[Code+#4]最短路 解题报告

Posted 2021-01-08 yjzoier

Luogu · 传送门

Orz THU众大佬，lct（注意不是link-cut-tree，是一个大佬）

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这道题很容易让人联想到 最短路，但是最短路需要先 建图；

暴力建出所有边的算法显然是不可行的，因为这样会建出 (O(n^2 + m)) 条边；

那么我们要考虑能不能 减少一些边 ，使边的数量可以接受。

从哪里入手减少边的数量呢？异或或许是一个不错的切入口。

举个栗子：

假设我们要从 (001_2) 到 (010_2)，我们要花费 (2^0 + 2^1) 的费用；
但是，最短路有一个 优越的性质，我们可以把边拆开来，可以先从 (001_2) 到 (000_2)，再从 (000_2) 到 (010_2)，费用是一样的。

这样我们对于每个点 (i)，只需要建 (i) 到 (i XOR 2^k) 的边，之后 Dijkstra 就可以了哈。

需要注意的是 边界情况：从 (i) 到 (j) 经过的中间点可能超过 (n)，对此有 2 种处理方法：

1. 建边和 Dijkstra 的范围调整为 ([0,n])
2. 建边和 Dijkstra 的范围调整为 ([1, 2^k-1]，k = min { k | n leq 2^k -1 })

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方法 1 的代码

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<queue>
using namespace std;
?
const int maxn = 100007;
const int maxm = 2500007;
int n, m, c;
int edgenum, head[maxn], nxt[maxm], vet[maxm], val[maxm];
inline void addedge(int u, int v, int w){
????++edgenum;
????vet[edgenum] = v;
????val[edgenum] = w;
????nxt[edgenum] = head[u];
????head[u] = edgenum;
}
?
inline int read(){
????int f = 1, val = 0; char ch = getchar();
????while ((ch < '0' || ch > '9') && (ch != '-')) ch = getchar();
????if (ch == '-') f = -1, ch = getchar();
????while (ch >= '0' && ch <= '9') val = (val << 3) + (val << 1) + ch - '0', ch = getchar();
????return val * f;
}
?
int dist[maxn];
bool vis[maxn];
#define pii pair<int, int>
priority_queue< pii, vector< pii >, greater< pii > > Qmin;
const int INF = 1000000007;
inline void Dijkstra(int s){
????for (int i = 0; i <= n; ++i){
????????vis[i] = false;
????????dist[i] = INF;
????}
????dist[s] = 0; Qmin.push( make_pair(0, s) );
????for (int i = 0; i <= n; ++i){
????????while (!Qmin.empty() && vis[Qmin.top().second]) Qmin.pop();
????????if (Qmin.empty()) break;
????????int u = Qmin.top().second; Qmin.pop();
????????vis[u] = true;
????????for (int e = head[u]; e; e = nxt[e]){
????????????int v = vet[e], cost = val[e];
????????????if (dist[v] > dist[u] + cost){
????????????????dist[v] = dist[u] + cost;
????????????????Qmin.push( make_pair(dist[v], v) );
????????????}
????????}
????}
}
?
int main(){
????n = read(); m = read(); c = read();
????for (int i = 1; i <= m; ++i){
????????int u = read(), v = read(), w = read();
????????addedge(u, v, w);
????}
?????
????for (int i = 0; i <= n; ++i){
????????for (int j = 0; j < 20; ++j){
????????????int to = i ^ (1 << j);
????????????if (to <= n) addedge(i, to, c * (1 << j));
????????}
????}
?????
????int A = read(), B = read();
????Dijkstra(A);
?????
????printf("%d
", dist[B]);
????return 0;
}

方法 2 的代码

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<queue>
#define pii pair<int,int>
using namespace std;

const int maxn = 200007;
const int maxm = 3000007;
int n, m, C, lgn, A, B;
int edgenum, hea[maxn], vet[maxm], nxt[maxm], val[maxm];
inline void addedge(int u, int v, int cost){
    ++edgenum;
    vet[edgenum] = v;
    val[edgenum] = cost;
    nxt[edgenum] = hea[u];
    hea[u] = edgenum;

    //printf("%d -> %d (%d)
", u, v, cost);
} 

inline int read(){
    int f=1, val=0; char ch=getchar();
    while ((ch<'0'||ch>'9')&&(ch!='-')) ch=getchar();
    if (ch=='-') f=-1,ch=getchar();
    while (ch>='0'&&ch<='9') val=(val<<3)+(val<<1)+ch-'0',ch=getchar();
    return val*f;
}

int dist[maxn];
bool vis[maxn];
priority_queue<pii, vector< pii >, greater< pii > > Qmin;
inline void Dijkstra(int s){
    for (int i = 1; i <= n; ++i){
        vis[i] = false;
        dist[i] = 1000000000;
    }
    dist[s] = 0; Qmin.push(make_pair(0, s));
    for (int i = 1; i <= n; ++i){
        while (!Qmin.empty() && vis[Qmin.top().second]) Qmin.pop();
        if (Qmin.empty()) break;
        int u = Qmin.top().second; Qmin.pop();
        vis[u] = true;
        for (int e = hea[u]; e; e = nxt[e]){
            int v = vet[e], cost = val[e];
            if (dist[v] > dist[u] + cost) dist[v] = dist[u] + cost, Qmin.push(make_pair(dist[v], v));
        }
    }
}

int main(){
    n = read(); m = read(); C = read();
    for (int i = 1; i <= m; ++i){
        int u = read(), v = read(), cost = read();
        addedge(u, v, cost);
    }
    lgn = floor(log2(n)) + 1;
    n = (1 << lgn) - 1;

    for (int i = 1; i <= n; ++i){
        for (int j = 0; j < lgn; ++j)
            addedge(i, i ^ (1 << j), (1 << j) * C);
    }

    A = read(); B = read();
    Dijkstra(A);

    printf("%d
", dist[B]);
    return 0;
}