poj2135--Farm Tour

Farm Tour

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 11495 Accepted: 4257

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

 

从1点走到n点，再走回1点，要求每条路，不会被走两次，因为路是无向的，所以要加两次边，将回转的转化为，从1到n走两次，求最小的花费值

#include <cstdio>#include <cstring>#include <queue>#include <algorithm>using namespace std;#define maxn 11000#define INF 0x3f3f3f3f#define LL __int64struct node{    LL v , w , s ;    LL next ;} p[maxn*30];LL head[maxn] , cnt , vis[maxn] , pre[maxn] , dis[maxn] ;queue <LL> q ;void add(LL u,LL v,LL w,LL s){    p[cnt].v = v ;    p[cnt].w = w ;    p[cnt].s = s ;    p[cnt].next = head[u] ;    head[u] = cnt++ ;    p[cnt].v = u ;    p[cnt].w = 0 ;    p[cnt].s = -s ;    p[cnt].next = head[v] ;    head[v] = cnt++ ;}void spfa(LL s,LL t){    LL u , v , i ;    memset(pre,-1,sizeof(pre));    memset(dis,INF,sizeof(dis));    memset(vis,0,sizeof(vis));    while( !q.empty() )        q.pop();    q.push(s) ;    vis[s] = 1 ;    dis[s] = 0 ;    while( !q.empty() )    {        u = q.front();        q.pop();        vis[u] = 0 ;        for(i = head[u] ; i != -1 ; i = p[i].next)        {            v = p[i].v ;            if( p[i].w && dis[v] > dis[u] + p[i].s )            {                dis[v] = dis[u] + p[i].s;                pre[v] = i ;                if( !vis[v] )                {                    vis[v] = 1 ;                    q.push(v) ;                }            }        }    }}void f(LL s,LL t){    LL i , j , m = 2 , ans = 0 , min1 ;    while(m--)    {        spfa(s,t) ;        min1 = INF ;        for(i = pre[t] ; i != -1 ; i = pre[ p[i^1].v ])            if( p[i].w < min1 )                min1 = p[i].w ;        for(i = pre[t] ; i != -1 ; i = pre[ p[i^1].v ])        {            p[i].w -= min1 ;            p[i^1].w += min1 ;            ans += p[i].s ;        }    }    printf("%I64d\n", ans);}int main(){    LL n , m , u , v , s ;    while(scanf("%I64d %I64d", &n, &m)!=EOF)    {        /*从1点走到n点再走回来，因为线是无向的，所以要建两次，同一条边只能走一次，把从1点走到n点再走回来，分解成从1到n走两次*/        memset(head,-1,sizeof(head));        cnt = 0 ;        while(m--)        {            scanf("%I64d %I64d %I64d", &u, &v, &s);            add(u,v,1,s);            add(v,u,1,s);        }        f(1,n);//费用流的模板。。。    }    return 0;}