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| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 13738 | Accepted: 6195 |
Description
One cow from each of N farms (1 ≤ N ≤ 1000) conveniently numbered 1..N is going to attend the big cow party to be held at farm #X (1 ≤ X ≤ N). A total of M (1 ≤ M ≤ 100,000) unidirectional (one-way roads connects pairs of farms; road i requires Ti (1 ≤ Ti ≤ 100) units of time to traverse.
Each cow must walk to the party and, when the party is over, return to her farm. Each cow is lazy and thus picks an optimal route with the shortest time. A cow‘s return route might be different from her original route to the party since roads are one-way.
Of all the cows, what is the longest amount of time a cow must spend walking to the party and back?
Input
Output
Sample Input
4 8 2 1 2 4 1 3 2 1 4 7 2 1 1 2 3 5 3 1 2 3 4 4 4 2 3
Sample Output
10
Hint
/** 题意：最短路中哪个走的路程最大 解法：SPFA **/ #include<iostream> #include<string.h> #include<stdio.h> #include<algorithm> #include<cmath> #include<queue> #include<vector> #define maxn 1000 + 10 #define INF 0x3f3f3f3f using namespace std; int vis[maxn]; int dist[maxn]; int dist1[maxn]; int flag[maxn]; struct Node { int v; int cost; Node(int _v,int _cost):v(_v),cost(_cost) {} }; vector<Node>edge[maxn]; vector<Node>edge1[maxn]; int n,m,p; void addedge(int u,int v,int w) { edge[u].push_back(Node(v,w)); edge1[v].push_back(Node(u,w)); } void SPFA(int start) { memset(vis,0,sizeof(vis)); memset(dist,INF,sizeof(dist)); queue<int>que; que.push(start); vis[start] = 1; dist[start] = 0; while(!que.empty()) { int tt = que.front(); que.pop(); vis[tt] = 0; for(int i=0; i<edge[tt].size(); i++) { int mm = edge[tt][i].v; if(dist[mm] > dist[tt] + edge[tt][i].cost) { dist[mm] = dist[tt] + edge[tt][i].cost; if(!vis[mm]) { vis[mm] = 1; que.push(mm); } } } } memset(vis,0,sizeof(vis)); memset(dist1,INF,sizeof(dist1)); while(!que.empty()) que.pop(); vis[start] = 1; dist1[start] = 0; que.push(start); while(!que.empty()) { int tt = que.front(); que.pop(); vis[tt] = 0; for(int i=0; i<edge1[tt].size(); i++) { int mm = edge1[tt][i].v; if(dist1[mm] > dist1[tt] + edge1[tt][i].cost) { dist1[mm] = dist1[tt] + edge1[tt][i].cost; if(!vis[mm]) { vis[mm] = 1; que.push(mm); } } } } return ; } int main() { //#ifndef ONLINE_JUDGE // freopen("in.txt","r",stdin); //#endif // ONLINE_JUDGE scanf("%d %d %d",&n,&m,&p); int u,v,w; for(int i=0; i<m; i++) { scanf("%d %d %d",&u,&v,&w); addedge(u,v,w); } SPFA(p); int mmax = -INF; for(int i=1; i<=n; i++) { if(dist[i] + dist1[i] > mmax && dist[i] != INF && dist1[i]!= INF ) mmax = dist[i] + dist1[i]; } printf("%d\n",mmax); return 0; }
/** 题意：最短路中哪个走的路程最大 解法：SPFA **/ #include<iostream> #include<string.h> #include<stdio.h> #include<algorithm> #include<cmath> #include<queue> #include<vector> #define maxn 1000 + 10 #define INF 0x3f3f3f3f using namespace std; int vis[maxn]; int dist[maxn]; int dist1[maxn]; int flag[maxn]; struct Node { int v; int cost; Node(int _v,int _cost):v(_v),cost(_cost) {} }; vector<Node>edge[maxn]; vector<Node>edge1[maxn]; int n,m,p; void addedge(int u,int v,int w) { edge[u].push_back(Node(v,w)); edge1[v].push_back(Node(u,w)); } void SPFA(int start) { memset(vis,0,sizeof(vis)); memset(dist,INF,sizeof(dist)); queue<int>que; que.push(start); vis[start] = 1; dist[start] = 0; while(!que.empty()) { int tt = que.front(); que.pop(); vis[tt] = 0; for(int i=0; i<edge[tt].size(); i++) { int mm = edge[tt][i].v; if(dist[mm] > dist[tt] + edge[tt][i].cost) { dist[mm] = dist[tt] + edge[tt][i].cost; if(!vis[mm]) { vis[mm] = 1; que.push(mm); } } } } memset(vis,0,sizeof(vis)); memset(dist1,INF,sizeof(dist1)); while(!que.empty()) que.pop(); vis[start] = 1; dist1[start] = 0; que.push(start); while(!que.empty()) { int tt = que.front(); que.pop(); vis[tt] = 0; for(int i=0; i<edge1[tt].size(); i++) { int mm = edge1[tt][i].v; if(dist1[mm] > dist1[tt] + edge1[tt][i].cost) { dist1[mm] = dist1[tt] + edge1[tt][i].cost; if(!vis[mm]) { vis[mm] = 1; que.push(mm); } } } } return ; } int main() { //#ifndef ONLINE_JUDGE // freopen("in.txt","r",stdin); //#endif // ONLINE_JUDGE scanf("%d %d %d",&n,&m,&p); int u,v,w; for(int i=0; i<m; i++) { scanf("%d %d %d",&u,&v,&w); addedge(u,v,w); } SPFA(p); int mmax = -INF; for(int i=1; i<=n; i++) { if(dist[i] + dist1[i] > mmax && dist[i] != INF && dist1[i]!= INF ) mmax = dist[i] + dist1[i]; } printf("%d\n",mmax); return 0; }
/** 题意：最短路中哪个走的路程最大 解法：SPFA **/ #include<iostream> #include<string.h> #include<stdio.h> #include<algorithm> #include<cmath> #include<queue> #include<vector> #define maxn 1000 + 10 #define INF 0x3f3f3f3f using namespace std; int vis[maxn]; int dist[maxn]; int dist1[maxn]; int flag[maxn]; struct Node { int v; int cost; Node(int _v,int _cost):v(_v),cost(_cost) {} }; vector<Node>edge[maxn]; vector<Node>edge1[maxn]; int n,m,p; void addedge(int u,int v,int w) { edge[u].push_back(Node(v,w)); edge1[v].push_back(Node(u,w)); } void SPFA(int start) { memset(vis,0,sizeof(vis)); memset(dist,INF,sizeof(dist)); queue<int>que; que.push(start); vis[start] = 1; dist[start] = 0; while(!que.empty()) { int tt = que.front(); que.pop(); vis[tt] = 0; for(int i=0; i<edge[tt].size(); i++) { int mm = edge[tt][i].v; if(dist[mm] > dist[tt] + edge[tt][i].cost) { dist[mm] = dist[tt] + edge[tt][i].cost; if(!vis[mm]) { vis[mm] = 1; que.push(mm); } } } } memset(vis,0,sizeof(vis)); memset(dist1,INF,sizeof(dist1)); while(!que.empty()) que.pop(); vis[start] = 1; dist1[start] = 0; que.push(start); while(!que.empty()) { int tt = que.front(); que.pop(); vis[tt] = 0; for(int i=0; i<edge1[tt].size(); i++) { int mm = edge1[tt][i].v; if(dist1[mm] > dist1[tt] + edge1[tt][i].cost) { dist1[mm] = dist1[tt] + edge1[tt][i].cost; if(!vis[mm]) { vis[mm] = 1; que.push(mm); } } } } return ; } int main() { //#ifndef ONLINE_JUDGE // freopen("in.txt","r",stdin); //#endif // ONLINE_JUDGE scanf("%d %d %d",&n,&m,&p); int u,v,w; for(int i=0; i<m; i++) { scanf("%d %d %d",&u,&v,&w); addedge(u,v,w); } SPFA(p); int mmax = -INF; for(int i=1; i<=n; i++) { if(dist[i] + dist1[i] > mmax && dist[i] != INF && dist1[i]!= INF ) mmax = dist[i] + dist1[i]; } printf("%d\n",mmax); return 0; }
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原文地址：http://www.cnblogs.com/chenyang920/p/4411693.html
