标签:type problem while pre getc cst node getch empty
发现数据范围很小, 考虑从这上面入手
不难发现, 如果我们把所有边的长度排序, 将每条边选与不选看作一个 01 串
假设最优路径长度为 L , 必然存在一个 \(K\) , 满足前 \(1 \to K\) 都是 1 , 其他的随便
考虑枚举这个 \(K\)
设 \(f[i][j][k]\) 满足到 \(i\) 点, 前 \(K\) 个中选了 \(j\) 条, 已经交换了 \(k\) 条
转移的话就是看这条边是不是可以换, 换不换就行
这里的转移方程和网上大部分的不太一样, 我把前 \(K\) 条的贡献提前加上去了
发现可以最短路转移, 跑就对了
#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
#include <queue>
const int N = 55;
const int M = 155;
using namespace std;
int head[N], cnt, n, m, K, sum[M], f[N][M][25], ans, U[M], V[M], W[M];
struct edge { int to, nxt, cost, id; } e[M << 1];
struct pir
{
int a, b;
pir(int _a = 0, int _b = 0) { a = _a, b = _b; }
bool operator < (const pir &p) const
{
return a < p.a;
}
} c[M << 1];
struct node
{
int a, b, c, d;
node(int _a = 0, int _b = 0, int _c = 0, int _d = 0) { a = _a, b = _b, c = _c, d = _d; }
bool operator < (const node &p) const
{
return d > p.d;
}
};
bool vis[N][M][25];
priority_queue<node> q;
template < typename T >
inline T read()
{
T x = 0, w = 1; char c = getchar();
while(c < '0' || c > '9') { if(c == '-') w = -1; c = getchar(); }
while(c >= '0' && c <= '9') { x = x * 10 + c - '0'; c = getchar(); }
return x * w;
}
inline void adde(int u, int v, int w, int id) { e[++cnt] = (edge) { v, head[u], w, id }, head[u] = cnt; }
void dijkstra(int lim)
{
for(int i = 1; i <= n; i++)
for(int j = 0; j <= lim; j++)
for(int k = 0; k <= K; k++)
f[i][j][k] = 0x3f3f3f3f, vis[i][j][k] = 0;
f[1][0][0] = sum[lim]; q.push(node(1, 0, 0, f[1][0][0]));
int u, j, k, w;
while(!q.empty())
{
node tmp = q.top(); q.pop();
u = tmp.a, j = tmp.b, k = tmp.c, w = tmp.d;
if(vis[u][j][k]) continue;
vis[u][j][k] = 1;
for(int v, i = head[u]; i; i = e[i].nxt)
{
v = e[i].to;
if(e[i].id <= lim)
{
if(j + 1 <= lim && w < f[v][j + 1][k])
{
f[v][j + 1][k] = w;
q.push(node(v, j + 1, k, f[v][j + 1][k]));
}
}
else
{
if(w + e[i].cost < f[v][j][k])
f[v][j][k] = w + e[i].cost, q.push(node(v, j, k, f[v][j][k]));
if(j < lim && k < K && w < f[v][j + 1][k + 1])
f[v][j + 1][k + 1] = w, q.push(node(v, j + 1, k + 1, f[v][j + 1][k + 1]));
}
}
}
for(int i = 0; i <= K; i++)
ans = min(ans, f[n][lim][i]);
}
int main()
{
n = read <int> (), m = read <int> (), K = read <int> ();
ans = 0x3f3f3f3f;
for(int i = 1; i <= m; i++)
U[i] = read <int> (), V[i] = read <int> (), W[i] = read <int> (), c[i] = pir(W[i], i);
sort(c + 1, c + m + 1);
for(int i = 1; i <= m; i++)
adde(U[c[i].b], V[c[i].b], W[c[i].b], i), adde(V[c[i].b], U[c[i].b], W[c[i].b], i);
for(int i = 1; i <= m; i++)
sum[i] = sum[i - 1] + c[i].a;
for(int i = 1; i <= m; i++)
dijkstra(i);
printf("%d\n", ans);
return 0;
}
标签:type problem while pre getc cst node getch empty
原文地址:https://www.cnblogs.com/ztlztl/p/12258427.html