uva 11008 - Antimatter Ray Clearcutting(状态dp)
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题目链接:11008 - Antimatter Ray Clearcutting
题目大意:有一篇森林,给出n棵树的坐标,现在有一种反物质光线,可以清楚直线上的树木,然后给出m,代表这片森林要除掉m棵树,问最少使用几次光线。
解题思路:状态压缩,看了别人的题解才知道要用位运算去做,平时有碰到记忆化搜索压缩状态都用数组可以解决的,今天写了位运算版,感觉对位运算有了进一步的了解,下面介绍一篇写的比较好的题解。
http://www.cnblogs.com/scau20110726/archive/2012/09/28/2707866.html
#include <stdio.h>#include <string.h>#define min(a,b) (a)<(b)?(a):(b)const int N = 70000;const int M = 30;const int MAX = 1 << 30;int x[M], y[M], g[M][M];;int n, m, dp[N], tmp;void init() {memset(dp, -1, sizeof(dp));memset(g, 0, sizeof(g));scanf("%d%d", &n, &m);tmp = (1 << n) - 1;for (int i = 0; i < n; i++)scanf("%d%d", &x[i], &y[i]);for (int i = 0; i < n; i++) {for (int j = 0; j < n; j++) {if (i == j) continue;for (int k = n - 1; k >= 0; k--) {g[i][j] <<= 1;if ((x[j] - x[i]) * (y[k] - y[i]) == (x[k] - x[i]) * (y[j] - y[i]))g[i][j]++;}}}}int handle(int k) {int cnt = 0;for (int i = 0; i < n; i++)if ((1 << i) & k) cnt++;return cnt;}int solve(int k) {int cnt = handle(k);int& ans = dp[k];if (cnt <= n - m) return ans = 0;else if (cnt == 1) return ans = 1;else if (ans > -1) return ans;ans = MAX;for (int i = 0; i < n; i++) {if ((1 << i) & k) {for (int j = i + 1; j < n; j++) {if ((1 << j) & k) {int t = k & (g[i][j] ^ tmp);ans = min(ans, solve(t) + 1);}}}}return ans;}int main () {int cas, t = 1;scanf("%d", &cas);while (cas--) {init();printf("Case #%d:\n%d\n", t++, solve(tmp));if (cas) printf("\n");}return 0;}
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