[Code+#4]最短路 解题报告
Orz THU众大佬,lct(注意不是link-cut-tree
,是一个大佬)
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:
- 建边和 Dijkstra 的范围调整为 \([0,n]\)
- 建边和 Dijkstra 的范围调整为 \([1, 2^k-1],k = min \{ k \ | \ n \leq 2^k -1 \}\)
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\n", 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)\n", 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\n", dist[B]);
return 0;
}
来自YJZoier的博客
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