题意：给定m<=50000的1-n有联通的图，求最多可以使K<=20条边变为0的情况下的最短路是多少。。
思路：简单的分层图最短路，对于每个点拆成K个点。。
然后求一边最短路。。
code：
1 /* 2 * Author: Yzcstc 3 * Created Time: 2014/11/5 19:25:47 4 * File Name: revamp.cpp 5 */ 6 #include<cstdio> 7 #include<iostream> 8 #include<cstring> 9 #include<cstdlib> 10 #include<cmath> 11 #include<algorithm> 12 #include<string> 13 #include<map> 14 #include<set> 15 #include<vector> 16 #include<queue> 17 #include<stack> 18 #include<ctime> 19 #define M0(x) memset(x, 0, sizeof(int) * (n+10)) 20 #define MP make_pair 21 #define x first 22 #define y second 23 using namespace std; 24 typedef long long ll; 25 typedef pair<ll, int> pii; 26 const int maxn = 1200000; 27 struct edge{ 28 int v, w, next; 29 } e[maxn * 5]; 30 int last[maxn], tot; 31 int n, k, m; 32 33 void add(const int &u,const int &v, const int& w){ 34 e[tot] = (edge){v, w, last[u]}; last[u] = tot++; 35 } 36 37 void init(){ 38 int u, v, w; 39 memset(last, -1, sizeof(int) * (n * k + 10)); 40 tot = 0; 41 for (int i = 0; i < m; ++i){ 42 scanf("%d%d%d", &u, &v, &w); 43 add(u, v, w); add(v, u, w); 44 for (int i = 1; i <= k; ++i){ 45 add(u + i * n, v + i * n, w); 46 add(v + i * n, u + i * n, w); 47 add(u + (i-1) * n, v + i * n, 0); 48 add(v + (i-1) * n, u + i * n, 0); 49 } 50 } 51 } 52 53 ll dis[maxn]; 54 int vis[maxn]; 55 void dij(){ 56 priority_queue<pii, vector<pii>, greater<pii> > q; 57 int nt = n * k + n; 58 memset(vis, 0, sizeof(vis)); 59 for (int i = 1; i <= nt; ++i) dis[i] = (1LL<<50); 60 pii tmp(0, 1); dis[1] = 0; 61 q.push(tmp); 62 int u, v; 63 while (!q.empty()){ 64 u = q.top().y; 65 q.pop(); 66 if (vis[u]) continue; 67 vis[u] = 1; 68 for (int p = last[u]; ~p; p = e[p].next){ 69 v = e[p].v; 70 if (dis[u] + e[p].w < dis[v]){ 71 dis[v] = dis[u] + e[p].w; 72 tmp.x = dis[v], tmp.y = v; 73 q.push(tmp); 74 } 75 } 76 } 77 } 78 79 80 void solve(){ 81 dij(); 82 ll ans = 1LL<<50; 83 for (int i = 0; i <= k; ++i) 84 ans = min(ans, dis[n+i*n]); 85 cout << ans << endl; 86 } 87 88 int main(){ 89 freopen("a.in", "r", stdin); 90 freopen("a.out", "w", stdout); 91 clock_t start, finish; 92 start = clock(); 93 while (scanf("%d%d%d", &n, &m, &k) != EOF){ 94 init(); 95 solve(); 96 } 97 finish = clock(); 98 double t = (double)(finish - start) / CLOCKS_PER_SEC; 99 // printf("time = %.6f ", t); 100 return 0; 101 }