本文共 1010 字,大约阅读时间需要 3 分钟。
,注意本题数据来到了10的9次方,保证两数相加不溢出,INF最好为1e20左右。
#include#include #include #include using namespace std;#define MAX 105#define INF (long long)1e20int n, m;long long d[MAX][MAX];void init(){ for (int i = 0; i < n; i++) for (int j = 0; j < n; j++) d[i][j] = (i == j ? 0 : INF);}void floyd(){ for (int k = 0; k < n; k++) for (int i = 0; i < n; i++) for (int j = 0; j < n; j++) if (d[i][k] < INF&&d[k][j] d[i][k] + d[k][j]) d[i][j] = d[i][k] + d[k][j]; }int main(){ int T; cin >> T; for (int num = 1; num <= T; num++) { long long L[4], C[4]; cin >> L[0] >> L[1] >> L[2] >> L[3] >> C[0] >> C[1] >> C[2] >> C[3]; cin >> n >> m; init(); int x[MAX]; for (int i = 0; i < n; i++) cin >> x[i]; for(int i=0;i > u >> v; u--; v--; if (d[u][v] != INF) printf("The minimum cost between station %d and station %d is %I64d.\n", u + 1, v + 1, d[u][v]); else printf("Station %d and station %d are not attainable.\n", u + 1, v + 1); } } system("pause");}
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