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杭电多校4 02

题目链接

题目大意

求从\(1-n\)的最短路,路径上有两种权值\(w1,w2\)要求在\(w1\)最短的前提下\(w2\)最大, 并且\(w1w2\)可能为\(0\),但是保证一定有解。

题解

图论大杂烩

\(w1\)\(w2\)都大于\(0\),我们只需要跑一遍最短路,同时维护\(w1\)\(w2\)的值即可,但本题可能出现\(0\)环的情况。

因此得考虑其他办法,我们发现求\(w2\)的最大值一定是在\(w1\)的最短路图上跑的,因此我们可以先处理出\(w1\)的最短路图

但由于\(0\)环的存在,导致最短路图不是个DAG图,因此我们对其缩点操作,最后用拓扑排序找到最长路即可

代码

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#include<bits/stdc++.h>
#define int long long
#define x first
#define y second
using namespace std;
const int N = 1e5 + 5, M = 3e5 + 5, inf = LLONG_MAX / 2;
typedef pair<int, int> PII;
int head1[N] ,e1[M], ne1[M], w1[M], w2[M], idx1 = 0;
int head2[N] ,e2[M], ne2[M], ww1[M], ww2[M], idx2 = 0;
bool vis[N];
int n, m;
int dis1[N], dis2[N];
int a, b, c, d;
void add1(int a, int b, int c, int d) {
    e1[idx1] = b;
    w1[idx1] = c;
    w2[idx1] = d;
    ne1[idx1] = head1[a];
    head1[a] = idx1 ++ ;
}

void add2(int a, int b, int c, int d) {
    e2[idx2] = b;
    ww1[idx2] = c;
    ww2[idx2] = d;
    ne2[idx2] = head2[a];
    head2[a] = idx2 ++ ;
}

void djs(int s, int t) {
    for(int i = 1; i <= n; i ++ ) dis1[i] = inf;
    memset(vis, false, sizeof(vis));
    dis1[s] = 0;
    priority_queue<PII, vector<PII>, greater<PII> > q;
    q.push({0, s});
    while(q.size()) {
        auto cur = q.top();
        q.pop();
        int pos = cur.y;
        if(vis[pos]) continue;
        vis[pos] = true;
        for(int i = head1[pos]; i != -1; i = ne1[i]) {
            int j = e1[i];
            if(dis1[j] > dis1[pos] + w1[i]) {
                dis1[j] = dis1[pos] + w1[i];
                q.push({dis1[j], j});
            }
        }
    }
}

int dfn[N], low[N], timestamp = 0;
bool is_loop[N];
int scc_cnt = 0;
int id[N];
int in[N];
int stk[N], tt = 0;
bool in_stk[N];
void tarjan(int u) { // tarjan缩点
    dfn[u] = low[u] = ++ timestamp;
    stk[++ tt] = u;
    in_stk[u] = true;
    for(int i = head2[u]; i != -1; i = ne2[i]) {
        int j = e2[i];
        if(!dfn[j]) {
            tarjan(j);
            low[u] = min(low[u], low[j]);
        } else if(in_stk[j]){
            low[u] = min(low[u], low[j]);
        }
    }

    if(dfn[u] == low[u]) {
        int y;
        scc_cnt ++ ;
        do {
            y = stk[tt -- ];
            in_stk[y] = false;
            id[y] = scc_cnt;
        } while(y != u);
    }
}

void clear() {
    memset(head2, -1, sizeof(head2));
    memset(head1, -1, sizeof(head1));
    memset(is_loop, false, sizeof(is_loop));
    memset(in_stk, false, sizeof(in_stk));
    memset(dfn, 0, sizeof(dfn));
    memset(low, 0, sizeof(low));
    memset(in, 0, sizeof(in));
    idx1 = 0 ;
    idx2 = 0;
    timestamp = 0;
    tt = 0;
    scc_cnt = 0;
}

void build() {
    for(int i = 1; i <= n; i ++ ) { // 建最短路图
        if(dis1[i] == inf) continue;
        for(int j = head1[i]; j != -1; j = ne1[j]) {
            int k = e1[j];
            if(dis1[k] == dis1[i] + w1[j]) {
                add2(i, k, w1[j], w2[j]);
            }
        }
    }
    for(int i = 1; i <= n; i ++ ) {
        if(!dfn[i]) {
            tarjan(i);
        }
    }
    memset(head1, -1, sizeof(head1));
    idx1 = 0;
    for(int i = 1; i <= n; i ++ ) {
        for(int j = head2[i]; j != -1; j = ne2[j]) {
            int k = e2[j];
            int a = id[i], b = id[k];
            if(a == b) {
                if(ww2[j] > 0)
                    is_loop[a] = true; // 找正环,题目保证不存在
            } else {
                add1(a, b, ww1[i], ww2[j]);
                in[b] ++ ;
            }
        }
    }
}

void dp() {
    for(int i = scc_cnt; i >= 1; i -- ) {
        dis2[i] = -inf;
    }

    int s = id[1];
    dis2[s] = 0;
    queue<int> q;
    for(int i = 1; i <= scc_cnt; i ++ ) {
        if(in[i] == 0) {
            q.push(i);
            in[i] = -1;
        }
    }
    while(q.size()) {
        int pos = q.front();
        q.pop();
        if(is_loop[pos]) dis2[pos] = inf;
        for(int i = head1[pos]; i != -1; i = ne1[i]) {
            int j = e1[i];
            dis2[j] = max(dis2[j], dis2[pos] + w2[i]);
            in[j] -- ;
            if(!in[j]) {
                q.push({j});
                in[j] = -1;
            }
        }
    }
}
void solve() {
    int res1 = 0, res2 = 0;
    cin >> n >> m;
    memset(head1, -1, sizeof(head1));
    memset(head2, -1, sizeof(head2));
    idx1 = 0, idx2 = 0;
    for(int i = 1; i <= m; i ++ ) {
        cin >> a >> b >> c >> d;
        add1(a, b, c, d);
    }

    djs(1, n);
    res1 = dis1[n];
    build();
    dp(); // 拓扑排序找最长路
    res2 = dis2[id[n]];
    cout << res1 << " " << res2 << '\n';
    clear();
}

signed main() {
    cin.tie(0);
    cout.tie(0);
    ios::sync_with_stdio(0);
    int t;
    cin >> t;
    while(t -- ) {
        solve();
    }
    return 0;
}