【日常训练】Volleyball（CodeForces-96D）

题意与分析

这题也是傻逼题，可是我当时打比赛的时候板子出问题了- -|||，怎么调也调不过。

不过思路是很清晰的：先做n次dijkstra然后重新建图，建完了以后根据新的单向图再跑一次dijkstra。

代码

#include <bits/stdc++.h>

#define ZERO(x) memset(x, 0, sizeof(x))
using namespace std;
using ll=long long;

struct Edge { int v, nxt; ll c; };
Edge w[20005];
vector<int>    que;
ll dist[1005], a[1005], b[1005];
int    e[1005][1005], ww[1005];
bool vis[1005];
int    N, M, x, y, W = 1;

void ShortestPath(int s)
{
    memset(dist, 60, sizeof(dist)); dist[s] = 0;
    ZERO(vis); vis[s]=true;
    que.clear();
    que.push_back(s);
    for (int fi = 0; fi < que.size(); ++ fi)
    {
        int    u = que[fi];
        for (int i = ww[u]; i; i = w[i].nxt)
        {
            int    v = w[i].v;
            if (dist[v] <= dist[u] + w[i].c) continue;
            dist[v] = dist[u] + w[i].c;
            if (vis[v]) continue;
            vis[v] = true;
            que.push_back(v);
        }
        vis[u] = false;
    }
    
    e[s][0] = 0;
    for (int i = 1; i <= N; ++ i)
        if (i != s && dist[i] <= a[s]) e[s][++ e[s][0]] = i;
}


int    main()
{
    cin >> N >> M;
    cin >> x >> y;
    for (int i = 1, u, v; i <= M; ++ i)
    {
        ll    c;
        cin >> u >> v >> c;
        w[++ W].v = v; w[W].c = c; w[W].nxt = ww[u]; ww[u] = W;
        w[++ W].v = u; w[W].c = c; w[W].nxt = ww[v]; ww[v] = W;
    }
    for (int i = 1; i <= N; ++ i) cin >> a[i] >> b[i];
    for (int i = 1; i <= N; ++ i) ShortestPath(i);
    
    memset(dist, 60, sizeof(dist)); dist[x] = 0;
    ZERO(vis); vis[x]=true;
    que.clear();
    que.push_back(x);
    for (int fi = 0; fi < que.size(); ++ fi)
    {
        int    u = que[fi];
        for (int i = 1; i <= e[u][0]; ++ i)
        {
            int    v = e[u][i];
            if (dist[v] <= dist[u] + b[u]) continue;
            dist[v] = dist[u] + b[u];
            if (vis[v]) continue;
            vis[v] = 1;
            que.push_back(v);
        }
        vis[u] = 0;
    }
    
    if (dist[y] > (1ll << 60)) dist[y] = -1;
    cout << dist[y] << endl;
        
    return 0;
}