UVaLive 4452 The Ministers' Major Mess (TwoSat)

Posted 2020-10-12 dwtfukgv

题意：有 m 个人对 n 个方案投票，每个人最多只能对其中的4个方案投票(其他的相当于弃权)，每一票要么支持要么反对。问是否存在一个最终决定，使得每个投票人都有超过一半的建议被采纳，在所有可能的最终决定中，哪些方案的态度是确定的。

析：注意这个题是超过一半，是TwoSat 算法，对于投小于三票的，他的票必须都成立（因为要超过一半），同理大于两票的最多一票不成立，这样考虑就是twosat的问题了，也就是说对于小于三票的，那么就相当于是标记了必须成立，对于大于两票的，那么就是如果有一个不成立，那么其他的必须全成立，那么就全连上边。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#include <list>
#include <assert.h>
#include <bitset>
#include <numeric>
#define debug() puts("++++")
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a, b, sizeof a)
#define sz size()
#define pu push_up
#define pd push_down
#define cl clear()
#define all 1,n,1
#define FOR(i,x,n)  for(int i = (x); i < (n); ++i)
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<LL, int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e17;
const double inf = 1e20;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 100 + 10;
const int maxm = 1e6 + 5;
const int mod = 10007;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, -1, 0, 1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c) {
  return r >= 0 && r < n && c >= 0 && c < m;
}

char ans[maxn];

struct TwoSat{
  int n;
  vector<int> G[maxn<<1];
  bool mark[maxn<<1];
  int S[maxn<<1], c;

  void init(int n){
    this-> n = n;
    for(int i = 0; i < (n<<1); ++i)  G[i].cl;
    ms(mark, 0);
  }

  void add_clause(int x, int xval, int y, int yval){
    x = x << 1 | xval;
    y = y << 1 | yval;
    G[x^1].pb(y);
    G[y^1].pb(x);
  }

  bool dfs(int x){
    if(mark[x^1])  return false;
    if(mark[x])  return true;
    mark[x] = true;
    S[c++] = x;
    for(int i = 0; i < G[x].sz; ++i)
      if(!dfs(G[x][i]))  return false;
    return true;
  }

  bool solve(){
    for(int i = 0; i < (n<<1); i += 2)
      if(!mark[i] && !mark[i^1]){
        c = 0;
        bool ok1 = dfs(i);
        while(c)  mark[S[--c]] = 0;
        bool ok2 = dfs(i^1);
        while(c)  mark[S[--c]] = 0;
        if(ok1 && ok2)  ans[i>>1] = ‘?‘;
        else if(ok1)  ans[i>>1] = ‘n‘;
        else if(ok2)  ans[i>>1] = ‘y‘;
        else  return false;
      }
    ans[n] = 0;
    return true;
  }
};
TwoSat twosat;

int a[10], val[10];

int main(){
  int kase = 0;
  while(scanf("%d %d", &n, &m ) == 2 && n+m){
    twosat.init(n);
    for(int i = 0; i < m; ++i){
      int num;  scanf("%d", &num);
      for(int j = 0; j < num; ++j){
        char ch;
        scanf("%d %c", a+j, &ch);
        --a[j];
        val[j] = ch == ‘y‘;
      }
      if(num < 3)  for(int j = 0; j < num; ++j)
        twosat.add_clause(a[j], val[j], a[j], val[j]);
      else  for(int j = 0; j < num; ++j)
        for(int k = j+1; k < num; ++k)
          twosat.add_clause(a[j], val[j], a[k], val[k]);
    }
    printf("Case %d: ", ++kase);
    if(!twosat.solve())  puts("impossible");
    else puts(ans);
  }
  return 0;
}