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
 */

 

posted @ 2019-07-05 20:10  wrjlinkkkkkk  阅读(150)  评论(0编辑  收藏  举报