zoukankan      html  css  js  c++  java
  • [NOI2017]游戏

    Sol

    然而我的代码在(UOJ)上被(hack)( TLE)
    但思路是没问题的(TAT)

    如果没有(x),就是个(2-SAT)
    我们爆搜(x)的地图是(a)还是(b)就好了
    不用枚举它是(c),枚举(a)(b)就能保证正好选(ABC)三种车

    我也不知道我的输出方案哪里学的
    拓扑排序+染色(QAQ)

    如果觉得自己的代码优秀就去(UOJ)上交吧(逃

    # include <bits/stdc++.h>
    # define RG register
    # define IL inline
    # define Fill(a, b) memset(a, b, sizeof(a))
    using namespace std;
    typedef long long ll;
    const int _(1e5 + 5);
    typedef int Arr[_];
    
    IL int Input(){
        RG int x = 0, z = 1; RG char c = getchar();
        for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;
        for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
        return x * z;
    }
    
    Arr dfn, low, S, vis, first, head, col, i1, i2, d, oth, cho;
    int n, m, D, cnt, pos[10], idx, num;
    struct Edge{
        int to, next;
    } edge[_ << 2], dag[_ << 2];
    char s[_], h1[_], h2[_], tp1[200], tp2[200];
    queue <int> Q;
    
    IL void Add(RG int u, RG int v){
        edge[cnt] = (Edge){v, first[u]}, first[u] = cnt++;
    }
    
    IL void Add_DAG(RG int u, RG int v){
        dag[cnt] = (Edge){v, head[u]}, head[u] = cnt++;
    }
    
    IL void Tarjan(RG int u){
        dfn[u] = low[u] = ++idx, vis[S[++S[0]] = u] = 1;
        for(RG int e = first[u]; e != -1; e = edge[e].next){
            RG int v = edge[e].to;
            if(!dfn[v]) Tarjan(v), low[u] = min(low[u], low[v]);
            else if(vis[v]) low[u] = min(low[u], dfn[v]);
        }
        if(low[u] == dfn[u]){
            RG int v = S[S[0]--]; vis[v] = 0, col[v] = ++num;
            while(v != u) vis[v = S[S[0]--]] = 0, col[v] = num;
        }
    }
    
    IL void Del(RG int u){
        cho[u] = 0;
        for(RG int e = head[u]; e != -1; e = dag[e].next)
            if(cho[u] == -1) Del(dag[e].to);
    }
    
    IL void Calc(){
        RG int tot = n + n; num = idx = cnt = 0;
        for(RG int i = 1; i <= tot; ++i)
            cho[i] = head[i] = first[i] = -1, d[i] = vis[i] = oth[i] = dfn[i] = low[i] = col[i] = 0;
        for(RG int i = 1; i <= m; ++i){
            if(s[i1[i]] == h1[i]) continue;
    		if(i1[i] == i2[i] && h1[i] == h2[i]) continue;
            RG int p1 = i1[i], p2 = i1[i] + n, p3 = i2[i], p4 = i2[i] + n;
            if(tp1[s[i1[i]]] != h1[i]) swap(p1, p2);
            if(tp1[s[i2[i]]] != h2[i]) swap(p3, p4);
            if(s[i2[i]] == h2[i]) Add(p1, p2);
            else Add(p1, p3), Add(p4, p2);
        }
        for(RG int i = 1; i <= tot; ++i) if(!dfn[i]) Tarjan(i);
        for(RG int i = 1; i <= n; ++i){
            if(col[i] == col[i + n]) return;
            oth[col[i]] = col[i + n], oth[col[i + n]] = col[i];
        }
        cnt = 0;
        for(RG int i = 1; i <= tot; ++i)
            for(RG int e = first[i]; e != -1; e = edge[e].next)
                if(col[edge[e].to] != col[i]) Add_DAG(col[edge[e].to], col[i]), ++d[col[i]];
        for(RG int i = 1; i <= num; ++i) if(!d[i]) Q.push(i);
        while(!Q.empty()){
            RG int u = Q.front(); Q.pop();
            if(cho[u] != -1) continue;
            cho[u] = 1, Del(oth[u]);
            for(RG int e = head[u]; e != -1; e = dag[e].next)
                if(!--d[dag[e].to]) Q.push(dag[e].to);
        }
        for(RG int i = 1; i <= n; ++i)
            (cho[col[i]] == 1) ? putchar(tp1[s[i]] - 'a' + 'A') : putchar(tp2[s[i]] - 'a' + 'A');
        puts(""); exit(0);
    }
    
    IL void Dfs(RG int x){
        if(x > D){
            Calc();
            return;
        }
        s[pos[x]] = 'a', Dfs(x + 1), s[pos[x]] = 'b', Dfs(x + 1);
    }
    
    int main(RG int argc, RG char *argv[]){
        tp1['a'] = 'b', tp1['b'] = 'c', tp1['c'] = 'a';
        tp2['a'] = 'c', tp2['b'] = 'a', tp2['c'] = 'b';
        n = Input(), D = Input(), scanf(" %s", s + 1), m = Input();
        for(RG int i = 1; i <= m; ++i)
            i1[i] = Input(), h1[i] = getchar() - 'A' + 'a', i2[i] = Input(), h2[i] = getchar() - 'A' + 'a';
        for(RG int i = 1, t = 0; i <= n; ++i) if(s[i] == 'x') pos[++t] = i;
        Dfs(1);
        puts("-1");
        return 0;
    }
    
    
  • 相关阅读:
    OCM_第十四天课程:Section6 —》数据库性能调优_各类索引 /调优工具使用/SQL 优化建议
    OCM_第十三天课程:Section6 —》数据库性能调优 _结果缓存 /多列数据信息采集统计/采集数据信息保持游标有效
    OCM_第十二天课程:Section6 —》数据库性能调优_ 资源管理器/执行计划
    使用NuGet时的一个乌龙
    .net调用存储过程碰到的一个问题
    数据库的备份与还原
    创建link server链接服务器碰到的问题及解决办法
    如何管理好项目的DLL
    项目中Enum枚举的使用
    .NET开发知识体系
  • 原文地址:https://www.cnblogs.com/cjoieryl/p/8666715.html
Copyright © 2011-2022 走看看