c编写 Battle Over Cities

Battle Over Cities - Hard Version   (20分)

It is vitally important to have all the cities connected by highways in a war. If a city is conquered by the enemy, all the highways from/toward that city will be closed. To keep the rest of the cities connected, we must repair some highways with the minimum cost. On the other hand, if losing a city will cost us too much to rebuild the connection, we must pay more attention to that city.

Given the map of cities which have all the destroyed and remaining highways marked, you are supposed to point out the city to which we must pay the most attention.

Input Specification:

Each input file contains one test case. Each case starts with a line containing 2 numbers N ($\le$500), and M, 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 whereCity1 and City2 are the numbers of the cities the highway connects (the cities are numbered from 1 to N), Cost 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.

Output Specification:

For each test case, just print in a line the city we must protest the most, that is, it will take us the maximum effort to rebuild the connection if that city is conquered by the enemy.

In case there is more than one city to be printed, output them in increasing order of the city numbers, separated by one space, but no extra space at the end of the line. In case there is no need to repair any highway at all, simply output 0.

Sample Input 1:

4 51 2 1 11 3 1 12 3 1 02 4 1 13 4 1 0

Sample Output 1:

1 2

Sample Input 2:

4 51 2 1 11 3 1 12 3 1 02 4 1 13 4 2 1

Sample Output 2:

0

/*主要考察了最小生成树的相关知识；（我采用了Kruskal算法；）本题测试点的最后两个对于大数据的测试，时间，空间要求比较严格；自己开始的几个写法都是最后两个点不能过；于是更改了多次算法；对于段错误：路径数组一定要开够；对于超时：个人认为时间主要损耗在于：1.对联通集find时若不路径压缩，可能会超时2.每次找最少的cost时，如果都遍历找最小的画，会超时于是最后更改写法为：输入cost和state时就对其经行排序（利用快排）；在findmin（）时直接取最左端的那一个（如果被占领则跳过，优先选取state=1的路径）；同时还要注意到，可能存在无法修复的情况，此时应直接赋一个最大cost给该站点；*/#include<stdio.h>#include<stdlib.h>int M,N;//记录city和highway总数int city1,city2;int count1,count2;int p1=0,p2=0;//记录state=1和state=0的highway数量struct v{int c1;int c2;int cost;} cost1[250000],cost2[250000];//分别记录state=1和state=0的路径int DisjSet[520];//联通集int costp[520]={-1};int Kruskal(int city);int find(int city);int findmin(int conquer);void output();int cmp1(const void *a,const void *b)  {      return ((struct v*)a)->cost-((struct v*)b)->cost;  } //快排比较函数int main(void){int i=0,j=0;int c1,c2;int cost=0,state=0;scanf("%d%d",&M,&N);for(i=0;i<N;i++){scanf("%d%d%d%d",&c1,&c2,&cost,&state);if(state==1){cost1[p1].c1=c1;cost1[p1].c2=c2;cost1[p1].cost=cost;p1++;}//如果state=1，记录于cost1else{cost2[p2].c1=c1;cost2[p2].c2=c2;cost2[p2].cost=cost;p2++;}//如果state=0,记录于cost2}qsort(cost1,p1,sizeof(struct v),cmp1);qsort(cost2,p2,sizeof(struct v),cmp1);//对cost1和cost2经行快排p1--;p2--;for(i=1;i<=M;i++){count1=count2=0;city1=city2=0;for(j=1;j<=M;j++)DisjSet[j]=0;costp[i]=Kruskal(i);//遍历所有city,找到每个city的cost}output();//输出函数return 0;}int Kruskal(int city){int V,W;int costs=0;int edge=0;int min;for(;;){if(edge==M-2)break;//如果联通，break；city1=city2=0;min=findmin(city);//找到最小cost的路径if(min==0x7FFFFFFF){return min;}//无法修复的情况V=find(city1);W=find(city2);//找到两个点的根节点（判断两个city是否已联通if(V!=W){edge++;if(min>0){   costs=costs+min;//如果需要修复，更改costs}DisjSet[W]=V;//联通}}return costs;}int findmin(int conquer){int i=0;int min=0x7FFFFFFF;//优先考虑state=1的highway；while(cost1[count1].c1==conquer||cost1[count1].c2==conquer){count1++;}//如果被攻占，不考虑，剔除；if(count1<=p1){city1=cost1[count1].c1;city2=cost1[count1].c2;min=cost1[count1].cost;count1++;//将已选择的点剔除return -1*min;}else{//若为联通，再考虑state=0的highwaywhile(cost2[count2].c1==conquer||cost2[count2].c2==conquer){count2++;}//若被攻占，不考虑，剔除if(count2<=p2){city1=cost2[count2].c1;city2=cost2[count2].c2;min=cost2[count2].cost;count2++;//将已选择的点剔除return min;}}//若还是未联通泽表明无法修复，返回一个大值return min;}int find(int city){if(DisjSet[city]<=0)return city;//找到根节点返回elsereturn DisjSet[city]=find(DisjSet[city]);//路径压缩}void output(){int max=0;int i=0;int flag=1;for(i=1;i<=M;i++){if(costp[i]>max)max=costp[i];}//找到最大costif(max==0)printf("0");//若不需修复，输出0elsefor(i=1;i<=M;i++){if(costp[i]==max){if(flag){printf("%d",i);flag=0;}elseprintf(" %d",i);}}}