题意：
给一张有向图，可以把$k$条边的边权变成$0$，求点$1$到$n$的最短路。
思路：
$dp$+$dijkstra$思想。
$dis[i][k]$表示点$1$到$i$实行了$k$次把边权变为$0$的操作之后的最短距离。转移为：
$dis[v][k]=min(dis[v][k],dis[u][k]+w)$
$dis[v][k+1]=min(dis[v][k+1],dis[u][k])$
代码：
1 //#include<bits/stdc++.h> 2 #include <set> 3 #include <map> 4 #include <stack> 5 #include <cmath> 6 #include <queue> 7 #include <cstdio> 8 #include <string> 9 #include <vector> 10 #include <cstring> 11 #include <iostream> 12 #include <algorithm> 13 14 #define ll long long 15 #define pll pair<ll,ll> 16 #define pii pair<int,int> 17 #define bug printf("********* ") 18 #define FIN freopen("input.txt","r",stdin); 19 #define FON freopen("output.txt","w+",stdout); 20 #define IO ios::sync_with_stdio(false),cin.tie(0) 21 #define ls root<<1 22 #define rs root<<1|1 23 #define pb push_back 24 25 using namespace std; 26 const int inf = 2e9 + 7; 27 const ll Inf = 1e18 + 7; 28 const int maxn = 2e5 + 5; 29 const int mod = 1e9 + 7; 30 31 struct node 32 { 33 int v, w, next; 34 }edge[maxn<<1]; 35 36 struct qnode 37 { 38 int u, k; 39 ll dis; 40 qnode(int u,int k,ll dis):u(u),k(k),dis(dis){} 41 bool operator <(const qnode& a)const { 42 return dis > a.dis; 43 } 44 }; 45 46 int head[maxn], tot = 0; 47 ll dis[maxn][20]; 48 int n, m, k; 49 50 void init() 51 { 52 tot = 0; 53 memset(head, -1, sizeof head); 54 } 55 56 void add(int u, int v, int w) 57 { 58 edge[tot].v = v, edge[tot].w = w; 59 edge[tot].next = head[u]; 60 head[u] = tot++; 61 } 62 63 ll dij() 64 { 65 priority_queue<qnode>q; 66 67 for (int i = 1; i <= n; ++i) 68 for (int j = 0; j <= k; ++j) dis[i][j] = Inf; 69 70 dis[1][0] = 0; 71 q.push(qnode(1, 0, 0)); 72 while (!q.empty()) 73 { 74 int u = q.top().u, kk = q.top().k; 75 if (u == n) return dis[u][kk];//已经全部更新结束 76 q.pop(); 77 for (int i = head[u]; i != -1; i = edge[i].next) 78 { 79 int v = edge[i].v, w = edge[i].w; 80 if (dis[u][kk] + w < dis[v][kk])//保留这条边 81 { 82 dis[v][kk] = dis[u][kk] + w; 83 q.push(qnode(v, kk, dis[v][kk])); 84 } 85 if (kk != k)//去除这条边 86 { 87 if (dis[v][kk + 1] > dis[u][kk]) 88 { 89 dis[v][kk + 1] = dis[u][kk]; 90 q.push(qnode(v, kk + 1, dis[v][kk + 1])); 91 } 92 } 93 } 94 } 95 return -1; 96 } 97 98 int main() 99 { 100 int T; 101 scanf("%d", &T); 102 while (T--) 103 { 104 init(); 105 scanf("%d %d %d", &n, &m, &k); 106 while (m--) 107 { 108 int u, v, w; 109 scanf("%d %d %d", &u, &v, &w); 110 add(u, v, w); 111 } 112 printf("%d ", dij()); 113 } 114 }