洛谷 P1194 买礼物 的题解

题目链接

题解

备战 CSP，复习一下最小生成树，祝大家 rp++！！！

经典的 $Kruskal$

把所有的物品都看作节点，物品之间的优惠或费用为边的权值，所以要做的就应该是把所有点连接起来的最小代价，也就是求该图的最小生成树。

然后正常建边，按边权跑一遍 $Kruskal$，就万事大吉了

AC 代码

#include <bits/stdc++.h>
#define lowbit(x) x & (-x)
#define endl "\n"
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
namespace fastIO {inline int read() {register int x = 0, f = 1;register char c = getchar();while (c < '0' || c > '9') {if(c == '-') f = -1;c = getchar();}while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();return x * f;}inline void write(int x) {if(x < 0) putchar('-'), x = -x;if(x > 9) write(x / 10);putchar(x % 10 + '0');return;}
}
using namespace fastIO;
int a, b, fa[300005], ans = 0, money, tot = 0;
struct node {int x, y, z;
}f[300005];
int find(int x) {return x == fa[x] ? x : fa[x] = find(fa[x]);}
bool cmp (node a, node b) {return a.z < b.z;}
int main() {//freopen(".in","r",stdin);//freopen(".out","w",stdout);a = read(), b = read();for(int i = 1; i <= b; i ++) {fa[i] = i; }for(int i = 1; i <= b; i ++) {tot ++;f[tot].x = 0;f[tot].y = i;f[tot].z = a;} for(int i = 1; i <= b; i ++) {for(int j = 1; j <= b; j ++) {money = read();if(money) {tot ++;f[tot].x = i;f[tot].y = j;f[tot].z = money;}}}sort(f + 1, f + tot + 1, cmp);for (int i = 1; i <= tot; i ++) {if (find(f[i].x) != find(f[i].y)) {fa[find(f[i].y)] = find(f[i].x);ans += f[i].z;}else continue;}write(ans);return 0;
}