【luogu P1186】玛丽卡

https://www.luogu.org/problem/show?pid=1186

考虑暴力，枚举图上每一条边删去后跑Dijkstra，取M次的最大值。

仔细想想就会发现删除最短路以外的边对最短路毫无影响，于是先跑出最短路，然后枚举最短路上的每一条边删去后跑Dijkstra，取这几次的最大值。

#include <iostream>
#include <vector>
#include <queue>
#include <cstring>
#define maxn 1005
using namespace std;
const int inf = 0x3f3f3f3f;
int n, m;
struct edge
{
    int from, to, weight;
};
vector<edge> e;
vector<int> g[maxn];

bool known[maxn];
int dist[maxn], front[maxn];
typedef pair<int, int> pin;
priority_queue<pin, vector<pin>, greater<pin>> pq;
void dijkstra()
{
    for (int i = 1; i <= n; ++i)
    {
        dist[i] = inf;
        known[i] = false;
        front[i] = -1;
    }
    while (!pq.empty())
        pq.pop();
    dist[1] = 0;
    pq.push(make_pair(dist[1], 1));
    while (!pq.empty() && !known[n])
    {
        int x = pq.top().second;
        pq.pop();
        if (known[x])
            continue;
        known[x] = true;
        for (int i = 0; i < g[x].size(); i++)
        {
            edge &ed = e[g[x][i]];
            if (!known[ed.to] && dist[ed.to] > dist[x] + ed.weight)
            {
                dist[ed.to] = dist[x] + ed.weight;
                front[ed.to] = g[x][i];
                pq.push(make_pair(dist[ed.to], ed.to));
            }
        }
    }
}
vector<int> path;

int main()
{
    cin >> n >> m;
    int u, v, w;
    for (int i = 1; i <= m; i++)
    {
        cin >> u >> v >> w;
        e.push_back((edge){u, v, w});
        g[u].push_back(e.size() - 1);
        e.push_back((edge){v, u, w});
        g[v].push_back(e.size() - 1);
    }
    dijkstra();
    for (int i = front[n]; i != -1; i = front[e[i].from])
        path.push_back(i);

    int ans = dist[n];
    for (int i = 0; i < path.size(); i++)
    {
        w = e[path[i]].weight;
        e[path[i]].weight = e[path[i] ^ 1].weight = inf;
        dijkstra();
        if (dist[n] != inf)
            ans = max(ans, dist[n]);
        e[path[i]].weight = e[path[i] ^ 1].weight = w;
    }
    cout << ans;
    return 0;
}

【luogu P1186】玛丽卡

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