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还是畅通工程[HDU1233]

2024-08-28 11:05:24   218人阅读

还是畅通工程

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 41897 Accepted Submission(s): 19126


Problem Description
某省调查乡村交通状况，得到的统计表中列出了任意两村庄间的距离。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），并要求铺设的公路总长度为最小。请计算最小的公路总长度。

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 ( < 100 )；随后的N(N-1)/2行对应村庄间的距离，每行给出一对正整数，分别是两个村庄的编号，以及此两村庄间的距离。为简单起见，村庄从1到N编号。
当N为0时，输入结束，该用例不被处理。

Output
对每个测试用例，在1行里输出最小的公路总长度。

Sample Input
3
1 2 1
1 3 2
2 3 4
4
1 2 1
1 3 4
1 4 1
2 3 3
2 4 2
3 4 5
0

Sample Output
3
5

Hint
Hint

Huge input, scanf is recommended.

技术分享
#include <stdio.h>#include <algorithm>#include <math.h>#include <string.h>using namespace std;class Union_Find_Set {#define MAX_UNION_FIND_SET_SIZE 105public:    int setSize;    int father[MAX_UNION_FIND_SET_SIZE];    Union_Find_Set() {        setSize = 0;    }    Union_Find_Set(int x) {        setSize = x;        clear(x);    }    void clear(int x) {        for (int i = 0; i < x; i++) {            father[i] = i;        }    }    int getFather(int x) {        if (x != father[x]) {            father[x] = getFather(father[x]);        }        return father[x];    }    bool merge(int a, int b) {        a = getFather(a);        b = getFather(b);        if (a != b) {            father[a] = b;            return true;        } else {            return false;        }    }    int countRoot() {        int ret = 0;        for (int i = 0; i < setSize; i++) {            if (father[i] = i) {                ret++;            }        }        return ret;    }};class Kruskal {#define Kruskal_MAXN 100#define Kruskal_MAXM 10005public:    Union_Find_Set ufs;    int x[Kruskal_MAXM], y[Kruskal_MAXM], w[Kruskal_MAXM], N, M;    Kruskal() {        N = 0;        M = 0;    }    void clear() {        N = 0;        M = 0;    }    void addEdge(int a, int b, int c) {        x[M] = a;        y[M] = b;        w[M] = c;        M++;        x[M] = b;        y[M] = a;        w[M] = c;        M++;    }    void sort(int l, int r) {        int i = l, j = r, m = w[(l + r) >> 1], t;        do {            while (w[i] < m) {                i++;            }            while (m < w[j]) {                j--;            }            if (i <= j) {                t = x[i];                x[i] = x[j];                x[j] = t;                t = y[i];                y[i] = y[j];                y[j] = t;                t = w[i];                w[i] = w[j];                w[j] = t;                i++;                j--;            }        } while (i <= j);        if (l < j) {            sort(l, j);        }        if (i < r) {            sort(i, r);        }    }    int MST() {        sort(0, M - 1);        ufs.clear(N + 1);        int cnt = 0, ret = 0;        for (int i = 0; i < M; i++) {            if (cnt == N - 1) {                return ret;            }            if (ufs.getFather(x[i]) != ufs.getFather(y[i])) {                ufs.merge(x[i], y[i]);                ret += w[i];                cnt++;            }        }        if (cnt == N - 1) {            return ret;        } else {            return -1;        }    }};Kruskal Kr;int main() {    int n;    while (scanf("%d", &n) != EOF) {        if (n == 0) {            break;        }        Kr.clear();        Kr.N=n;        for (int i = 1; i < n; i++)            for (int j = i + 1; j <= n; j++) {                int a, b, c;                scanf("%d%d%d", &a, &b, &c);                Kr.addEdge(a, b, c);            }        printf("%d\n", Kr.MST());    }    return 0;}
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还是畅通工程[HDU1233]


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