BZOJ 2199 [Usaco2011 Jan]奶牛议会

BZOJ_2199

    这个题目可以枚举每个方案是Y还是N，然后用2-SAT判断是否可以the bills are to be passed or not in such a way that every cow gets her way on at least one of her votes，整体复杂度是O(N*M)的。

#include<stdio.h>
#include<string.h>
#include<algorithm>
#define MAXN 2010
#define MAXM 8010
int N, M, one[MAXM], two[MAXM], first[MAXN], e, next[MAXM], v[MAXM];
int ins[MAXN], s[MAXN], color[MAXN], dfn[MAXN], low[MAXN], cnt, top, col;
void init()
{
    int x;
    char b[5];
    N *= 2;
    for(int i = 0; i < M; i ++)
    {
        scanf("%d%s", &x, b);
        if(b[0] == 'Y') one[i] = 2 * (x - 1) + 1;
        else one[i] = 2 * (x - 1);
        scanf("%d%s", &x, b);
        if(b[0] == 'Y') two[i] = 2 * (x - 1) + 1;
        else two[i] = 2 * (x - 1);
    }    
}
void add(int x, int y)
{
    v[e] = y;
    next[e]    = first[x], first[x] = e ++;
}
void build()
{
    for(int i = 0; i < M; i ++)
        add(one[i] ^ 1, two[i]), add(two[i] ^ 1, one[i]);
}
void tarjan(int x)
{
    dfn[x] = low[x] = ++ cnt;
    s[top ++] = x, ins[x] = 1;
    for(int i = first[x]; i != -1; i = next[i])
    {
        if(dfn[v[i]] == 0)
        {
            tarjan(v[i]);
            low[x] = std::min(low[x], low[v[i]]);
        }
        else if(ins[v[i]] && dfn[v[i]] < low[x])
            low[x] = dfn[v[i]];
    }
    if(low[x] == dfn[x])
    {
        ++ col;
        for(s[top] = -1; s[top] != x;)
            color[s[--top]] = col, ins[s[top]] = 0;
    }
}
int judge()
{
    top = cnt = col = 0;
    memset(dfn, 0, sizeof(dfn[0]) * (N + 1));
    memset(ins, 0, sizeof(ins[0]) * (N + 1));
    for(int i = 0; i < N; i ++)
        if(!dfn[i]) tarjan(i);
    for(int i = 0; i < N; i += 2)
        if(color[i] == color[i ^ 1]) return 0;
    return 1;
}
int report[MAXN];
void solve()
{
    for(int i = 0; i < N; i += 2)
    {
        int flag = 0;
        report[i / 2] = 0;
        memset(first, -1, sizeof(first[0]) * (N + 1)), e = 0;
        add(i, i ^ 1), build();
        if(judge()) ++ report[i / 2], flag = 1;
        memset(first, -1, sizeof(first[0]) * (N + 1)), e = 0;
        add(i ^ 1, i), build();
        if(judge()) -- report[i / 2], flag = 1;
        if(flag == 0)
        {
            printf("IMPOSSIBLE\n");
            return ;    
        }
    }
    for(int i = 0; i < N; i += 2)
    {
        if(report[i / 2] == -1) putchar('N');
        else if(report[i / 2] == 0) putchar('?');
        else putchar('Y');
    }
    putchar('\n');
}
int main()
{
    while(scanf("%d%d", &N, &M) == 2)
    {
        init();
        solve();    
    }
    return 0;    
}