10 
 11 void init(){
 12     xn = p1 + p2;
 13     bgcnt = 0;
 14     for(int i = 0; i <= xn; i++) {
 15         fa[i] = i;
 16         bagele[i][0].clear();
 17         bagele[i][1].clear();
 18     }
 19     memset(val, 0, sizeof(val));
 20     memset(fb, 0, sizeof(fb));
 21     memset(bagcnt, 0, sizeof(bagcnt));
 22     memset(vis, 0, sizeof(vis));
 23     memset(dp, 0, sizeof(dp));
 24     memset(pre, 0, sizeof(pre));
 25 }
 26 int find(int x){
 27     if(x ==fa[x]) return x;
 28     int tmp = find(fa[x]);
 29     val[x] += val[fa[x]];
 30     val[x] %= 2;
 31     return fa[x] = tmp;
 32 }
 33 
 34 int main(){
 35     while(scanf("%d %d %d", &n, &p1, &p2) && !(n == 0 && p1 == 0 && p2 == 0)){
 36         init();
 37         for(int i = 1; i <= n; i++){
 38             char str[5];
 39             int a, b;
 40             scanf("%d %d %s", &a, &b, str);
 41             int tmp = str[0] == ‘y‘? 0: 1;
 42             int rb = find(b), ra = find(a);
 43             if(ra != rb){
 44                 fa[ra] = rb;
 45                 val[ra] = (val[b] - val[a] + tmp + 2) % 2;
 46             }
 47         }
 48         if(p1 == p2){
 49             puts("no");
 50             continue;
 51         }
 52         for(int i = 1; i <= xn; i++){
 53             int ri = find(i);
 54             if(fb[ri] == 0) fb[ri] = ++bgcnt;
 55             bagele[fb[ri]][val[i]].push_back(i);
 56             bagcnt[fb[ri]][val[i]]++;
 57         }
 58         
 59 //        test
 60 //        puts("ancestors:");
 61 //        for(int i = 1; i <= xn; i++) printf("%d:%d\n", i, fa[i]);
 62 //        for(int i = 1; i <= bgcnt; i++){
 63 //            printf("bag %d:\ncnt0:%d cnt1:%d\n", i, bagcnt[i][0], bagcnt[i][1]);
 64 //            for(int j = 0; j < bagele[i][0].size(); j++) printf("%d ", bagele[i][0][j]);
 65 //            puts("");
 66 //            for(int j = 0; j < bagele[i][1].size(); j++) printf("%d ", bagele[i][1][j]);
 67 //            puts("");
 68 //        }
 69         
 70         dp[0][0] = 1;
 71         for(int i = 1; i <= bgcnt; i++){
 72             for(int j = 0; j <= p1; j++){
 73                 if(j >= bagcnt[i][0] && dp[i - 1][j - bagcnt[i][0]]){
 74                     dp[i][j] += dp[i - 1][j - bagcnt[i][0]];
 75                     pre[i][j] = j - bagcnt[i][0];
 76                 }
 77                 if(j >= bagcnt[i][1] && dp[i - 1][j - bagcnt[i][1]]){
 78                     dp[i][j] += dp[i - 1][j - bagcnt[i][1]];
 79                     pre[i][j] = j - bagcnt[i][1];
 80                 }
 81             }
 82         }
 83 //        test
 84 //        printf("dp: %d\n", dp[bgcnt][p1]);
 85 //        int j = p1, tmp = p1;
 86 //        for(int i = bgcnt; i > 0; i--){
 87 //            printf("pre %d %d: %d\n", i, j,tmp - pre[i][j]);
 88 //            tmp = pre[i][j];
 89 //            j = pre[i][j];
 90 //        } 
 91         if(dp[bgcnt][p1] != 1){
 92             puts("no");
 93             continue;
 94         }
 95         
 96         anss.clear();
 97         int tmp = p1, j = p1;
 98         for(int i = bgcnt; i > 0; i--){
 99             int tmx = tmp - pre[i][j];
100             if(bagcnt[i][0] == tmx){
101                 for(int k = 0; k < bagele[i][0].size(); k++) 
102                     anss.push_back(bagele[i][0][k]);
103             }
104             else{
105                 for(int k = 0; k < bagele[i][1].size(); k++)
106                     anss.push_back(bagele[i][1][k]);
107             }
108             tmp = pre[i][j];
109             j = pre[i][j];
110         }
111         sort(anss.begin(), anss.end());
112         for(int i = 0; i < anss.size(); i++){
113             printf("%d\n", anss[i]);
114         }
115         puts("end");
116     }
117 }

题目：

True Liars

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 3677		Accepted: 1186

Description

After having drifted about in a small boat for a couple of days, Akira Crusoe Maeda was finally cast ashore on a foggy island. Though he was exhausted and despaired, he was still fortunate to remember a legend of the foggy island, which he had heard from patriarchs in his childhood. This must be the island in the legend. In the legend, two tribes have inhabited the island, one is divine and the other is devilish, once members of the divine tribe bless you, your future is bright and promising, and your soul will eventually go to Heaven, in contrast, once members of the devilish tribe curse you, your future is bleak and hopeless, and your soul will eventually fall down to Hell.

In order to prevent the worst-case scenario, Akira should distinguish the devilish from the divine. But how? They looked exactly alike and he could not distinguish one from the other solely by their appearances. He still had his last hope, however. The members of the divine tribe are truth-tellers, that is, they always tell the truth and those of the devilish tribe are liars, that is, they always tell a lie.

He asked some of them whether or not some are divine. They knew one another very much and always responded to him "faithfully" according to their individual natures (i.e., they always tell the truth or always a lie). He did not dare to ask any other forms of questions, since the legend says that a devilish member would curse a person forever when he did not like the question. He had another piece of useful informationf the legend tells the populations of both tribes. These numbers in the legend are trustworthy since everyone living on this island is immortal and none have ever been born at least these millennia.

You are a good computer programmer and so requested to help Akira by writing a program that classifies the inhabitants according to their answers to his inquiries.

Input

The input consists of multiple data sets, each in the following format :

n p1 p2
xl yl a1
x2 y2 a2
...
xi yi ai
...
xn yn an

The first line has three non-negative integers n, p1, and p2. n is the number of questions Akira asked. pl and p2 are the populations of the divine and devilish tribes, respectively, in the legend. Each of the following n lines has two integers xi, yi and one word ai. xi and yi are the identification numbers of inhabitants, each of which is between 1 and p1 + p2, inclusive. ai is either yes, if the inhabitant xi said that the inhabitant yi was a member of the divine tribe, or no, otherwise. Note that xi and yi can be the same number since "are you a member of the divine tribe?" is a valid question. Note also that two lines may have the same x‘s and y‘s since Akira was very upset and might have asked the same question to the same one more than once.

You may assume that n is less than 1000 and that p1 and p2 are less than 300. A line with three zeros, i.e., 0 0 0, represents the end of the input. You can assume that each data set is consistent and no contradictory answers are included.

Output

For each data set, if it includes sufficient information to classify all the inhabitants, print the identification numbers of all the divine ones in ascending order, one in a line. In addition, following the output numbers, print end in a line. Otherwise, i.e., if a given data set does not include sufficient information to identify all the divine members, print no in a line.

Sample Input

2 1 1
1 2 no
2 1 no
3 2 1
1 1 yes
2 2 yes
3 3 yes
2 2 1
1 2 yes
2 3 no
5 4 3
1 2 yes
1 3 no
4 5 yes
5 6 yes
6 7 no
0 0 0

Sample Output

no
no
1
2
end
3
4
5
6
end

Source

Japan 2002 Kanazawa

POJ 1417 True Liars 并查集+背包

标签：测试数据 print future 一个 scan always down 统计 ict

原文地址：http://www.cnblogs.com/bolderic/p/7298280.html

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