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Connect the Cities

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

1

题意：大约就是给出了连通的和不连通的岛屿 不连通的给出了 维修所需的价格 最后求用最小的代价 得到所有的连通 

题解：比较简单的最小生成树的问题  Prim和Kruskal都可以 但是Kruskal用C++提交会超时 应该是自己剪枝不够 Prime跑了500+ms 感觉还好吧 后面总结两种算法的时候我会专门提到两种方法的选择问题 先给出两种方法的AC代码 

//看HDU上面大牛的耗时都是200-ms 可惜不清楚是怎么实现的

Kruskal：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

const int maxn = 511;
int pi[maxn];
struct node{
    int p, q, c;
}no[maxn*maxn];

bool cmp(node a, node b);
int Find(int x);

int main(){
    int cas;
    scanf("%d", &cas);
    while(cas--){
        for(int i = 1; i <= 500; ++i){
            pi[i] = i;
        }
        int n, m, k;
        scanf("%d%d%d", &n, &m, &k);
        for(int i = 1; i <= m; ++i){
            scanf("%d%d%d", &no[i].p, &no[i].q, &no[i].c);
        }
        sort(no+1, no+m+1, cmp);

        int num = 0;
        int t, root, tmp;
        for(int i = 0; i < k; ++i){
            scanf("%d%d", &t, &root);
            for(int j = 1; j < t; ++j){
                scanf("%d", &tmp);
                int x = Find(tmp);
                int y = Find(root);
                if(x != y){
                    pi[x] = y;
                    ++num;
                }
            }
        }

        if(num == n-1){
            printf("0\n");
            continue;
        }
        //bool flag = false;
        int ans = 0;
        for(int i = 1; i <= m; ++i){
            int x = Find(no[i].p);
            int y = Find(no[i].q);
            if(x == y){
                continue;
            }
            pi[x] = y;
            ans += no[i].c;
            ++num;

            if(num == n-1){
                //flag = true;
                break;
            }
        }

        if(num == n-1){
            printf("%d\n", ans);
        }
        else{
            printf("-1\n");
        }
    }
    return 0;
}

bool cmp(node a, node b){
    return a.c < b.c;
}

int Find(int x){
    int r=x;
    while(pi[r]!=r)
        r=pi[r];
    return r;
}

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;

const int maxn = 511;
const int inf = 0x7ffffff;
int dist[maxn][maxn];
int low[maxn];
bool vis[maxn];
int v[maxn];

void Prim(int n);

int main(){
    int cas;
    scanf("%d", &cas);
    while(cas--){
        int n, m, k;
        scanf("%d%d%d", &n, &m, &k);

        for(int i = 1; i <= n; ++i){
            for(int j = 1; j <= n; ++j){
                if(i == j){
                    dist[i][j] = 0;
                }
                else{
                    dist[i][j] = inf;
                }
            }
        }

        while(m--){
            int p, q, c;
            scanf("%d%d%d", &p, &q, &c);
            dist[p][q] = dist[q][p] = min(dist[p][q], c);
        }

        while(k--){
            int t;
            scanf("%d", &t);
            for(int i = 0; i < t; ++i){
                scanf("%d", &v[i]);
            }
            for(int i = 0; i < t; ++i){
                for(int j = 0; j < t; ++j){
                    dist[v[i]][v[j]] = 0;
                }
            }
        }

        Prim(n);
    }
    return 0;
}

void Prim(int n){
    int ans = 0;

    for(int i = 1; i <= n; ++i){
        low[i] = dist[1][i];
        vis[i] = false;
    }

    vis[1] = true;

    int flag;

    for(int i = 2; i <= n; ++i){
        int mi = inf;
        flag = -1;
        for(int j = 2; j <= n; ++j){
            if(mi > low[j] && !vis[j]){
                mi = low[j];
                flag = j;
            }
        }

        if(mi >= inf){
            flag = -1;
            break;
        }

        ans += mi;
        vis[flag] = true;

        for(int j = 2; j <= n; ++j){
            if(dist[flag][j] < low[j] && !vis[j]){
                low[j] = dist[flag][j];
            }
        }
    }

    if(flag == -1){
        printf("-1\n");
    }
    else{
        printf("%d\n", ans);
    }

}

【HDU3371】Connect the Cities

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原文地址：http://blog.csdn.net/u012431590/article/details/46053311