Connect the Cities
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 7338 Accepted Submission(s): 2093
Problem Description
In 2100, since the sea level rise, most of the cities disappear. Though some survived cities are still connected with others, but most of them become disconnected. The government wants to build some roads to connect all of these cities again, but they don’t want to take too much money.
Input
The first line contains the number of test cases. Each test case starts with three integers: n, m and k. n (3 <= n <=500) stands for the number of survived cities, m (0 <= m <= 25000) stands for the number of roads you can choose to connect the cities and k (0 <= k <= 100) stands for the number of still connected cities. To make it easy, the cities are signed from 1 to n. Then follow m lines, each contains three integers p, q and c (0 <= c <= 1000), means it takes c to connect p and q. Then follow k lines, each line starts with an integer t (2 <= t <= n) stands for the number of this connected cities. Then t integers follow stands for the id of these cities.
Output
For each case, output the least money you need to take, if it’s impossible, just output -1.
Sample Input
1
6 4 3
1 4 2
2 6 1
2 3 5
3 4 33
2 1 2
2 1 3
3 4 5 6
Sample Output
1
这道题我的思路是：先处理k个数，然后再用并查集求剩下没连通的。之前要排序，最后检查是否连通。。。。
╮(╯▽╰)╭ 自己超时了好多次，还有就是，题意马虎了，又WA了几次，，，悲剧。
#include <iostream> #include <cstdlib> #include<algorithm> using namespace std; int father[560],Q; struct sum { int a; int b; int c; }num[50005]; //路线数 bool cmp(const sum &x,const sum &y) //按长度从小到大快排, { return x.c<y.c;//原理有待研究 } int Find(int x) //找出祖先 { while(x!=father[x]) x=father[x]; return x; } void Union(int a,int b,int i) { if(a!=b) { father[a]=b; Q+=num[i].c; //并入家族且把长度加上来 } } int main() { int T,k,n,m,i,j,l,p,q,c,t; int ss[505]; scanf("%d",&T); while(T--) { scanf("%d%d%d",&n,&m,&k); for(i=1;i<=n;i++) father[i]=i; for(i=0;i<m;i++) scanf("%d%d%d",&num[i].a,&num[i].b,&num[i].c); memset(ss,0,sizeof(ss)); for(l=0;l<k;l++) { scanf("%d",&t); for(j=0;j<t;j++) scanf("%d",&ss[j]); for(j=1;j<t;j++) { if(Find(ss[0])!=Find(ss[j])) father[Find(ss[j])]=Find(ss[0]); } memset(ss,0,sizeof(ss)); } sort(num,num+m,cmp);//排序 for(i=0,Q=0;i<m;i++) { Union(Find(num[i].a),Find(num[i].b),i); } for(i=1,t=0;t<2&&i<=n;i++) if(father[i]==i) t++; if(t==2) printf("-1 "); else printf("%d ",Q); } return 0; } /* 5 6 4 3 1 4 2 2 6 1 2 3 5 3 4 33 2 1 2 2 1 3 3 4 5 6 6 4 3 1 4 3 2 6 2 6 2 1 3 4 33 2 1 2 2 1 3 3 4 5 6 6 4 3 1 4 2 2 6 1 2 3 5 3 4 33 2 1 2 2 2 3 2 4 5 6 4 3 1 4 3 2 6 2 6 2 1 3 4 33 2 1 2 2 1 3 2 4 6*/
在网上找了Kruskal函数， 有待学习！
#include<stdio.h> //Kruskal函数 #include<algorithm> using namespace std; typedef struct{ int u; int v; int w; }Edge; const int EdgeNum=50010; const int PointNum=510; Edge E[EdgeNum]; int P[PointNum]; int union_find(int x) // 并查集 { return P[x]==x? x : P[x]=union_find(P[x]); } bool cmp(Edge a,Edge b){ return a.w<b.w; } int MST_Kruskal(int n,int m) // 传入点数和边数 { int i,j,x,y,k=1,sum=0; for(i=0;i<n;++i) { for(j=0;j<n;++j) { if(i==j) continue; if(P[j]==i) { k++; // printf("P[%d]=%d ",j,i); } } } // printf("k=%d ",k); sort(E,E+m,cmp); for(i=0;k<n&&i<m;++i) { x=union_find(E[i].u); y=union_find(E[i].v); if(x!=y) { sum+=E[i].w; P[x]=y; k++; } } if(k<n) return -1; return sum; } int f[510]; int solve(int n,int m,int k) { int i,j,t; for(i=0;i<n;++i) { // 初始化并查集 P[i]=i; } for(i=0;i<k;++i){ scanf("%d",&t); for(j=0;j<t;++j) { scanf("%d",&f[j]); f[j]--; } for(j=1;j<t;++j) P[union_find(f[j])]=union_find(f[j-1]); } return MST_Kruskal(n,m); } int main() { int t,n,m,k,i,p,q,c,ans; scanf("%d",&t); while(t--) { scanf("%d%d%d",&n,&m,&k); for(i=0;i<m;++i) { scanf("%d%d%d",&p,&q,&c); p--;q--; E[i].u=p;E[i].v=q;E[i].w=c; } ans=solve(n,m,k); printf("%d ",ans); } return 0; }