problem
在一个无向图\(G=(V,E)\)中，可以改变\(k\)条边的权值为\(\Delta w\)，求单源最短路径。
solution
分层图的想法就是如果有\(k\)条边就建\(k+1\)层图
这个图实际上是这样的，对于每层中相连的点\((u,v)\)连权值为\(w\)的无向边，在本层图中相连的点\((u,v)\)由上层点\(u\)向下层点\(v\)以及上层点\(v\)向下层点\(u\)连权值为\(\Delta w\)的有向边，方向是从上层向下层。
这样构造完成一张分层图后，从第\(1\)层的起始点\(s\)求单源最短路径，最终第\(k + 1\)层的终点\(t\)的单源最短路径值即为答案所求。
原理其实很简单，如果从上层图到下层图，有向边\((u,v)\)是一条\(\Delta w\)权边，相当于把\(w\)权边变成了\(\Delta w\)权边，这样如果有\(k+1\)层图的话，相当于进行了\(k\)次这种操作，自然就实现了在\(k+1\)层图中\(k\)次改变边权的目标。
这是最简单的一种分层图，如果学习了更难的构图方法和题目，会再补上。
T1
P2939 [USACO09FEB]改造路Revamping Trails
#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;
const int N = 1e4 + 5, M = 5e4 + 5, K = 105;
int n, m, k;
struct Edge {
int Next, to, dis;
}e[M * K * 2];
int head[N * K], num;
void add(int from, int to, int dis)
{
e[++num].Next = head[from];
e[num].to = to;
e[num].dis = dis;
head[from] = num;
}
int dist[N * K], vis[N * K];
struct node {
int u, d;
bool operator < (const node &x) const {
return d > x.d;
}
};
priority_queue<node> q;
void dijk()
{
memset(dist, 0x3f, sizeof(dist));
dist[1] = 0;
q.push((node){1, 0});
while(!q.empty())
{
int u = q.top().u; q.pop();
if(vis[u]) continue;
vis[u] = 1;
for(int i = head[u]; i; i = e[i].Next)
{
int v = e[i].to, w = e[i].dis;
if(dist[v] > dist[u] + w)
{
dist[v] = dist[u] + w;
if(!vis[v]) q.push((node){v, dist[v]});
}
}
}
}
int main()
{
scanf("%d%d%d", &n, &m, &k);
for(int i = 1, u, v, z; i <= m; i++)
{
scanf("%d%d%d", &u, &v, &z);
add(u, v, z); add(v, u, z);
for(int j = 1; j <= k; j++)
{
add(j * n + u, j * n + v, z);
add(j * n + v, j * n + u, z);
add((j - 1) * n + u, j * n + v, 0);
add((j - 1) * n + v, j * n + u, 0);
}
}
dijk();
int ans = 0x3fffffff;
for(int i = 1; i <= k + 1; i++)
ans = min(dist[i * n], ans);
printf("%d\n", ans);
return 0;
}T2
#include <iostream>
#include <cstdio>
#include <queue>
#include <cstring>
using namespace std;
const int N = 1e4 + 5, M = 5e4 + 5, K = 12;
int n, m, k;
int s, t;
struct _edge {
int Next, v, w;
}e[M * K * 4 + M * 2];
int head[N * K], num;
void add(int from, int to, int dis)
{
e[++num].Next = head[from];
e[num].v = to;
e[num].w = dis;
head[from] = num;
}
int dist[N * K], vis[N * K];
struct node {
int u, d;
bool operator < (const node &x) const {
return d > x.d;
}
};
priority_queue<node> q;
void dijk(int x)
{
memset(dist, 0x3f, sizeof(dist));
dist[x] = 0;
q.push((node){x, 0});
while(!q.empty())
{
node tp = q.top(); q.pop();
int u = tp.u;
if(u == t + k * n) break;
if(vis[u]) continue;
vis[u] = 1;
for(int i = head[u]; i; i = e[i].Next)
{
int v = e[i].v, w = e[i].w;
if(dist[v] > dist[u] + w)
{
dist[v] = dist[u] + w;
if(!vis[v]) q.push((node){v, dist[v]});
}
}
}
}
int main()
{
scanf("%d%d%d", &n, &m, &k);
scanf("%d%d", &s, &t); s++, t++;
for(int i = 1, u, v, z; i <= m; i++)
{
scanf("%d%d%d", &u, &v, &z); u++, v++;
add(u, v, z); add(v, u, z);
for(int j = 1; j <= k; j++)
{
add(j * n + u, j * n + v, z);
add(j * n + v, j * n + u, z);
add((j - 1) * n + u, j * n + v, 0);
add((j - 1) * n + v, j * n + u, 0);
}
}
dijk(s);
int ans = 1e9;
for(int i = 0; i <= k; i++)
ans = min(ans, dist[t + i * n]);
printf("%d\n", ans);
return 0;
}T3
#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;
const int N = 55, M = 1005, K = 55;
int n, m, k;
struct Edge {
int Next, to, dis;
}e[M * K * 10];
int head[N * K * 10], num;
void add(int from, int to, int dis)
{
e[++num].Next = head[from];
e[num].to = to;
e[num].dis = dis;
head[from] = num;
}
struct node {
int u, d;
bool operator < (const node& x) const {
return d > x.d;
}
};
priority_queue<node> q;
int dist[N * K * 10], vis[N * K * 10];
void dijk()
{
memset(dist, 0x3f, sizeof(dist));
dist[1] = 0;
q.push((node){1, 0});
while(!q.empty())
{
int u = q.top().u; q.pop();
if(vis[u]) continue;
vis[u] = 1;
for(int i = head[u]; i; i = e[i].Next)
{
int v = e[i].to, w = e[i].dis;
if(dist[v] > dist[u] + w)
{
dist[v] = dist[u] + w;
if(!vis[v]) q.push((node){v, dist[v]});
}
}
}
}
int main()
{
scanf("%d%d%d", &n, &m, &k);
for(int i = 1, u, v, z; i <= m; i++)
{
scanf("%d%d%d", &u, &v, &z);
add(u, v, z); add(v, u, z);
for(int j = 1; j <= k; j++)
{
add(j * n + u, j * n + v, z);
add(j * n + v, j * n + u, z);
add((j - 1) * n + u, j * n + v, z >> 1);
add((j - 1) * n + v, j * n + u, z >> 1);
}
}
dijk();
int ans = 0x7fffffff;
for(int i = 1; i <= k + 1; i++)
ans = min(ans, dist[i * n]);
printf("%d\n", ans);
return 0;
}