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題目傳送門

題意：給出n個長度為7的字符串，一個字符串到另一個的距離為不同的字符數，問所有連通的最小代價是多少

分析：Kuskal/Prim: 先用并查集做，簡單好寫，然而效率并不高，稠密圖應該用Prim而且要用鄰接矩陣，鄰接表的效率也不高。裸題但題目有點坑爹:(

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Kruskal：

#include <cstdio> #include <cstring> #include <string> #include <algorithm> #include <iostream> #include <cmath> using namespace std;const int MAXN = 2e3 + 10; const int INF = 0x3f3f3f3f; struct UF {int rt[MAXN], rk[MAXN];void init(void) {memset (rt, -1, sizeof (rt));memset (rk, 0, sizeof (rk));}int Find(int x) {return (rt[x] == -1) ? x : rt[x] = Find (rt[x]);}void Union(int x, int y) {x = Find (x), y = Find (y);if (x == y) return ;if (rk[x] > rk[y]) {rt[y] = x; rk[x] += rk[y] + 1;}else {rt[x] = y; rk[y] += rk[x] + 1;}}bool same(int x, int y) {return Find (x) == Find (y);} }uf; char s[MAXN][10]; struct Node {int u, v, w; }node[MAXN*MAXN/2]; int n, tot;bool cmp(Node x, Node y) {return x.w < y.w;}void get_w(int x) {for (int i=1; i<x; ++i){int res = 0;for (int j=0; j<7; ++j){if (s[i][j] != s[x][j]) res++;}node[++tot].u = i; node[tot].v = x; node[tot].w = res;} }int main(void) //POJ 1789 Truck History {// freopen ("POJ_1789.in", "r", stdin);while (scanf ("%d", &n) == 1){if (n == 0) break;tot = 0;for (int i=1; i<=n; ++i){scanf ("%s", s[i]);get_w (i);}sort (node+1, node+1+tot, cmp);int ans = 0; uf.init ();for (int i=1; i<=tot; ++i){int u = node[i].u; int v = node[i].v; int w = node[i].w;if (!uf.same (u, v)) {uf.Union (u, v); ans += w;}}printf ("The highest possible quality is 1/%d.\n", ans);}return 0; }

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Prim：

/*模版搞錯了，糾結半天。。。算法還是想清楚才行啊 */ #include <cstdio> #include <algorithm> #include <cstring> #include <string> #include <iostream> #include <vector> #include <queue> using namespace std;const int MAXN = 2e3 + 10; const int INF = 0x3f3f3f3f; int d[MAXN], w[MAXN][MAXN]; bool vis[MAXN]; int n; char s[MAXN][10];int Prim(int s) {memset (vis, false, sizeof (vis));memset (d, INF, sizeof (d)); d[s] = 0;int ret = 0;for (int i=1; i<=n; ++i) {int mn = INF, u = -1;for (int i=1; i<=n; ++i) {if (!vis[i] && d[i] < mn) mn = d[u=i];}if (u == -1) return -1;vis[u] = true; ret += d[u];for (int i=1; i<=n; ++i) {if (!vis[i] && d[i] > w[u][i]) {d[i] = w[u][i];}}}return ret; }void get_w(int x) {for (int i=1; i<x; ++i) {int res = 0;for (int j=0; j<7; ++j) {if (s[i][j] != s[x][j]) res++;}w[i][x] = w[x][i] = res;} }int main(void) {while (scanf ("%d", &n) == 1) {if (n == 0) break;memset (w, INF, sizeof (w));for (int i=1; i<=n; ++i) {scanf ("%s", s[i]); get_w (i);}printf ("The highest possible quality is 1/%d.\n", Prim (1));}return 0; }

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轉載于:https://www.cnblogs.com/Running-Time/p/4573735.html

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