UVA 690 Pipeline Scheduling (搜索+位运算+剪枝）

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
, wn;
char str
;

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 s1, int s2, int s3, 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;
}