HDU-1385-Minimum Transport Cost

描述

题解

代码

#include <iostream>
#include <cstring>

using namespace std;

/*
*  Floyd算法，求从任意节点i到任意节点j的最短路径
*  cost[][]:初始化为INF（cost[i][i]：初始化为0）
*/
const int MAXN = 110;
const int INF = 0x1f1f1f1f;

int val[MAXN];
int cost[MAXN][MAXN];
int lowcost[MAXN][MAXN];
int path[MAXN][MAXN];

void Floyd(int n)
{
memcpy(lowcost, cost, sizeof(cost));
for (int i = 0; i <= n; i++)
{
for (int j = 0; j <= n; j++)
{
path[i][j] = j;
}
}

for (int k = 1; k <= n; k++)
{
for (int i = 1; i <= n; i++)
{
for (int j = 1; j <= n; j++)
{
int temp = lowcost[i][k] + lowcost[k][j] + val[k];
if (lowcost[i][j] > temp)
{
lowcost[i][j] = temp;
path[i][j] = path[i][k];
}
else if (lowcost[i][j] == temp && path[i][j] > path[i][k])
{
path[i][j] = path[i][k];
}
}
}
}
return ;
}

int main(int argc, const char * argv[])
{
int N;
while (cin >> N, N != 0)
{
for (int i = 1; i <= N; i++)
{
for (int j = 1; j <= N; j++)
{
scanf("%d", cost[i] + j);
if (cost[i][j] == -1)
{
cost[i][j] = INF;
}
}
}

for (int i = 1; i <= N; i++)
{
scanf("%d", val + i);
}

Floyd(N);

int st, ed;
while (cin >> st >> ed)
{
if (st + ed == -2)
{
break;
}
printf("From %d to %d :\nPath: ", st, ed);
int u = st;
printf("%d", st);
while (u != ed)
{
printf("-->%d", path[u][ed]);
u = path[u][ed];
}
printf("\nTotal cost : %d\n\n", lowcost[st][ed]);
}
}

return 0;
}

参考