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 }    

  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

  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 }

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.

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.