题目链接：http://codeforces.com/contest/1076/problem/D
题意：给一个n个点,m条边的无向图。要求保留最多k条边，使得其他点到1点的最短路剩余最多。
思路：当找到单源最短路后，将其转换为一个所有点到点1都是最短路的树状结构，利用贪心确定所要保留的K条边（找离根最近的边，利用BFS）。
代码：
1 #include <iostream> 2 #include <cstring> 3 #include <algorithm> 4 #include <cmath> 5 #include <cstdio> 6 #include <vector> 7 #include <queue> 8 #include <set> 9 #include <map> 10 #include <stack> 11 #define ll long long 12 //#define local 13 14 using namespace std; 15 16 const int MOD = 1e9+7; 17 const int inf = 0x3f3f3f3f; 18 const double PI = acos(-1.0); 19 const int maxn = 3e5 + 10; 20 const int maxedge = 3e5 + 10; 21 22 struct qnode { 23 int v; 24 ll c; 25 qnode(int v=0,ll c=0) : v(v), c(c) {} 26 bool operator < (const qnode &r)const {//priority_queue 默认从大到小排列 27 return c > r.c; 28 } 29 }; 30 31 struct Edge { 32 int v, w, pre; 33 } edge[maxedge*2]; 34 35 int point[maxn]; //point[i] = -1 36 int cnt; 37 bool vis[maxn]; // already init in the dijkstra 38 ll dist[maxn]; // already init in the dijkstra, overflow! 39 vector <int> v1[maxn]; 40 vector <int> v2[maxn]; 41 struct Path { 42 int u, e; 43 }path[maxn]; 44 45 void AddEdge(int u, int v, int w) { 46 edge[cnt].v = v; 47 edge[cnt].w = w; 48 edge[cnt].pre = point[u]; 49 point[u] = cnt++; 50 } 51 52 void Dijkstra(int n,int start) { 53 memset(vis,false,sizeof(vis)); 54 memset(dist, 0x3f, sizeof(dist)); 55 priority_queue <qnode> que; 56 while(!que.empty()) que.pop(); 57 dist[start] = 0; 58 que.push(qnode(start, 0)); 59 qnode tmp; 60 while(!que.empty()) { 61 tmp = que.top(); que.pop(); 62 int u = tmp.v; 63 if(vis[u]) continue; 64 vis[u] = true; 65 for(int i = point[u]; i != -1; i = edge[i].pre) { 66 int v = edge[i].v; 67 int w = edge[i].w; 68 if(!vis[v] && dist[v]>dist[u]+w) { 69 dist[v] = dist[u]+w; 70 que.push(qnode(v, dist[v])); 71 path[v].u = u; 72 path[v].e = i; 73 } 74 } 75 } 76 } 77 78 void init() { 79 cnt = 0; 80 memset(point, -1, sizeof(point)); 81 } 82 83 int main() { 84 #ifdef local 85 if(freopen("/Users/Andrew/Desktop/data.txt", "r", stdin) == NULL) printf("can't open this file! "); 86 #endif 87 88 int n, m, k; 89 scanf("%d%d%d", &n, &m, &k); 90 init(); 91 for (int i = 0; i < m; ++i) { 92 int u, v, w; 93 scanf("%d%d%d", &u, &v, &w); 94 AddEdge(u, v, w); 95 AddEdge(v, u, w); 96 } 97 Dijkstra(n, 1); 98 for (int i = 1; i <= n; ++i) { 99 v1[path[i].u].push_back(i); 100 v2[path[i].u].push_back(path[i].e); 101 } 102 queue<int> q; 103 q.push(1); 104 int ans = 0; 105 int Ans[maxn]; 106 while (q.size()) { 107 int tmp = q.front(); 108 q.pop(); 109 for (int j = 0; j < v1[tmp].size() && ans < k; ++j) { 110 q.push(v1[tmp][j]); 111 Ans[ans++] = v2[tmp][j]/2; 112 } 113 } 114 printf("%d ", ans); 115 for (int i = 0; i < ans; ++i) { 116 if (i) 117 printf(" %d", Ans[i]+1); 118 else printf("%d", Ans[i]+1); 119 } 120 printf(" "); 121 #ifdef local 122 fclose(stdin); 123 #endif 124 return 0; 125 }