2012-10-06 13:32:19Morris
[UVA][TSP] 10944 - Nuts for nuts..
Problem B: Nuts for nuts..
So as Ryan and Larry decided that they don't really taste so good, they
realized that there are some nuts located in certain places
of the island.. and they love them! Since they're lazy, but greedy,
they want to know the shortest tour that they can use to gather
every single nut!Can you help them?
Input
You'll be given x, and y, both less than 20, followed by x lines of y characters each as a map of the area, consisting sorely of ".", "#", and "L". Larry and Ryan are currently located in "L", and the nuts are represented by "#". They can travel in all 8 adjacent direction in one step. See below for an example. There will be at most 15 places where there are nuts, and "L" will only appear once.Output
On each line, output the minimum amount of steps starting from "L", gather all the nuts, and back to "L".Sample Input
5 5 L.... #.... #.... ..... #.... 5 5 L.... #.... #.... ..... #....
Sample Output
8 8
TSP 問題
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define oo 32767
#define min(x, y) ((x) < (y) ? (x) : (y))
int g[20][20], dp[1<<18][18], t2[1<<18];
void tsp(int state, int last) {
if(dp[state][last] != oo) return;
int i, j, tmp;
dp[state][last]--;
for(i = state; i; i -= j) {
j = i&(-i);
tsp(state-j, t2[j]);
tmp = dp[state-j][t2[j]]+g[t2[j]][last];
dp[state][last] = min(dp[state][last], tmp);
}
}
int main() {
int n, m, i, j, k, x[20], y[20];
char s[30];
for(i = 0; i < 18; i++) t2[1<<i] = i;
while(scanf("%d %d", &n, &m) == 2) {
int nut = 1;
for(i = 0; i < n; i++) {
scanf("%s", s);
for(j = 0; s[j]; j++) {
if(s[j] == '#')
x[nut] = i, y[nut++] = j;
if(s[j] == 'L')
x[0] = i, y[0] = j;
}
}
for(i = 0; i < nut; i++) {
for(j = 0; j < nut; j++) {
for(k = 0; ; k++)
if(x[i]+k >= x[j] && x[i]-k <= x[j] &&
y[i]+k >= y[j] && y[i]-k <= y[j]) {
g[i][j] = k;
break;
}
}
}
for(i = 1<<nut; i >= 0; i--)
for(j = 0; j < nut; j++)
dp[i][j] = oo;
dp[0][0] = 0;
tsp((1<<nut)-1, 0);
printf("%d\n", dp[(1<<nut)-1][0]);
}
return 0;
}