P1462 通往奥格瑞玛的道路 (二分+最短路)

Posted 2021-10-29 lykkk

题目

P1462 通往奥格瑞玛的道路
给定\(n\)个点\(m\)条边，每个点上都有点权\(f[i]\)，每条边上有边权，找一条道路，使边权和小于给定的数\(b\)，并使最大点权最小。

解析

二分一下钱，然后跑最短路，判断一下如果只有这么多钱的话能不能到终点（最短路边权和是不是不超过\(b\)），套个最短路板子，套个二分板子，没了。

代码

//二分+最短路 
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10;
const int INF = 0x3f3f3f3f;
int n, m, b, num;
int l, r, mid = INF;
int dis[N], head[N], f[N];
bool vis[N];

class edge {
    public :
        int v, nx, w;
}e[N];

class node {
    public:
        int id, dis;
        bool operator < (const node &oth) const {
            return this->dis > oth.dis;
        }
};

template<class T>inline void read(T &x) {
    x = 0; int f = 0; char ch = getchar();
    while (!isdigit(ch)) f |= (ch == '-'), ch = getchar();
    while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
    x = f ? -x : x;
    return;
}

inline void add(int u, int v, int w) {
    e[++num].nx = head[u], e[num].v = v, e[num].w = w, head[u] = num;
}

priority_queue<node>q;
bool dijkstra() {
    memset(dis, INF, sizeof dis);
    memset(vis, 0, sizeof vis);
    dis[1] = 0;
    q.push((node){1, 0}) ;
    while (!q.empty()) {
        node d = q.top(); q.pop();
        int u = d.id;
        if (vis[u]) continue;
        vis[u] = 1;
        for (int i = head[u]; ~i; i = e[i].nx) {
            int v = e[i].v;
            if (dis[u] + e[i].w < dis[v] && f[v] <= mid) {
                dis[v] = dis[u] + e[i].w;
                q.push((node){v, dis[v]});
            }
        }
    }
    return (dis[n] > b);
}

int main() {
    memset(head, -1, sizeof (head));
    read(n), read(m), read(b);
    for (int i = 1; i <= n; ++i) read(f[i]), r = max(r, f[i]);
    for (int i = 1, x, y, z; i <= m; ++i) {
        read(x), read(y), read(z);
        add(x, y, z), add(y, x, z);
    }
    if (dijkstra()) {
        printf("AFK\n");
        return 0;
    }
    while (l <= r) {
        mid = (l + r) >> 1;
        if (dijkstra()) l = mid + 1;
        else r = mid - 1;
    }
    printf("%d\n", l);
    return 0; 
}