标签:blog sse com har 方案 大于 bool main comm
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; }
UVaLive 4452 The Ministers' Major Mess (TwoSat)
标签:blog sse com har 方案 大于 bool main comm
原文地址:http://www.cnblogs.com/dwtfukgv/p/7739280.html