*51nod 1815

从若干个数中选出最大的任意两数取模之后的结果

严格次大值

对于此题

首先缩点

然后拓扑排序

维护到达每个点的最大值与严格次大值

感觉思路与代码都OK啊

then....

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <string>
#include <queue>
#include <ctime>

using namespace std;

#define LL long long

#define gc getchar()
inline int read() {int x = 0; char c = gc; while(c < '0' || c > '9') c = gc;
while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = gc; return x;}
inline LL read_LL() {LL x = 0; char c = gc; while(c < '0' || c > '9') c = gc;
while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = gc; return x;}
#undef gc

const int N = 4e5 + 10, M = 2e6 + 10;
int n, m, S, Q;
int A[N];

int Dfn[N], Low[N], Belong[N], Beljs, Tim;
bool vis[N];
int Stack[N], topp;

pair <int, int> Pair[N];
struct Node {int u, v, nxt;} G[M];
int head1[N], cnt;

inline void Link(int u, int v) {G[++ cnt].v = v; G[cnt].nxt = head1[u]; head1[u] = cnt;}

void Tarjan(int u) {
    Dfn[u] = Low[u] = ++ Tim;
    vis[u] = 1;
    Stack[++ topp] = u;
    for(int i = head1[u]; ~ i; i = G[i].nxt) {
        int v = G[i].v;
        if(!Dfn[v]) {
            Tarjan(v);
            Low[u] = min(Low[u], Low[v]);
        } else if(vis[v]) Low[u] = min(Low[u], Low[v]);
    }
    if(Low[u] == Dfn[u]) {
        int Max = 0, CMax = 0;
        Max = A[u];
        vis[u] = 0; Belong[u] = ++ Beljs;
        Pair[Beljs].first = A[u]; Pair[Beljs].second = -1;
        while(Stack[topp] != u) {
            int a  = Stack[topp];
            vis[a] = 0;
            Belong[a] = Beljs;
            topp --;
            if(A[a] > Max) {CMax = Max, Max = A[a];}
            else if(A[a] != Max) CMax = max(CMax, A[a]);
            Pair[Beljs] = make_pair(Max, CMax);
        } topp --;
    }
}

Node E[M];
int head2[N], Du[N];

inline void ReLink(int u, int v) {Du[v] ++; E[++ cnt].v = v; E[cnt].nxt = head2[u]; head2[u] = cnt;}

int use[N];

void Rebuild() {
    for(int i = 1; i <= Beljs; i ++) head2[i] = -1;
    cnt = 0;
    for(int u = 1; u <= n; u ++)
        for(int i = head1[u]; ~ i; i = G[i].nxt) {
            int v = G[i].v;
            if(Belong[u] != Belong[v] && use[Belong[v]] != u) {
                use[Belong[v]] = u;
//                cout << Belong[u] << " " << Belong[v] << "\n";
                ReLink(Belong[u], Belong[v]);    
            }
        }
}

int Que[N];

bool beuse[N];

void Topsort() {
    int h = 1, t = 0;
//    for(int i = 1; i <= Beljs; i ++) if(Du[i] == 0) Que[++ t] = i;
    Que[++ t] = Belong[S];
    while(h <= t) {
        int topu = Que[h ++];
        beuse[topu] = 1;
        for(int i = head2[topu]; ~ i; i = E[i].nxt) {
            int v = E[i].v;
            Du[v] --;
            if(Du[v] == 0) Que[++ t] = v;
            int B[5];
//            cout << Pair[topu].first << " " << Pair[topu].second << " " << Pair[v].first << " " << Pair[v].second << "\n";
            B[1] = Pair[topu].first, B[2] = Pair[topu].second, B[3] = Pair[v].first, B[4] = Pair[v].second;
            sort(B + 1, B + 5);
            Pair[v].first = B[4];
            int j = 3;
            while(B[j] == B[4]) j --;
            Pair[v].second = B[j];
        }
    }
}

int main() {
//    srand(time(0) - 99999);
//    freopen("gg.in", "r", stdin);
//    freopen("gg.out", "w", stdout);
    n = read(), m = read(), Q = read(), S = read();
    for(int i = 1; i <= n; i ++) A[i] = read();
    for(int i = 1; i <= n; i ++) head1[i] = -1;
    for(int i = 1; i <= m; i ++) {
        int u = read(), v = read();
        Link(u, v);
    }
    for(int i = 1; i <= n; i ++) 
        if(!Dfn[i])
            Tarjan(i);
//    for(int i = 1; i <= Beljs; i ++) {
//        cout << Pair[i].first << " " << Pair[i].second << "\n";
//    }
//    return 0;
    Rebuild();
//    for(int i = 1; i <= Beljs; i ++) {
//        for(int j = head2[i]; ~ j; j = E[j].nxt) {
//            int v = E[j].v;
//            cout << i << " " << v << "\n";
//        }
//    }
    Topsort();
    for(int i = 1; i <= Q; i ++) {
        int a = read();
        if(beuse[Belong[a]] == 0) cout << -1 << " ";
        else cout << Pair[Belong[a]].second << " ";
    //    if(ans == 0) cout << -1 << " ";
//        cout << ans << " ";
    }
    return 0;
}
/*
8 10 5 1
5 5 5 7 5 5 5 5
1 2
2 3
3 1
3 4
5 1
6 5
1 6
6 8
8 7
7 6
1 2 3 4 5
*/ 

 

posted @ 2018-09-05 20:39  xayata  阅读(203)  评论(0编辑  收藏  举报