| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 2713 | Accepted: 868 |
Description
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
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
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
1
2
end
3
4
5
6
end
#include<set>
#include<map>
#include<queue>
#include<stack>
#include<cmath>
#include<string>
#include<time.h>
#include<vector>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define INF 1000000001
#define ll long long
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
using namespace std;
const int MAXN = ;
int pa[MAXN],n,p1,p2,rel[MAXN],vis[MAXN];
vector<int>b[MAXN][];
int a[MAXN][],dp[MAXN][MAXN],pre[MAXN][MAXN];
void Init()
{
for(int i = ; i <= p1 + p2; i++){
pa[i] = i;
b[i][].clear();
b[i][].clear();
a[i][] = ;
a[i][] = ;
}
memset(rel,,sizeof(rel));
}
int find(int x)
{
if(x != pa[x]){
int fx = find(pa[x]);
rel[x] = (rel[x] + rel[pa[x]]) % ;
pa[x] = fx;
}
return pa[x];
}
int main()
{
while(~scanf("%d%d%d",&n,&p1,&p2)){
if(!n && !p1 && !p2)break;
Init();
int x,y,z;
char s[];
for(int i = ; i < n; i++){
scanf("%d %d %s",&x,&y,s);
if(s[] == 'y'){
z = ;
}
else {
z = ;
}
int fx = find(x);
int fy = find(y);
if(fx != fy){
pa[fx] = fy;
rel[fx] = ( - rel[x] + z + rel[y]) % ;
}
}
memset(vis,,sizeof(vis));
int cnt = ;
for(int i = ; i <= p1 + p2; i++){
if(!vis[i]){
int tp = find(i);
for(int j = i; j <= p1 + p2; j++){
int fp = find(j);
if(fp == tp){
vis[j] = ;
b[cnt][rel[j]].push_back(j);
a[cnt][rel[j]] ++;
}
}
//cout<<a[cnt][0]<<' '<<a[cnt][1]<<endl;
cnt ++;
}
}
memset(vis,,sizeof(vis));
memset(dp,,sizeof(dp));
dp[][] = ;
for(int i = ; i < cnt; i++){
for(int j = p1; j >= ; j--){
if(j - a[i][] >= && dp[i-][j - a[i][]]){
dp[i][j] += dp[i-][j-a[i][]];
pre[i][j] = j - a[i][];
}
if(j - a[i][] >= && dp[i-][j - a[i][]]){
dp[i][j] += dp[i-][j-a[i][]];
pre[i][j] = j - a[i][];
}
}
}
if(dp[cnt-][p1] != ){
printf("no\n");
}
else {
vector<int>ans;
int l = p1;
for(int i = cnt - ; i >= ; i--){
int tp = l - pre[i][l];
if(tp == a[i][]){
for(int j = ; j < b[i][].size(); j++){
ans.push_back(b[i][][j]);
}
}
else {
for(int j = ; j < b[i][].size(); j++){
ans.push_back(b[i][][j]);
}
}
l = pre[i][l];
}
sort(ans.begin(),ans.end());
for(int i = ; i < ans.size(); i++){
printf("%d\n",ans[i]);
}
printf("end\n");
}
}
return ;
}