思路:题目要求所有点到1号点最小距离不变,所以我们采用单源最短路。考虑spfa,对于每一个点来说都是被更新,我们只要记住在这些更新该点的边里
dist最短且cost尽量小的,而且这张图到最后一定是一棵树,我们只需要记录下每个cost[i],最后整体相加就行了。
详见代码:
#include<bits/stdc++.h> using namespace std; const int maxn = 10005; const int F = 0x3f; const int INF = 0x3f3f3f3f; struct edge{ int to,dist,cost; edge(int a = 0,int b = 0,int c = 0){ to = a,dist = b,cost = c; } }; vector<edge> graph[maxn]; int dist[maxn],len[maxn],N,M; bool inque[maxn]; int spfa(int s){ memset(dist,F,sizeof(dist)); memset(len,F,sizeof(len)); memset(inque,false,sizeof(inque)); dist[s] = len[s] = 0; queue<int> que; que.push(s); inque[s] = true; while(que.size()){ int temp = que.front(); que.pop(); inque[temp] = false; for(int i = 0;i < graph[temp].size();++i){ edge& e = graph[temp][i]; int to = e.to,d = e.dist,c = e.cost; if(dist[to] > dist[temp] + d || (dist[to] == dist[temp] + d && len[to] > c)){ dist[to] = dist[temp] + d; len[to] = c; if(!inque[to]){ inque[to] = true; que.push(to); } } } } int ret = 0; for(int i = 1;i <= N;++i){ ret += len[i]; } return ret; } int main(){ while(scanf("%d%d",&N,&M) == 2 && (N + M)){ for(int i = 1;i <= N;++i) graph[i].clear(); int u,v,d,c; for(int i = 0;i < M;++i){ scanf("%d%d%d%d",&u,&v,&d,&c); graph[u].push_back(edge(v,d,c)); graph[v].push_back(edge(u,d,c)); } int ans = spfa(1); printf("%d ",ans); } return 0; }