首页 > ACM题库 > HDU-杭电 > hdu 3710 Battle over Cities待解决[解题报告]C++
2015
02-21

hdu 3710 Battle over Cities待解决[解题报告]C++

Battle over Cities

问题描述 :

It is vitally important to have all the cities connected by highways in a war, but some of them are destroyed now because of the war. Furthermore,if a city is conquered, all the highways from/toward that city will be closed by the enemy, and we must repair some destroyed highways to keep other cities connected, with the minimum cost if possible.
Given the map of cities which have all the destroyed and remaining highways marked, you are supposed to tell the cost to connect other cities if each city is conquered by the enemy.

输入:

The input contains multiple test cases. The first line is the total number of cases T (T ≤ 10). Each case starts with a line containing 2 numbers N (0 < N ≤ 20000), and M (0 ≤ M ≤ 100000), which are the total number of cities, and the number of highways, respectively. Then M lines follow, each describes a highway by 4 integers: City1 City2 Cost Status where City1 and City2 are the numbers of the cities the highway connects (the cities are numbered from 1 to N), Cost (0 < Cost ≤ 20000) is the effort taken to repair that highway if necessary, and Status is either 0, meaning that highway is destroyed, or 1, meaning that highway is in use.
Note: It is guaranteed that the whole country was connected before the war and there is no duplicated high ways between any two cities.

输出:

The input contains multiple test cases. The first line is the total number of cases T (T ≤ 10). Each case starts with a line containing 2 numbers N (0 < N ≤ 20000), and M (0 ≤ M ≤ 100000), which are the total number of cities, and the number of highways, respectively. Then M lines follow, each describes a highway by 4 integers: City1 City2 Cost Status where City1 and City2 are the numbers of the cities the highway connects (the cities are numbered from 1 to N), Cost (0 < Cost ≤ 20000) is the effort taken to repair that highway if necessary, and Status is either 0, meaning that highway is destroyed, or 1, meaning that highway is in use.
Note: It is guaranteed that the whole country was connected before the war and there is no duplicated high ways between any two cities.

样例输入:

3
4 5
1 2 1 1
1 3 1 1
2 3 1 0
2 4 1 1
3 4 2 0
4 5
1 2 1 1
1 3 1 1
2 3 1 0
2 4 1 1
3 4 1 0
3 2
1 2 1 1
1 3 1 1

样例输出:

1
2
0
0
1
1
0
0
inf
0
0


  1. 约瑟夫也用说这么长……很成熟的一个问题了,分治的方法解起来o(n)就可以了,有兴趣可以看看具体数学的第一章,关于约瑟夫问题推导出了一系列的结论,很漂亮

  2. 第二个方法挺不错。NewHead代表新的头节点,通过递归找到最后一个节点之后,就把这个节点赋给NewHead,然后一直返回返回,中途这个值是没有变化的,一边返回一边把相应的指针方向颠倒,最后结束时返回新的头节点到主函数。

  3. 代码不对,仔细对比一下输入输出, 怎么会有‘‘printf("The winning moves are:");’’这种输出呢?

  4. “再把所有不和该节点相邻的节点着相同的颜色”,程序中没有进行不和该节点相邻的其他节点是否相邻进行判断。再说求出来的也不一样是颜色数最少的