題意：
????? 就是從1走到n然后再走回來，一條邊只能走一次，要求路徑最短。

思路：
????? 比較水，可以直接一遍費用流，不解釋了，具體的看看代碼，敲這個題就是為了練

練手，好久不敲了，怕比賽手生。

?

#include<queue>
#include<stdio.h>
#include<string.h>

#define N_node 1000 + 10
#define N_edge 40000 + 20
#define INF 100000000

using namespace std;

typedef struct
{
?? int from ,to ,cost ,flow ,next;
}STAR;

STAR E[N_edge];
int list[N_node] ,tot;
int mer[N_edge];
int s_x[N_node];

void add(int a ,int b ,int c ,int d)
{
???? E[++tot].from = a;
???? E[tot].to = b;
???? E[tot].cost = c;
???? E[tot].flow = d;
???? E[tot].next = list[a];
???? list[a] = tot;
????
???? E[++tot].from = b;
???? E[tot].to = a;
???? E[tot].cost = -c;
???? E[tot].flow = 0;
???? E[tot].next = list[b];
???? list[b] = tot;
}

bool Spfa(int s ,int t ,int n)
{
??? int mark[N_node] = {0};
??? for(int i = 0 ;i <= n ;i ++) s_x[i] = INF;
??? mark[s] = 1 ,s_x[s] = 0;
??? queue<int>q;
??? q.push(s);
??? memset(mer ,255 ,sizeof(mer));
??? while(!q.empty())
??? {
??????? int xin ,tou = q.front();
??????? q.pop();
??????? mark[tou] = 0;
??????? for(int k = list[tou] ;k ;k = E[k].next)
??????? {
??????????? xin = E[k].to;
??????????? if(s_x[xin] > s_x[tou] + E[k].cost && E[k].flow)
??????????? {
??????????????? s_x[xin] = s_x[tou] + E[k].cost;
??????????????? mer[xin] = k;
??????????????? if(!mark[xin])
??????????????? {
??????????????????? mark[xin] = 1;
??????????????????? q.push(xin);
??????????????? }
??????????? }
??????? }
??? }
??? return mer[t] != -1;
}

int M_C_Flow(int s ,int t ,int n)
{
?? int maxflow = 0 ,mincost = 0 ,minflow;
?? while(Spfa(s ,t ,n))
?? {
?????? minflow = INF;
?????? for(int i = mer[t] ;i + 1 ;i = mer[E[i].from])
?????? if(minflow > E[i].flow) minflow = E[i].flow;
?????? for(int i = mer[t] ;i + 1 ;i = mer[E[i].from])
?????? {
?????????? E[i].flow -= minflow;
?????????? E[i^1].flow += minflow;
?????????? mincost += minflow * E[i].cost;
?????? }
?????? maxflow += minflow;
?? }
?? return mincost;
}

int main ()
{
??? int n ,m ,i ,a ,b ,c;
??? while(~scanf("%d %d" ,&n ,&m))
??? {
?????? memset(list ,0 ,sizeof(list)) ,tot = 1;
?????? for(i = 1 ;i <= m ;i ++)
?????? {
????????? scanf("%d %d %d" ,&a ,&b ,&c);
????????? add(a ,b ,c ,1);
????????? add(b ,a ,c ,1);
?????? }
?????? add(0 ,1 ,0 ,2);
?????? add(n ,n + 1 ,0 ,2);
?????? printf("%d\n" ,M_C_Flow(0 ,n + 1 ,n + 1));
??? }
??? return 0;
}
?????????

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