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BZOJ4144: [AMPPZ2014]Petrol(最短路 最小生成树)

2019年01月01日  | 移动技术网IT编程  | 我要评论

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题意

题目链接

sol

做的时候忘记写题解了

可以参考

#include<bits/stdc++.h>
#define pair pair<int, int>
#define mp make_pair
#define fi first
#define se second 
using namespace std;
const int maxn = 2e6 + 10;
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}
int n, s, m, q, c[maxn], vis[maxn], ga[maxn], ans[maxn], fa[maxn], dis[maxn];
vector<pair> v[maxn];
struct edge {
    int u, v, w;
    bool operator < (const edge &rhs) const {
        return w < rhs.w;
    }
}e[maxn];
struct query {
    int x, y, b, id;
    bool operator < (const query &rhs) const {
        return b < rhs.b;
    }
}q[maxn];
void dij() {
    memset(dis, 0x3f, sizeof(dis));
    priority_queue<pair> q;
    for(int i = 1; i <= s; i++) ga[c[i]] = c[i], dis[c[i]] = 0, q.push(mp(0, c[i]));
    while(!q.empty()) {
        if(vis[q.top().se]) {q.pop(); continue;}
        int p = q.top().se; q.pop(); vis[p] = 1;
        for(int i = 0; i < v[p].size(); i++) {
            int to = v[p][i].fi, w = v[p][i].se;
            if(dis[to] > dis[p] + w) 
                dis[to] = dis[p] + w, q.push(mp(-dis[to], to)), ga[to] = ga[p];
        }
    }
}
int find(int x) {
    return fa[x] == x ? fa[x] : fa[x] = find(fa[x]);
}
void unionn(int x, int y) {
    fa[find(x)] = find(y);
}
void build() {
    for(int i = 1; i <= n; i++) fa[i] = i;
    int tmp = 0;
    for(int i = 1; i <= m; i++) {
        int x = e[i].u, y = e[i].v;
        if(ga[x] == ga[y]) continue;
        e[++tmp] = (edge) {ga[x], ga[y], dis[x] + dis[y] + e[i].w};
    }
    sort(e + 1, e + tmp + 1);
    sort(q + 1, q + q + 1);
    int cur = 1;
    for(int i = 1; i <= q; i++) {
        while(cur <= tmp && e[cur].w <= q[i].b) unionn(e[cur].u, e[cur].v), cur++;
        ans[q[i].id] = (bool)(find(q[i].x) == find(q[i].y));
    }
}
int main() {
    n = read(); s = read(); m = read();
    for(int i = 1; i <= s; i++) c[i] = read();
    for(int i = 1; i <= m; i++) {
        int x = read(), y = read(), z = read();
        e[i] = (edge) {x, y, z};
        v[x].push_back(mp(y, z));
        v[y].push_back(mp(x, z));
    }
    dij();
    q = read();
    for(int i = 1; i <= q; i++) q[i].x = read(), q[i].y = read(), q[i].b = read(), q[i].id = i;
    build();    
    for(int i = 1; i <= q; i++) puts(ans[i] ? "tak" : "nie");
    return 0;
}
/*
6 4 5
1 5 2 6
1 3 1
2 3 2
3 4 3
4 5 5
6 4 5
4
1 2 4
2 6 9
1 5 9
6 5 8

*/

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