BFS+枚举 Codeforces666B World Tour

传送门：点击打开链接

题意：有向边n个点，m条边(n < 3000, m < 4000)，现在选出不相同的4个点，要求a->b->c->d的路径最大，从u->v走的路径必须是最短路。

题目保证存在4个点可以走到。

思路：先BFS预处理出所有的从u到v的最短路径，并保存其他点到u的最短路径和次短路径，并保存好方案

保存u到其他点的最短路径和次短路径，并保存好方案

之后枚举b和c，通过之前保存的最短路和次短路组合起来，取最大值，就搞定了

#include <map>
#include <set>
#include <cmath>
#include <ctime>
#include <stack>
#include <queue>
#include <cstdio>
#include <cctype>
#include <bitset>
#include <string>
#include <vector>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <functional>
#define fuck(x) cout<<"["<<x<<"]";
#define FIN freopen("input.txt","r",stdin);
#define FOUT freopen("output.txt","w+",stdout);
//#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
typedef long long LL;
typedef pair<int, int> PII;

const int MX = 3e3 + 5;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7;

struct Edge {
int v, nxt;
} E[20000];
int Head[MX], rear;
void edge_init() {
rear = 0;
memset(Head, -1, sizeof(Head));
}
void edge_add(int u, int v) {
E[rear].v = v;
E[rear].nxt = Head[u];
Head[u] = rear++;
}

int n, m, d[MX], G[MX][MX];
int to[2][MX], from[2][MX];
int toid[2][MX], fromid[2][MX];

bool check(int a, int b, int c, int d) {
int s[] = {a, b, c, d};
sort(s, s + 4);
int sz = unique(s, s + 4) - s;
return sz == 4;
}
int main() {
edge_init(); //FIN;
scanf("%d%d", &n, &m);
for(int i = 1; i <= m; i++) {
int u, v;
scanf("%d%d", &u, &v);
edge_add(u, v);
}
for(int i = 1; i <= n; i++) {
memset(d, INF, sizeof(d));
queue<int> Q;
d[i] = 0; Q.push(i);
while(!Q.empty()) {
int u = Q.front(); Q.pop();
for(int j = Head[u]; ~j; j = E[j].nxt) {
int v = E[j].v;
if(d[v] == INF) {
d[v] = d[u] + 1;
Q.push(v);
}
}
}
for(int j = 1; j <= n; j++) {
if(d[j] != INF) {
G[i][j] = d[j];
if(from[0][j] < d[j]) {
from[1][j] = from[0][j]; fromid[1][j] = fromid[0][j];
from[0][j] = d[j]; fromid[0][j] = i;
} else if(from[1][j] < d[j]) {
from[1][j] = d[j]; fromid[1][j] = i;
}
if(to[0][i] < d[j]) {
to[1][i] = to[0][i]; toid[1][i] = toid[0][i];
to[0][i] = d[j]; toid[0][i] = j;
} else if(to[1][i] < d[j]) {
to[1][i] = d[j]; toid[1][i] = j;
}
}
}
}

int a, b, c, d, Max = 0;
for(int u = 1; u <= n; u++) {
for(int v = 1; v <= n; v++) {
for(int i = 0; i <= 1; i++) {
for(int j = 0; j <= 1; j++) {
if(G[u][v] && from[i][u] && to[j][v]) {
if(check(u, v, fromid[i][u], toid[j][v]) && Max < G[u][v] + from[i][u] + to[j][v]) {
Max = G[u][v] + from[i][u] + to[j][v];
a = fromid[i][u]; b = u; c = v; d = toid[j][v];
}
}
}
}
}
}
printf("%d %d %d %d\n", a, b, c, d);
return 0;
}