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計(jì)算整個(gè)汽車演化過程中所要改變的最少的字符數(shù)。改變的字符數(shù)的計(jì)算方式為1-7位不同位的個(gè)數(shù)。

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代碼如下：

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#include <cstdlib> #include <cstring> #include <algorithm> #include <cstdio> using namespace std;int N, pos, set[2005];char s[2005][10];struct Node {int x, y, diff;bool operator < (Node t) const{return diff < t.diff;} }e[4000005];int diff(int x, int y) {int cnt = 0;for (int i = 0; i < 7; ++i) {if (s[x][i] != s[y][i]) {++cnt;}}return cnt; }int find(int x) {return set[x] = x == set[x] ? x : find(set[x]); }void merge(int x, int y) {set[x] = y; }int main() {int a, b, ans, cnt;while (scanf("%d", &N), N) {pos = ans = cnt = 0;for (int i = 1; i <= N; ++i) {scanf("%s", s[i]); set[i] = i;}for (int i = 1; i <= N; ++i) {for (int j = i+1; j <= N; ++j) { ++pos;e[pos].x = i, e[pos].y = j;e[pos].diff = diff(i, j);;}}sort(e+1, e+1+pos);for (int i = 1; i <= pos; ++i) {a = find(e[i].x), b = find(e[i].y);if (a != b) {printf("%d %d\n", e[i].x, e[i].y);++cnt;merge(a, b);ans += e[i].diff;if (cnt == N-1) {break;}}}printf("The highest possible quality is 1/%d.\n", ans);}return 0; }

轉(zhuǎn)載于:https://www.cnblogs.com/Lyush/archive/2012/07/01/2572080.html

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