我们有以下Oracle表：
CREATE TABLE graph
( company VARCHAR2(10)
, from VARCHAR2(15)
, to VARCHAR2(15)
, cost NUMBER(18,2));
使用以下数据：
INSERT INTO graph VALUES('Lufthansa', 'San Francisco', 'Denver', 1000);
INSERT INTO graph VALUES('Lufthansa', 'San Francisco', 'Dallas', 10000);
INSERT INTO graph VALUES('Lufthansa', 'Denver', 'Dallas', 500);
INSERT INTO graph VALUES('Lufthansa', 'Denver', 'Chicago', 2000);
INSERT INTO graph VALUES('Lufthansa', 'Dallas', 'Chicago', 600);
INSERT INTO graph VALUES('Lufthansa', 'Dallas', 'New York', 2000);
INSERT INTO graph VALUES('Lufthansa', 'Chicago', 'New York', 3000);
INSERT INTO graph VALUES('Lufthansa', 'Chicago', 'Denver', 2000);
我们被要求创建一个表格：
CREATE TABLE paths
( from VARCHAR2(15)
, to VARCHAR2(15)
, minimal_cost NUMBER);`
表paths中的一行应指示“我可以X Y以及C中的最低费用。
到目前为止我尝试过：
INSERT INTO paths ("from", "to")
SELECT DISTINCT CONNECT_BY_ROOT "from" AS "From", "to" AS "To"
FROM graph
START WITH "from" in (SELECT "from" FROM graph)
CONNECT BY NOCYCLE PRIOR "to" = "from";
（我已经从我的语言重命名了表名和属性，因此如果您尝试直接运行语句，可能会出现语法错误。）
以上语句几乎在所有方面都会创建重复项。另外，我不知道应该如何计算cost。我是否需要创建PL/SQL程序？ Oracle是否支持这种类型的队列？
答案 0 :(得分：4)
通过对SYS_CONNECT_BY_PATH的一点滥用以及使用正则表达式从列表中提取数字的函数，您可以这样做：
Oracle 11g R2架构设置：
CREATE TABLE graph
( company VARCHAR2(10)
, "from" VARCHAR2(15)
, "to" VARCHAR2(15)
, cost NUMBER(18,2))
/
BEGIN
INSERT INTO graph VALUES('Lufthansa', 'San Francisco', 'Denver', 1000);
INSERT INTO graph VALUES('Lufthansa', 'San Francisco', 'Dallas', 10000);
INSERT INTO graph VALUES('Lufthansa', 'Denver', 'Dallas', 500);
INSERT INTO graph VALUES('Lufthansa', 'Denver', 'Chicago', 2000);
INSERT INTO graph VALUES('Lufthansa', 'Dallas', 'Chicago', 600);
INSERT INTO graph VALUES('Lufthansa', 'Dallas', 'New York', 2000);
INSERT INTO graph VALUES('Lufthansa', 'Chicago', 'New York', 3000);
INSERT INTO graph VALUES('Lufthansa', 'Chicago', 'Denver', 2000);
END;
/
CREATE TABLE paths
( "from" VARCHAR2(15)
, "to" VARCHAR2(15)
, minimal_cost NUMBER)
/
CREATE OR REPLACE FUNCTION sum_costs (
vals VARCHAR2
) RETURN NUMBER
AS
num_vals SIMPLE_INTEGER := REGEXP_COUNT( vals, '\d+' );
total graph.cost%TYPE := 0;
BEGIN
FOR i IN 1 .. num_vals LOOP
total := total + TO_NUMBER( REGEXP_SUBSTR( vals, '\d+', 1, i ) );
END LOOP;
RETURN total;
END;
/
查询1 ：
WITH costs AS (
SELECT CONNECT_BY_ROOT "from" AS "from",
"to",
sum_costs( SYS_CONNECT_BY_PATH ( cost, ',' ) ) AS total_cost
FROM graph
WHERE CONNECT_BY_ROOT "from" <> "to"
CONNECT BY NOCYCLE PRIOR "to" = "from"
)
SELECT "from",
"to",
MIN( total_cost )
FROM costs
GROUP BY "from", "to"
<强> Results ：
| FROM | TO | MIN(TOTAL_COST) |
|---------------|----------|-----------------|
| Dallas | Chicago | 600 |
| Chicago | New York | 3000 |
| Chicago | Denver | 2000 |
| San Francisco | Denver | 1000 |
| Dallas | New York | 2000 |
| Denver | Chicago | 1100 |
| San Francisco | New York | 3500 |
| Denver | Dallas | 500 |
| Dallas | Denver | 2600 |
| Denver | New York | 2500 |
| San Francisco | Dallas | 1500 |
| San Francisco | Chicago | 2100 |
| Chicago | Dallas | 2500 |
这也为每对目的地获得了最佳路线：
查询2 ：
WITH costs AS (
SELECT CONNECT_BY_ROOT "from" AS "from",
"to",
SUBSTR( SYS_CONNECT_BY_PATH ( "from", ',' ), 2 ) || ',' || "to" AS route,
sum_costs( SYS_CONNECT_BY_PATH ( cost, ',' ) ) AS total_cost
FROM graph
WHERE CONNECT_BY_ROOT "from" <> "to"
CONNECT BY NOCYCLE PRIOR "to" = "from"
)
SELECT
"from",
"to",
MIN( route ) KEEP ( DENSE_RANK FIRST ORDER BY total_cost ) AS optimal_route,
MIN( total_cost ) AS minimum_cost
FROM costs
GROUP BY "from", "to"
<强> Results ：
| FROM | TO | OPTIMAL_ROUTE | MINIMUM_COST |
|---------------|----------|--------------------------------------|--------------|
| Dallas | Denver | Dallas,Chicago,Denver | 2600 |
| Dallas | Chicago | Dallas,Chicago | 600 |
| Dallas | New York | Dallas,New York | 2000 |
| Denver | Dallas | Denver,Dallas | 500 |
| Denver | Chicago | Denver,Dallas,Chicago | 1100 |
| Denver | New York | Denver,Dallas,New York | 2500 |
| Chicago | Dallas | Chicago,Denver,Dallas | 2500 |
| Chicago | Denver | Chicago,Denver | 2000 |
| Chicago | New York | Chicago,New York | 3000 |
| San Francisco | Dallas | San Francisco,Denver,Dallas | 1500 |
| San Francisco | Denver | San Francisco,Denver | 1000 |
| San Francisco | Chicago | San Francisco,Denver,Dallas,Chicago | 2100 |
| San Francisco | New York | San Francisco,Denver,Dallas,New York | 3500 |
纯SQL解决方案：
查询3 ：
WITH Routes AS (
SELECT CONNECT_BY_ROOT "from" AS "from",
"to",
SUBSTR( SYS_CONNECT_BY_PATH ( "from", ',' ), 2 ) || ',' || "to" AS route,
cost
FROM graph
WHERE CONNECT_BY_ROOT "from" <> "to"
CONNECT BY NOCYCLE PRIOR "to" = "from"
),
costs AS (
SELECT r."from",
r."to",
r.route,
SUM( s.cost ) AS total_cost
FROM Routes r
INNER JOIN
Routes s
ON ( r."from" = s."from"
AND LENGTH( r.route ) >= LENGTH( s.route )
AND SUBSTR( r.route, 1, LENGTH( s.route ) ) = s.route )
GROUP BY r."from", r."to", r.route
)
SELECT "from",
"to",
MIN( route ) KEEP ( DENSE_RANK FIRST ORDER BY total_cost ) AS optimal_route,
MIN( total_cost )
FROM costs
GROUP BY "from", "to"
<强> Results ：
| FROM | TO | OPTIMAL_ROUTE | MIN(TOTAL_COST) |
|---------------|----------|--------------------------------------|-----------------|
| Dallas | Denver | Dallas,Chicago,Denver | 2600 |
| Dallas | Chicago | Dallas,Chicago | 600 |
| Dallas | New York | Dallas,New York | 2000 |
| Denver | Dallas | Denver,Dallas | 500 |
| Denver | Chicago | Denver,Dallas,Chicago | 1100 |
| Denver | New York | Denver,Dallas,New York | 2500 |
| Chicago | Dallas | Chicago,Denver,Dallas | 2500 |
| Chicago | Denver | Chicago,Denver | 2000 |
| Chicago | New York | Chicago,New York | 3000 |
| San Francisco | Dallas | San Francisco,Denver,Dallas | 1500 |
| San Francisco | Denver | San Francisco,Denver | 1000 |
| San Francisco | Chicago | San Francisco,Denver,Dallas,Chicago | 2100 |
| San Francisco | New York | San Francisco,Denver,Dallas,New York | 3500 |