BZOJ 1880: [Sdoi2009]Elaxia的路线

分别求出以 \(s_1\)、\(t_1\)、\(s_2\)、\(t_2\) 为起点的最短路
然后 \(O(n^2)\) 枚举公共路径的两个端点
根据四个最短路的关系判断是否都在最短路上

#include <bits/stdc++.h>
#define pb push_back
#define fi first
#define se second
#define pii pair<int, int>
#define pli pair<ll, int>
#define lp p << 1
#define rp p << 1 | 1
#define mid ((l + r) >> 1)
#define lowbit(i) ((i) & (-i))
#define ll long long
#define ull unsigned long long
#define db double
#define rep(i,a,b) for(int i=a;i<b;i++)
#define per(i,a,b) for(int i=b-1;i>=a;i--)
#define Edg int ccnt=1,head[N],to[E],ne[E];void addd(int u,int v){to[++ccnt]=v;ne[ccnt]=head[u];head[u]=ccnt;}void add(int u,int v){addd(u,v);addd(v,u);}
#define Edgc int ccnt=1,head[N],to[E],ne[E],c[E];void addd(int u,int v,int w){to[++ccnt]=v;ne[ccnt]=head[u];c[ccnt]=w;head[u]=ccnt;}void add(int u,int v,int w){addd(u,v,w);addd(v,u,w);}
#define es(u,i,v) for(int i=head[u],v=to[i];i;i=ne[i],v=to[i])
const int MOD = 1e9 + 7;
void M(int &x) {if (x >= MOD)x -= MOD; if (x < 0)x += MOD;}
int qp(int a, int b = MOD - 2) {int ans = 1; for (; b; a = 1LL * a * a % MOD, b >>= 1)if (b & 1)ans = 1LL * ans * a % MOD; return ans % MOD;}
template<class T>T gcd(T a, T b) { while (b) { a %= b; std::swap(a, b); } return a; }
template<class T>bool chkmin(T &a, T b) { return a > b ? a = b, 1 : 0; }
template<class T>bool chkmax(T &a, T b) { return a < b ? a = b, 1 : 0; }
char buf[1 << 21], *p1 = buf, *p2 = buf;
inline char getc() {
    return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2) ? EOF : *p1++;
}
inline int _() {
    int x = 0, f = 1; char ch = getc();
    while (ch < '0' || ch > '9') { if (ch == '-') f = -1; ch = getc(); }
    while (ch >= '0' && ch <= '9') { x = x * 10ll + ch - 48; ch = getc(); }
    return x * f;
}

const int N = 2222, E = 3e6 + 7;
const ll INF = 0x3f3f3f3f3f3f3f3f;
Edgc
ll dis[N][4];
int n, m, s1, s2, t1, t2;
bool inq[N];
void spfa(int s, int opt) {
    rep (i, 1, n + 1) dis[i][opt] = INF, inq[i] = 0;
    std::queue<int> que;
    que.push(s);
    dis[s][opt] = 0;
    inq[s] = 1;
    while (!que.empty()) {
        int u = que.front(); que.pop();
        inq[u] = 0;
        es (u, i, v) {
            if (chkmin(dis[v][opt], dis[u][opt] + c[i]))
                if (!inq[v]) inq[v] = 1, que.push(v);
        }
    }
}

int main() {
#ifdef LOCAL
    freopen("ans.out", "w", stdout);
#endif
    n  = _(), m = _();
    s1 = _(), t1 = _(), s2 = _(), t2 = _();
    rep (i, 0, m) {
        int u = _(), v = _(), x = _();
        add(u, v, x);
    }
    spfa(s1, 0);
    spfa(t1, 1);
    spfa(s2, 2);
    spfa(t2, 3);
    ll ans = 0, l1 = dis[t1][0], l2 = dis[t2][2];
    rep (i, 1, n + 1) {
        rep (j, 1, n + 1) {
            if (i == j) continue;
            ll cur = l1 - dis[i][0] - dis[j][1];
            if (cur == l2 - dis[i][2] - dis[j][3] || cur == l2 - dis[j][2] - dis[i][3])
                chkmax(ans, cur);
        }
    }
    printf("%lld\n", ans);
    return 0;
}