继续畅通工程
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 14309 Accepted Submission(s): 6228
Problem Description
省政府“畅通工程”的目标是使全省任何两个村庄间都可以实现公路交通(但不一定有直接的公路相连,只要能间接通过公路可达即可)。现得到城镇道路统计表,表中列出了任意两城镇间修建道路的费用,以及该道路是否已经修通的状态。现请你编写程序,计算出全省畅通需要的最低成本。
Input
测试输入包含若干测试用例。每个测试用例的第1行给出村庄数目N ( 1< N < 100 );随后的 N(N-1)/2 行对应村庄间道路的成本及修建状态,每行给4个正整数,分别是两个村庄的编号(从1编号到N),此两村庄间道路的成本,以及修建状态:1表示已建,0表示未建。
当N为0时输入结束。
Output
每个测试用例的输出占一行,输出全省畅通需要的最低成本。
Sample Input
3 1 2 1 0 1 3 2 0 2 3 4 0 3 1 2 1 0 1 3 2 0 2 3 4 1 3 1 2 1 0 1 3 2 1 2 3 4 1 0
Sample Output
3 1 0#include <stdio.h> #include<string.h> #include <algorithm> using namespace std; typedef struct Road { int c1, c2, cost, state; }Road; bool myCompare(const Road &a, const Road &b) { if(a.cost < b.cost) return 1; return 0; } Road road[5051]; int city[101]; int Find(int n) { if(city[n] == -1) return n; return city[n] = Find(city[n]); } bool Merge(int s1, int s2) { int r1 = Find(s1); int r2 = Find(s2); if(r1 == r2) return 0; if(r1 < r2) city[r2] = r1; else city[r1] = r2; return 1; } int main() { //freopen("input.txt", "r", stdin); int n; while(scanf("%d", &n) && n) { int m = n*(n-1)/2; memset(city, -1, sizeof(city)); int count = 0; for(int i=0; i<m; ++i) { scanf("%d %d %d %d", &road[i].c1, &road[i].c2, &road[i].cost, &road[i].state); if(road[i].state == 1) { count ++; Merge(road[i].c1, road[i].c2); } } sort(road, road+m, myCompare); int sum = 0; for(int i=0; i<m; ++i) { if(Merge(road[i].c1, road[i].c2) && road[i].state == 0) { count ++; sum += road[i].cost; } if(count == n-1) break; } printf("%d\n", sum); } return 0; }#include <stdio.h> #include<string.h> #include <algorithm> int city[5000]; using namespace std; struct Road { int c1; int c2; int cost; int state; }; Road road[110]; bool cmp(Road a,Road b) { if(a.cost<b.cost) return 1; return 0; } int find(int a) { if(city[a]==-1) return a; return city[a]=find(city[a]); } bool merge(int x, int y) { int fx,fy; fx = find(x); fy = find(y); if(fx == fy) return 0; else if(fx < fy) //不能省去 city[fy] = fx; else city[fx] = fy; return 1; } int main() { int n, m; while(scanf("%d",&n),n) { m=n*(n-1)/2; int i; int count = 0; int sum = 0; memset(city,-1,sizeof(city)); for(i=0; i<m; i++) { scanf("%d %d %d %d",&road[i].c1, &road[i].c2, &road[i].cost, &road[i].state); if(road[i].state == 1) { count ++; merge(road[i].c1, road[i].c2); } } sort(road, road+m,cmp); for(i=0; i<m; i++) { if(merge(road[i].c1, road[i].c2) && road[i].state==0) { count ++; sum += road[i].cost; } if(count == n-1) break; } printf("%d\n",sum); } return 0; }
时间: 2024-10-19 02:07:29