Farm Tour
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 13509 | Accepted: 5125 |
Description
When FJ's friends visit him on the farm, he likes to show them around. His farm comprises N (1 <= N <= 1000) fields numbered 1..N, the first of which contains his house and the Nth of which contains the big barn. A total M (1 <= M <= 10000) paths that connect the fields in various ways. Each path connects two different fields and has a nonzero length smaller than 35,000. To show off his farm in the best way, he walks a tour that starts at his house, potentially travels through some fields, and ends at the barn. Later, he returns (potentially through some fields) back to his house again. He wants his tour to be as short as possible, however he doesn't want to walk on any given path more than once. Calculate the shortest tour possible. FJ is sure that some tour exists for any given farm.
Input
* Line 1: Two space-separated integers: N and M. * Lines 2..M+1: Three space-separated integers that define a path: The starting field, the end field, and the path's length.
Output
A single line containing the length of the shortest tour.
Sample Input
4 51 2 12 3 13 4 11 3 22 4 2
Sample Output
6
Source
本体题意为有n个农场，某人·1要从一个厂子走到另一个厂子，问最短的距离是多少
可以建点，建一个超级远点，加盟一个超级会点，分别设为0.和n+1l设超级原点到1点的流量为2，n点到超级会点的流量为2，
其余个边的流量都设为1，可以从超级原点到超级会点跑最小费用最大流，各个变的费即为该变所在的；路径长度，
我用的框斌的模板，一定涛注意在村边的时候addedge的时候需要正反两条边都要存进去
别的地方不需要动
还有返回的flow是流量，并不是最小费用，最小费用的是传参数传过去的cost的值，所以在输出的时候应该输出cost的值
下面附上本人搓搓的代码
#include#include #include #include #include #include using namespace std;//最小费用最大流，求最大费用只需要取相反数，结果取相反数即可。//点的总数为 N，点的编号 0~N-1const int MAXN = 1005;const int MAXM = 10005;const int INF = 0x3f3f3f3f;struct Edge{ int to,next,cap,flow,cost;} edge[MAXM*4];int head[MAXN],tol;int pre[MAXN],dis[MAXN];bool vis[MAXN];int N;//节点总个数，节点编号从0~N-1void init(int n){ N = n; tol = 0; memset(head,-1,sizeof (head));}void addedge (int u,int v,int cap,int cost){ edge[tol].to = v; edge[tol].cap = cap; edge[tol].cost = cost; edge[tol].flow = 0; edge[tol].next = head[u]; head[u] = tol++; edge[tol].to = u; edge[tol].cap = 0; edge[tol].cost = -cost; edge[tol].flow = 0; edge[tol].next = head[v]; head[v] = tol++;}bool spfa(int s,int t){ queue q; for(int i = 0; i < N; i++) { dis[i] = INF; vis[i] = false; pre[i] = -1; } dis[s] = 0; vis[s] = true; q.push(s); while(!q.empty()) { int u = q.front(); q.pop(); vis[u] = false; for(int i = head[u]; i != -1; i = edge[i]. next) { int v = edge[i]. to; if(edge[i].cap > edge[i].flow && dis[v] > dis[u] + edge[i]. cost ) { dis[v] = dis[u] + edge[i]. cost; pre[v] = i; if(!vis[v]) { vis[v] = true; q.push(v); } } } } if(pre[t] == -1)return false; else return true;}//返回的是最大流，cost存的是最小费用int minCostMaxflow(int s,int t,int &cost){ int flow = 0; cost = 0; while(spfa(s,t)) { int Min = INF; for(int i = pre[t]; i != -1; i = pre[edge[i^1].to]) { if(Min > edge[i].cap - edge[i]. flow) Min = edge[i].cap - edge[i].flow; } for(int i = pre[t]; i != -1; i = pre[edge[i^1].to]) { edge[i].flow += Min; edge[i^1].flow -= Min; cost += edge[i]. cost * Min; } flow += Min; } return flow;}int main(){ int n,m,sta; while(scanf("%d%d",&n,&m)!=EOF){ memset(pre,0,sizeof(pre)); memset(dis,0,sizeof(dis)); memset(vis,false,sizeof(vis)); memset(edge,0,sizeof(edge)); init(n+2); int u,v,w; for(int i=0;i