poj2090

题意：给一个数列，两个栈，要求数列从后往前依次入栈，问能否使出栈序列是不减的。（双栈排序）

分析：利用二分图染色法。

首先观察那些牌绝对不能压入同一个栈，若两个不能入同一栈则连一条边，然后根据二分图染色，看是否能构成二分图。如果不能直接输出impossible

两张牌i,j不能入同一栈的充要条件是，i>j>k（i最先入栈） && stock[k]<stock[i]<stock[j]

然后根据每个点所染的颜色决定把每个牌压入哪个栈。然后模拟即可。

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
usingnamespace std;

#define maxn 250

struct Edge
{
    int v, next;
} edge[maxn * maxn];

int n;
int stock[maxn], f[maxn];
int head[maxn], ecount, color[maxn], q[maxn];
intout[maxn];
bool ok;
int stk1[maxn], stk2[maxn];
int top1, top2, step;

void input()
{
    for (int i =0; i < n; i++)
    {
        scanf("%d", &stock[i]);
        out[i] = stock[i];
    }
}

void addedge(int&a, int&b)
{
    edge[ecount].next = head[a];
    edge[ecount].v = b;
    head[a] = ecount++;
}

void bfs(int&s)
{
    int front =0;
    int rear =1;
    q[0] = s;
    color[s] =1;
    while (front != rear)
    {
        int a = q[front++];
        for (int i = head[a]; i !=-1; i = edge[i].next)
        {
            int b = edge[i].v;
            if (!color[b])
            {
                q[rear++] = b;
                color[b] =3- color[a];
            }
            elseif (color[a] == color[b])
            {
                ok =false;
                return;
            }
        }
    }
}

void make(int i)
{
    int a = stock[i];
    bool did =true;
    while (did)
    {
        did =false;
        if (top1 >0&& stk1[top1 -1] ==out[step])
        {
            top1--;
            printf("pop 1\n");
            step++;
            did =true;
        }
        if (top2 >0&& stk2[top2 -1] ==out[step])
        {
            top2--;
            printf("pop 2\n");
            step++;
            did =true;
        }
    }
    if (i <0)
        return;
    if (color[i] ==1)
    {
        stk1[top1++] = a;
        printf("push 1\n");
    }
    else
    {
        stk2[top2++] = a;
        printf("push 2\n");
    }
}

void print()
{
    top1 = top2 =0;
    step =0;
    for (int i = n -1; i >=0; i--)
        make(i);
    make(-1);
}

void work()
{
    memset(head, -1, sizeof(head));
    f[0] = stock[0];
    for (int i =1; i < n; i++)
        f[i] = min(f[i -1], stock[i]);
    ecount =0;
    for (int i =1; i < n -1; i++)
    {
        for (int j = i +1; j < n; j++)
        {
            if (stock[j] >= stock[i])
                continue;
            if (stock[j] > f[i -1])
            {
                addedge(i, j);
                addedge(j, i);
            }
        }
    }
    ok =true;
    memset(color, 0, sizeof(color));
    for (int i =0; i < n; i++)
    {
        if (!color[i])
            bfs(i);
        if (!ok)
        {
            printf("impossible\n");
            return;
        }
    }
    print();
}

int main()
{
    //freopen("t.txt", "r", stdin);
int t =0;
    while (scanf("%d", &n), n)
    {
        t++;
        printf("#%d\n", t);
        input();
        sort(out, out+ n);
        work();
    }
    return0;
}