CF666B. World Tour

CF666B. World Tour

题意:

给定一张边权为 1 的有向图,求四个不同点 A, B, C, D
使得 dis(A, B) + dis(B, C) + dis(C, D) 取最大值,dis表示最短路距离 1 ≤ n ≤ 3000, 1 ≤ m ≤ 5000


我又写了假做法呜呜呜

首先当然\(O(nm)\)预处理两点之间最短路

然后预处理出从每个点开始前3个dis最大的点

假做法:

贪心,枚举A,贪心选择dis较大的BCD,就是3*3*3搜索选出B、C、D

假的原因:

贪心不成立,因为前面会影响后面,可以先差,再优、优

真做法:

再预处理到达每个点的前3个dis最大的点

枚举B、C,再3*3选出A、D,选A、D不会像之前那样互相干扰了

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;
const int N = 1e4+5, inf = 1e9;

int n, m;
struct edge {int v, ne;} e[N];
int cnt, h[N], d[N][N];
inline void ins(int u, int v) {
    e[++cnt] = (edge) {v, h[u]}; h[u] = cnt;
}

int q[N], head, tail, vis[N];
#define fir first
#define sec second
pair<int, int> a[N], f[N][5], g[N][5];
void bfs(int s) {
    memset(vis, 0, sizeof(vis));
    head = tail = 1;
    q[tail++] = s; vis[s] = 1;
    memset(d[s], 0x3f, sizeof(d[s]));
    int *dis = d[s];
    dis[s] = 0;
    while(head != tail) {
        int u = q[head++];
        for(int i=h[u]; i; i=e[i].ne) {
            int v = e[i].v;
            if(vis[v]) continue;
            dis[v] = dis[u] + 1;
            vis[v] = 1;
            q[tail++] = v;
        }
    }
}
void solve(int s) {
    int cnt = 0;
    for(int i=1; i<=n; i++) if(i!=s && d[s][i] < inf) a[++cnt] = make_pair(d[s][i], i);
    for(int i=cnt+1; i<=3; i++) a[i] = make_pair(0, 0);
    sort(a+1, a+cnt+1, greater<pair<int, int> >());
    f[s][1] = a[1], f[s][2] = a[2], f[s][3] = a[3];

    cnt = 0;
    for(int i=1; i<=n; i++) if(i!=s && d[i][s] < inf) a[++cnt] = make_pair(d[i][s], i);
    for(int i=cnt+1; i<=3; i++) a[i] = make_pair(0, 0);
    sort(a+1, a+cnt+1, greater<pair<int, int> >());
    g[s][1] = a[1], g[s][2] = a[2], g[s][3] = a[3];
    //if(s == 1) printf("hi %d %d %d\n", a[1].sec, a[2].sec, a[3].sec);


    //printf("sssssssssssssssssssssssssssss %d\n", s);
    //for(int i=1; i<=3; i++) printf("f %d   %d %d\n", i, f[s][i].fir, f[s][i].sec);
}

int ans, li[10];
int main() {
    //freopen("in", "r", stdin);
    ios::sync_with_stdio(false); cin.tie(); cout.tie();
    cin >> n >> m;
    for(int i=1; i<=m; i++) {
        int u, v;
        cin >> u >> v;
        if(u != v && !d[u][v]) ins(u, v), d[u][v] = 1;
    }
    for(int i=1; i<=n; i++) bfs(i);
    for(int i=1; i<=n; i++) solve(i);

    for(int b=1; b<=n; b++)
        for(int c=1; c<=n; c++) if(d[b][c] < inf && c != b) {
            int now = d[b][c];
            for(int i=1; i<=3; i++) {
                pair<int, int> &ra = g[b][i];
                int a = ra.sec;
                if(a == b || a == c || !ra.fir) continue;
                now += ra.fir;
                for(int j=1; j<=3; j++) {
                    pair<int, int> &rd = f[c][j];
                    int d = rd.sec;
                    if(d ==a || d == b || d == c || !rd.fir) continue;
                    now += rd.fir; 
                    if(now > ans) {
                        ans = now;
                        li[1] = a; li[2] = b; li[3] = c; li[4] = d;
                    }
                    now -= rd.fir;
                }
                now -= ra.fir;
            }
        }
    for(int i=1; i<=4; i++) cout << li[i] << ' ';
}
posted @ 2018-07-10 11:28 Candy? 阅读(...) 评论(...) 编辑 收藏