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  • bzoj4144

    最短路+最小生成树

    因为所有东西都只跟加油站有关,那么我们只留加油站就行了。

    先预处理出每个点到加油站的最短距离,这个用多源最短路,就是把所有加油站放到堆里,然后就是重构图,我们不用直接连边,计算出每条边的贡献,然后跑最小生成树,这里直接离线查询就行了。

    #include<bits/stdc++.h>
    using namespace std;
    const int N = 2e5 + 5;
    namespace IO 
    {
        const int Maxlen = N * 50;
        char buf[Maxlen], *C = buf;
        int Len;
        inline void read_in()
        {
            Len = fread(C, 1, Maxlen, stdin);
            buf[Len] = '';
        }
        inline void fread(int &x) 
        {
            x = 0;
            int f = 1;
            while (*C < '0' || '9' < *C) { if(*C == '-') f = -1; ++C; }
            while ('0' <= *C && *C <= '9') x = (x << 1) + (x << 3) + *C - '0', ++C;
            x *= f;
        }
        inline void fread(long long &x) 
        {
            x = 0;
            long long f = 1;
            while (*C < '0' || '9' < *C) { if(*C == '-') f = -1; ++C; }
            while ('0' <= *C && *C <= '9') x = (x << 1) + (x << 3) + *C - '0', ++C;
            x *= f;
        }
        inline void read(int &x)
        {
            x = 0;
            int f = 1; char c = getchar();
            while(c < '0' || c > '9') { if(c == '-') f = -1; c = getchar(); }
            while(c >= '0' && c <= '9') { x = (x << 1) + (x << 3) + c - '0'; c = getchar(); }
            x *= f;
        }
        inline void read(long long &x)
        {
            x = 0;
            long long f = 1; char c = getchar();
            while(c < '0' || c > '9') { if(c == '-') f = -1; c = getchar(); }
            while(c >= '0' && c <= '9') { x = (x << 1ll) + (x << 3ll) + c - '0'; c = getchar(); }
            x *= f;
        } 
    } using namespace IO;
    int rd()
    {
        int x = 0, f = 1; char c = getchar();
        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, m, Q, cnt = 1, s;
    int head[N], ans[N], c[N], fa[N], d[N], rank[N];
    struct edge {
        int nxt, to, w;
    } e[N << 1];
    struct Edge {
        int u, v, w;
        Edge() {}
        Edge(int u, int v, int w) : u(u), v(v), w(w) {}
        bool friend operator < (const Edge &a, const Edge &b) {
            return a.w < b.w;
        }
    } E[N];
    struct Query {
        int u, v, id, b;
        bool friend operator < (const Query &a, const Query &b) {
            return a.b < b.b;
        }
    } que[N]; 
    void link(int u, int v, int w) 
    {
        e[++cnt].nxt = head[u];
        head[u] = cnt;
        e[cnt].to = v;
        e[cnt].w = w;
    }
    int find(int x)
    {
        return x == fa[x] ? x : fa[x] = find(fa[x]);
    }
    int main()
    {
        read_in();
        fread(n);
        fread(s);
        fread(m);
        memset(d, 0x7f7f, sizeof(d));
        priority_queue<pair<int, int>, vector<pair<int, int> >, greater<pair<int, int> > > q;
        for(int i = 1; i <= s; ++i) 
        {
            fread(c[i]);
            d[c[i]] = 0;
            q.push(make_pair(0, c[i]));
        }
        for(int i = 1; i <= m; ++i) 
        {
            int u, v, w;
            fread(u);
            fread(v);
            fread(w);
            E[i] = Edge(u, v, w);
            link(u, v, w);
            link(v, u, w);  
        }   
        fread(Q);
        for(int i = 1; i <= Q; ++i) 
        {
            fread(que[i].u);
            fread(que[i].v);
            fread(que[i].b);
            que[i].id = i;
        }
        sort(que + 1, que + Q + 1);
        while(!q.empty()) 
        {
            pair<int, int> o = q.top();
            q.pop();
            int u = o.second;
            if(o.first > d[u]) continue;
            for(int i = head[u]; i; i = e[i].nxt) if(d[e[i].to] > d[u] + e[i].w)
            {
                d[e[i].to] = d[u] + e[i].w;
                q.push(make_pair(d[e[i].to], e[i].to)); 
            }   
        }
        for(int i = 1; i <= m; ++i) E[i].w += d[E[i].u] + d[E[i].v];
        sort(E + 1, E + m + 1);
        for(int i = 1; i <= n; ++i) fa[i] = i;
        for(int i = 1, j = 0; i <= Q; ++i) 
        {
            while(j + 1 <= m && E[j + 1].w <= que[i].b) 
            {
                ++j;
                int u = find(E[j].u), v = find(E[j].v);
                if(u == v) continue;
                if(rank[u] < rank[v]) swap(u, v);
                rank[u] += rank[v];
                fa[u] = v;  
            }
            ans[que[i].id] = find(que[i].u) == find(que[i].v);
        }
        for(int i = 1; i <= Q; ++i) puts(ans[i] ? "TAK" : "NIE");
        return 0;
    }
    
    View Code
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  • 原文地址:https://www.cnblogs.com/19992147orz/p/7911808.html
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