【POJ3613】Cow Relays
题目链接:https://www.acwing.com/problem/content/347/
题目大意:给定一张无向带权图 , 找出从 \(s\) 到 \(e\) 恰好经过 \(n\) 条边的最短路径
solution
不难发现 , 顶点数可能很多 , 但边数不超过 100 , 因此对于节点编号应该先离散化一下
用矩阵 \(G^r\) , 表示恰好经过 \(r\) 条边的最短路 , 其中 \(G^r[i][j]\) 表示从 \(i\) 到 \(j\) 恰好经过 \(r\) 条边的最短路径 , 显然可以得出以下推论 :
如果\(r = p + q\) , 则$$G^r[i][j] = min_{k \leq n}\left{G^p[i][k] + G^q[k][j]\right}(n为节点数量)$$
同时发现 , 这个式子满足结合律 , 因此可以用矩阵快速幂快速求出 \(G^r\) , \(G^n[s][e]\)即为题目所求
时间复杂度: \((T^3logn)\)
code
#include<bits/stdc++.h>
using namespace std;
template <typename T> inline void read(T &FF) {
int RR = 1; FF = 0; char CH = getchar();
for(; !isdigit(CH); CH = getchar()) if(CH == '-') RR = -RR;
for(; isdigit(CH); CH = getchar()) FF = FF * 10 + CH - 48;
FF *= RR;
}
inline void file(string str) {
freopen((str + ".in").c_str(), "r", stdin);
freopen((str + ".out").c_str(), "w", stdout);
}
const int N = 205;
int n, m, s, e, path[N][3], ni, num[N];
struct matrix{
int g[N][N];
matrix() {
memset(g, 0x3f, sizeof(g));
//for(int i = 1; i <= n; i++) g[i][i] = 0;
}
friend matrix operator * (const matrix &ai, const matrix &bi) {
matrix res;
for(int k = 1; k <= ni; k++)
for(int i = 1; i <= ni; i++)
for(int j = 1; j <= ni; j++)
res.g[i][j] = min(res.g[i][j], ai.g[i][k] + bi.g[k][j]);
return res;
}
}base;
void cpy(matrix &ai, const matrix &bi) {
for(int i = 1; i <= ni; i++)
for(int j = 1; j <= ni; j++)
ai.g[i][j] = bi.g[i][j];
}
matrix Qpow(int ki) {
if(ki == 1) return base;
matrix hi = Qpow(ki / 2); cpy(hi, hi * hi);
return ki & 1 ? hi * base : hi;
}
int main() {
//file("");
read(n), read(m), read(s), read(e);
for(int i = 1; i <= m; i++) {
read(path[i][0]), read(path[i][1]), read(path[i][2]);
num[++ni] = path[i][1], num[++ni] = path[i][2];
}
sort(num + 1, num + ni + 1);
ni = unique(num + 1, num + ni + 1) - num - 1;
//for(int i = 1; i <= n; i++) base.g[i][i] = 0;
for(int i = 1; i <= m; i++) {
path[i][1] = lower_bound(num + 1, num + ni + 1, path[i][1]) - num;
path[i][2] = lower_bound(num + 1, num + ni + 1, path[i][2]) - num;
base.g[path[i][1]][path[i][2]] = base.g[path[i][2]][path[i][1]] = path[i][0];
}
s = lower_bound(num + 1, num + ni + 1, s) - num;
e = lower_bound(num + 1, num + ni + 1, e) - num;
matrix ans = Qpow(n);
cout << ans.g[s][e] << endl;
return 0;
}