• BZOJ 1601 [Usaco2008 Oct]灌水 (建图+mst)


    题意:

    300个坑,每个坑能从别的坑引水,或者自己出水,i从j饮水有个代价,每个坑自己饮水也有代价,问让所有坑都有谁的最少代价

    思路:

    先建一个n的完全图,然后建一个超级汇点,对每个点连w[i],跑mst,这样就能保证所有坑联通,并且至少有一个坑有水

    代码:

    #include<iostream>
    #include<cstdio>
    #include<algorithm>
    #include<cmath>
    #include<cstring>
    #include<string>
    #include<stack>
    #include<queue>
    #include<deque>
    #include<set>
    #include<vector>
    #include<map>
    
    #define fst first
    #define sc second
    #define pb push_back
    #define mem(a,b) memset(a,b,sizeof(a))
    #define lson l,mid,root<<1
    #define rson mid+1,r,root<<1|1
    #define lc root<<1
    #define rc root<<1|1
    //#define lowbit(x) ((x)&(-x)) 
    
    using namespace std;
    
    typedef double db;
    typedef long double ldb;
    typedef long long ll;
    typedef unsigned long long ull;
    typedef pair<int,int> PI;
    typedef pair<ll,ll> PLL;
    
    const db eps = 1e-6;
    const int mod = 1e9+7;
    const int maxn = 2e6+100;
    const int maxm = 2e6+100;
    const int inf = 0x3f3f3f3f;
    const db pi = acos(-1.0);
    
    int p[333][333];
    int w[333];
    int n;
    int f[maxn];
    int find(int x){
        return f[x]==x?x:f[x]=find(f[x]);
    }
    struct node{
        int x, y;
        int w;
    }edge[maxn];
    bool cmp(node a, node b){
        return a.w<b.w;
    }
    int tot;
    int add(int x, int y, int w){
        edge[++tot].x=x;
        edge[tot].y=y;
        edge[tot].w=w;
    }
    int main(){
        scanf("%d", &n);
        for(int i = 1; i <= n; i++){
            scanf("%d", &w[i]);
            add(0,i,w[i]);
        }
        for(int i = 1; i <= n; i++){
            for(int j = 1; j <= n; j++){
                scanf("%d", &p[i][j]);
                add(i,j,p[i][j]);
            }
        }
        for(int i = 0; i <= n; i++)f[i]=i;
        sort(edge+1, edge+1+tot,cmp);
        int cnt = 0;
        int ans = 0;
        for(int i = 1; i <= tot; i++){
            int x = edge[i].x;
            int y = edge[i].y;
            int w = edge[i].w;
            int t1 = find(x);
            int t2 = find(y);
            if(t1 != t2){
                f[t1] = t2;
                ans+=w;
                cnt++;
            }
            if(cnt==n)break;
        }
        printf("%d",ans);
        return 0;        
    }
    
    /*
    4
    5
    4
    4
    3
    0 2 2 2
    2 0 3 3
    2 3 0 4
    2 3 4 0
     */
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  • 原文地址:https://www.cnblogs.com/wrjlinkkkkkk/p/11140540.html
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