洛谷P1462 通往奥格瑞玛的道路（二分+spfa，二分+Dijkstra）

洛谷P1462 通往奥格瑞玛的道路

二分费用。

用血量花费建图，用单源最短路判断 (1) 到 (n) 的最短路花费是否小于 (b) 。二分时需要不断记录合法的 (mid) 值。

这里建议使用while(l <= r)，会避免出现答案为 (r) 时和答案AFK搞混，样例就是这种情况。

复杂度为 (O(log n) imes) 最短路的复杂度。

• 二分 + Dijkstra版本，复杂度为 (O(log n imes Elog E)) 。

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<queue>

using namespace std;

const long long inf = 1e9+7;
const int maxn = 10005;
const int maxm = 50005;
int n, m, l, r, ai, bi, tot, head[maxn], vis[maxn], f[maxn];
long long b, ci, dist[maxn];
struct node{
    int to, nex;
    long long cost;
}edge[2 * maxm];
struct G{
    int id;
    long long cost;
    G(){}
    G(int _id, long long _cost){
        id = _id; cost = _cost;
    }
    /******/
    bool operator < (const G & _G) const {
        return cost > _G.cost;
    }
}now;

inline void addedge(int u, int v, long long w){
    edge[++tot].to = v;
    edge[tot].cost = w;
    edge[tot].nex = head[u];
    head[u] = tot;
}
int dijkstra(int c)
{
    for(int i = 1; i <= n; i++) vis[i] = 0, dist[i] = inf;
    priority_queue<G> q;
    while(!q.empty()) q.pop();
    q.push(G(1, 0));
    dist[1] = 0;
    while(!q.empty()){
        now = q.top(); q.pop();
        if(vis[now.id] == 1) continue;
        vis[now.id] = 1;
        for(int j = head[now.id]; j > 0; j = edge[j].nex){
            int tmp = edge[j].to;
            if(vis[tmp] == 0 && dist[tmp] > dist[now.id] + edge[j].cost && f[tmp] <= c){
                dist[tmp] = dist[now.id] + edge[j].cost;
                q.push(G(tmp, dist[tmp]));
            }
        }
    }
    if(dist[n] < b) return 1;
    else return 0;
}
int main()
{
    scanf("%d%d%lld", &n, &m, &b);
    for(int i = 1; i <= n; i++){
        scanf("%d", &f[i]);
        r = max(r, f[i]);
    }
    for(int i = 1; i <= m; i++){
        scanf("%d%d%lld", &ai, &bi, &ci);
        addedge(ai, bi, ci);
        addedge(bi, ai, ci);
    }
    int ans = 0;
    while(l <= r){
        int mid = (l + r) >> 1;
        if(dijkstra(mid)) ans = mid, r = mid - 1;
        else l = mid + 1;
    }
    if(ans == 0) printf("AFK
");
    else printf("%d
", ans);
    return 0;
}

• 二分 + spfa版本，复杂度为 (O(log n imes kE)) ，(k) 通常为 (2) 。

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<queue>

using namespace std;

const long long inf = 1e9+7;
const int maxn = 10005;
const int maxm = 50005;
int n, m, l, r, ai, bi, tot, head[maxn], vis[maxn], f[maxn];
long long b, ci, dist[maxn];
struct node{
    int to, nex;
    long long cost;
}edge[2 * maxm];
queue<int> q;

inline void addedge(int u, int v, long long w){
    edge[++tot].to = v;
    edge[tot].cost = w;
    edge[tot].nex = head[u];
    head[u] = tot;
}
int spfa(int c){
    for(int i = 1; i <= n; i++) dist[i] = inf, vis[i] = 0;
    q.push(1); vis[1] = 1; dist[1] = 0;
    while(!q.empty()){
        int now = q.front();
        q.pop();
        vis[now] = 0;
        for(int i = head[now]; i > 0; i = edge[i].nex){
            int v = edge[i].to;
            if(dist[v] > dist[now] + edge[i].cost && f[v] <= c){
                dist[v] = dist[now] + edge[i].cost;
                if(vis[v] == 0){
                    /******/
                    vis[v] = 1;
                    q.push(v);
                }
            }
        }
    }
    if(dist[n] < b) return 1;
    else return 0;
}
int main()
{
    scanf("%d%d%lld", &n, &m, &b);
    for(int i = 1; i <= n; i++){
        scanf("%d", &f[i]);
        r = max(r, f[i]);
    }
    for(int i = 1; i <= m; i++){
        scanf("%d%d%lld", &ai, &bi, &ci);
        addedge(ai, bi, ci);
        addedge(bi, ai, ci);
    }
    int ans = 0;
    while(l <= r){
        int mid = (l + r) >> 1;
        if(spfa(mid)) ans = mid, r = mid - 1;
        else l = mid + 1;
    }
    if(ans == 0) printf("AFK
");
    else printf("%d
", ans);
    return 0;
}