UVa 567 - Risk
传送门
题意
求点到点的最短路
思路
Floyd模板题
代码
#include <bits/stdc++.h>
using namespace std;
#define LL long long
#pragma comment(linker, "/STACK:102400000,102400000")
const int MAXN = 20 + 1;
const int INF = 0x3f3f3f3f;
int dis[MAXN][MAXN];
void Floyd()
{
for (int k = 1; k < MAXN; k++)
for (int i = 1; i < MAXN; i++)
for (int j = 1; j < MAXN; j++)
dis[i][j] = min(dis[i][k] + dis[k][j], dis[i][j]);
}
int main()
{
//freopen("input.txt", "r", stdin);
int n, i, j, a, cases = 0, st, ed;
while (~scanf("%d", &n))
{
//read data
for (i = 1; i < MAXN; i++)
for (j = 1; j < MAXN; j++)
dis[i][j] = (i == j ? 0 : INF);
for (i = 0; i < n; i++)
{
scanf("%d", &a);
dis[1][a] = dis[a][1] = 1;
}
for (i = 2; i < 20; i++)
{
scanf("%d", &n);
for (j = 0; j < n; j++)
{
scanf("%d", &a);
dis[i][a] = dis[a][i] = 1;
}
}
Floyd();
scanf("%d", &n);
printf("Test Set #%d\n", ++cases);
for (i = 0; i < n; i++)
{
scanf("%d%d", &st, &ed);
printf( "%2d to %2d: %d\n", st, ed, dis[st][ed]);
}
printf("\n");
}
return 0;
}