Luogu 1194买礼物

题意:买n个东西,这些东西有个原价m,若买东西i,则东西j降到a[i][j]元,求买n个东西最少的钱.

做法:若买i后买j的价钱a[i][j]比m小,那么连一条无向边，跑一波最小生成树.

看剩下还有几个点没连进去的就直接按原价买..

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <algorithm>
 4 #include <cstring>
 5 #include <cmath>
 6 #include <queue>
 7 #include <stack>
 8 #include <map>
 9 #define ll long long
10 #define out(a) printf("%d",a)
11 #define ln printf("\n")
12 const int N=3e5+50050;
13 const int MOD=1e9+7;
14 using namespace std;
15 int n,m,x;
16 int tot=0,ans=0,cnt=0;
17 int pre[N];
18 struct node
19 {
20     int x,y,cost;
21 }a[N];
22 int read()
23 {
24     int s=0,t=1; char c;
25     while (c<‘0‘||c>‘9‘){if (c==‘-‘) t=-1; c=getchar();}
26     while (c>=‘0‘&&c<=‘9‘){s=s*10+c-‘0‘; c=getchar();}
27     return s*t;
28 }
29 ll readl()
30 {
31     ll s=0,t=1; char c;
32     while (c<‘0‘||c>‘9‘){if (c==‘-‘) t=-1; c=getchar();}
33     while (c>=‘0‘&&c<=‘9‘){s=s*10+c-‘0‘; c=getchar();}
34     return s*t;
35 }
36 bool cmp(node a,node b)
37 {
38     return a.cost<b.cost;
39 }
40 int find(int x)
41 {
42     if (pre[x]==x) return x;
43     else return pre[x]=find(pre[x]);
44 }
45 void Kruskal()
46 {
47     for (int i=1;i<=tot;i++){
48       int x=find(a[i].x),y=find(a[i].y);
49       if (x!=y){
50         ans+=a[i].cost;
51         pre[x]=y; cnt++;
52       }
53       if (cnt==n-1) return;
54     }
55 }
56 int main()
57 {
58     m=read(),n=read();
59     for (int i=1;i<=n;i++){
60       pre[i]=i;
61       for (int j=1;j<=n;j++){
62         x=read();
63         if (x!=0&&x<m) a[++tot].x=i,a[tot].y=j,a[tot].cost=x;
64       }
65     }
66     sort(a+1,a+tot+1,cmp);
67     Kruskal();
68     ans+=(n-cnt)*m;
69     out(ans);
70     return 0;
71 }

原文地址：https://www.cnblogs.com/Kaleidoscope233/p/9807293.html