Constructing Roads

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 17353    Accepted Submission(s): 6600

Problem Description There are N villages, which are numbered from 1 to N, and you should build some roads such that every two villages can connect to each other. We say two village A and B are connected, if and only if there is a road between A and B, or there exists a village C such that there is a road between A and C, and C and B are connected.

We know that there are already some roads between some villages and your job is the build some roads such that all the villages are connect and the length of all the roads built is minimum.

Input The first line is an integer N (3 <= N <= 100), which is the number of villages. Then come N lines, the i-th of which contains N integers, and the j-th of these N integers is the distance (the distance should be an integer within [1, 1000]) between village i and village j.

Then there is an integer Q (0 <= Q <= N * (N + 1) / 2). Then come Q lines, each line contains two integers a and b (1 <= a < b <= N), which means the road between village a and village b has been built.

Output You should output a line contains an integer, which is the length of all the roads to be built such that all the villages are connected, and this value is minimum.

Sample Input 3 0 990 692 990 0 179 692 179 0 1 1 2
Sample Output 179

最短路模板題

普利姆演算法直接的模板,克魯斯卡爾也可寫,附上兩種程式碼

prim:

#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
#define INF 0x3f3f3f3f
#define N 110
int dis[N][N];
int n;

void prim(){
	int min,distance[N]; 
	int vis[N],k;
	memset(vis,0,sizeof(vis));
	for(int i=1;i<=N;i++)
		distance[i]=dis[1][i];
	distance[1]=0;
	vis[1]=1;
	for(int v=1;v<n;v++){
		min=INF;
		k=1;
		for(int i=1;i<=n;i++){
			if(!vis[i]&&distance[i]<min){
				min=distance[i];
				k=i;
			}
		}
		vis[k]=1;
		for(int i=1;i<=n;i++){
			if(!vis[i]&&distance[i]>dis[k][i]) 
				distance[i]=dis[k][i];
		}
	}
	int sum=0;
	for(int i=2;i<=n;i++){//
		sum+=distance[i];
//		printf("%d ",distance[i]);
	}
	printf("%d\n",sum);
	return;
}

int main(){
	while(scanf("%d",&n)!=EOF){
		for(int i=1;i<=n;i++){
			for(int j=1;j<=n;j++){
				scanf("%d",&dis[i][j]);
			}
		}

		int k,a,b;
		scanf("%d",&k);
		while(k--){
			scanf("%d%d",&a,&b);
			dis[a][b]=dis[b][a]=0;
		}
//		for(int i=1;i<=n;i++){
//			for(int j=1;j<=n;j++){
//				printf("%d ",dis[i][j]);
//			}
//			putchar('\n');
//		}
		prim();
	}
	return 0;
}
 

寫克魯斯卡爾的程式碼時遇見了最坑爹的問題,在杭電上C++AC而G++WA;後將呼叫<algorithm>的sort刪去,改為qsort就神奇的AC了,原因至今不明:

問題程式碼:

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int per[110000],N,M;
struct str{
	int a,b,x;
}edge[1001000];

int p[1100][1100];

int cmp(str a,str b){
	return a.x<b.x;
}

void init(){
	for(int i=1;i<=110000;i++)
	per[i]=i;
}

int find(int x){
	if(x==per[x])
	return x;
	return per[x]=find(per[x]);
}

bool join(int x,int y){
	int i=find(x);
	int j=find(y);
	if(i==j)
		return false;
	per[i]=j;
	return true;
}

int main(){
	int N,a,b;
	while(scanf("%d",&N)!=EOF){
		for(int i=1;i<=N;i++)
			for(int j=1;j<=N;j++){
				scanf("%d",&p[i][j]);
			}
		scanf("%d",&M);
		for(int m=1;m<=M;m++){
			scanf("%d%d",&a,&b);
			p[a][b]=0;
			p[b][a]=0;
		}
//		for(int i=1;i<=N;i++){
//			for(int j=1;j<=N;j++){
//				printf("%d ",p[i][j]);
//			}
//			putchar('\n');
//		}
		int k=1;
		for(int i=1;i<=N;i++)
			for(int j=1;j<=N;j++){
				edge[k].a=i;
				edge[k].b=j;
				edge[k].x=p[i][j];
//				printf("--%d %d %d - %d %d %d\n",edge[k].a,edge[k].b,edge[k].x,i,j,p[i][j]);
				k++;
			}
		M=k;
		sort(edge,edge+M,cmp);
//				for(int m=1;m<M;m++){
//			printf("--%d %d %d\n",edge[m].a,edge[m].b,edge[m].x);
//		}
		init();
		int sum=0;
		for(int m=1;m<M;m++){
			if(join(edge[m].a,edge[m].b)){
				sum+=edge[m].x;
			}
		}
		printf("%d\n",sum);		
	}
}
 

AC程式碼:

#include<stdio.h>
#include<string.h>
#include <stdlib.h>
int per[110000],N,M;
struct str{
	int a,b,x;
}edge[1001000];

int p[1100][1100];


int cmp(const void *a,const void *b)  
{  
    struct str *aa=(struct str *)a;  
    struct str *bb=(struct str *)b;  
    if(aa->x != bb->x)  
        return aa->x - bb->x;  
    else  
        return aa->a - bb->a;  
}  

void init(){
	for(int i=0;i<=110000;i++)
	per[i]=i;
}

int find(int x){
	if(x==per[x])
	return x;
	return per[x]=find(per[x]);
}

bool join(int x,int y){
	int i=find(x);
	int j=find(y);
	if(i==j)
		return false;
	per[i]=j;
	return true;
}

int main(){
	int N,a,b;
	while(scanf("%d",&N)!=EOF){
		for(int i=1;i<=N;i++)
			for(int j=1;j<=N;j++){
				scanf("%d",&p[i][j]);
			}
		scanf("%d",&M);
		for(int m=1;m<=M;m++){
			scanf("%d%d",&a,&b);
			p[a][b]=0;
			p[b][a]=0;
		}
		int k=1;
		for(int i=1;i<=N;i++)
			for(int j=1;j<=N;j++){
				edge[k].a=i;
				edge[k].b=j;
				edge[k].x=p[i][j];
				k++;
			}
		M=k;
		qsort(&edge[1],M,sizeof(edge[1]),cmp);
		init();
		int sum=0;
		for(int m=1;m<M;m++){
			if(join(edge[m].a,edge[m].b)){
				sum+=edge[m].x;
			}
		}
		printf("%d\n",sum);		
	}
}