690 - Pipeline Scheduling
题意:10个任务,5个通道,要求每个通道都能放下10个任务且不冲突,然后每个通道的放的方式间隔都是一样的,问最短需要时间。思路:利用位运算保存每个通道的放置方法,然后去深搜,要加剪枝。详细见代码
代码:
- #include <stdio.h>
- #include <string.h>
- #define min(a,b) ((a)<(b)?(a):(b))
- #define max(a,b) ((a)>(b)?(a):(b))
- const int N = 35;
- int n,p[5],ans,w[N],wn;
- char str[N];
- bool judge(int s0,int s1,int s2,int s3,int s4) {
- return ((s0&p[0]) == 0
- && (s1&p[1]) == 0
- && (s2&p[2]) == 0
- && (s3&p[3]) == 0
- && (s4&p[4]) == 0);
- }
- void init() {
- ans = 10 * n; wn = 0;
- for (int i = 0; i < 5; i++) {
- p[i] = 0;
- scanf("%s",str);
- for (int j = n - 1; j >= 0; j--) {
- p[i] = p[i] * 2 + (str[j] == 'X');
- }
- }
- int s0 = p[0],s1 = p[1],s2 = p[2],s3 = p[3],s4 = p[4];
- for (int k = 0; k <= n; k++) {
- if (judge((s0>>k),(s1>>k),(s2>>k),(s3>>k),(s4>>k))) {
- w[wn++] = k;//剪枝,开w数组把不能放的位置剔除掉
- }
- }
- }
- void dfs(int s0,int s4,int d,int len) {
- if (len + w[0] * (10 - d) > ans) return;//关键剪枝
- if (d == 10) {
- ans = min(ans,len);
- return;
- }
- for (int i = 0; i < wn; i++) {
- int ss0 = (s0>>w[i]),ss1 = (s1>>w[i]),ss2 = (s2>>w[i]),ss3 = (s3>>w[i]),ss4 = (s4>>w[i]);
- if (judge(ss0,ss1,ss2,ss3,ss4)) {
- dfs(ss0^p[0],ss1^p[1],ss2^p[2],ss3^p[3],ss4^p[4],d + 1,len + w[i]);
- }
- }
- }
- int main() {
- while (~scanf("%d",&n) && n) {
- init();
- dfs(p[0],p[1],p[2],p[3],p[4],1,n);
- printf("%d\n",ans);
- }
- return 0;
- }