51Nod-1530-稳定方块

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ACM模版

描述

题解

很巧妙的利用两个堆来搞事情，一个大顶堆，一个小顶堆，就是优先队列，维护每次能够拆除的方块儿，每拆除一个向周围扩展一次，并且删除拆掉的这个方块儿，这里用 map 处理。说白了，就是一个有趣的贪心问题，用到两种数据结构罢了，STL ……

代码

#include <iostream>#include <cstdio>#include <queue>#include <map>using namespace std;typedef long long ll;const int MAXM = 1e5 + 10;const ll MOD = 1e9 + 9;int m;bool vis[MAXM];int x[MAXM];int y[MAXM];ll ans[MAXM];map<pair<int, int>, int> mpi;bool judge(int a, int b){    if (mpi[make_pair(a, b + 1)])    {        if (mpi[make_pair(a - 1, b)] == 0 && mpi[make_pair(a + 1, b)] == 0)        {            return 0;        }    }    if (mpi[make_pair(a + 1, b + 1)])    {        if (mpi[make_pair(a + 1, b)] == 0 && mpi[make_pair(a + 2, b)] == 0)        {            return 0;        }    }    if (mpi[make_pair(a - 1, b + 1)])    {        if (mpi[make_pair(a - 2, b)] == 0 && mpi[make_pair(a - 1, b)] == 0)        {            return 0;        }    }    return 1;}template <class T>inline bool scan_d(T &ret){    char c;    int sgn;    if (c = getchar(), c == EOF)    {        return 0; //EOF    }    while (c != '-' && (c < '0' || c > '9'))    {        c = getchar();    }    sgn = (c == '-') ? -1 : 1;    ret = (c == '-') ? 0 : (c - '0');    while (c = getchar(), c >= '0' && c <= '9')    {        ret = ret * 10 + (c - '0');    }    ret *= sgn;    return 1;}int main(){    scan_d(m);    for (int i = 1; i <= m; i++)    {        scan_d(x[i]), scan_d(y[i]);        mpi[make_pair(x[i], y[i])] = i;    }    priority_queue<int> mx;                             //  大顶堆    priority_queue<int, vector<int>, greater<int> > mn; //  小顶堆    for (int i = 1; i <= m; i++)    {        if (judge(x[i], y[i]))        {            mx.push(mpi[make_pair(x[i], y[i])]);            mn.push(mpi[make_pair(x[i], y[i])]);        }    }    int a, b;    for (int i = 1; i <= m; i++)    {        if (i & 1)        {            ans[i] = mx.top();            mx.pop();        }        else        {            ans[i] = mn.top();            mn.pop();        }        if (vis[ans[i]])        {            i--;            continue;        }        if (!judge(x[ans[i]], y[ans[i]]))        {            i--;            continue;        }        vis[ans[i]] = 1;        a = x[ans[i]];        b = y[ans[i]];        mpi.erase(make_pair(a, b));        if (mpi[make_pair(a - 1, b - 1)])        {            if (judge(a - 1, b - 1))            {                mx.push(mpi[make_pair(a - 1, b - 1)]);                mn.push(mpi[make_pair(a - 1, b - 1)]);            }        }        if (mpi[make_pair(a, b - 1)])        {            if (judge(a, b - 1))            {                mx.push(mpi[make_pair(a, b - 1)]);                mn.push(mpi[make_pair(a, b - 1)]);            }        }        if (mpi[make_pair(a + 1, b - 1)])        {            if (judge(a + 1, b - 1))            {                mx.push(mpi[make_pair(a + 1, b - 1)]);                mn.push(mpi[make_pair(a + 1, b - 1)]);            }        }    }    ll tmp = 1;    ll res = 0;    for (int i = m; i >= 1; i--)    {        res += (ans[i] - 1) * tmp;        res %= MOD;        tmp *= m;        tmp %= MOD;    }    cout << res << endl;    return 0;}

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