UVA116-Unidirectional TSP(动态规划基础)
Accept: 7167 Submit: 56893
Time Limit: 3000 mSec
Problem Description
Input
The input consists of a sequence of matrix specifications. Each matrix specification consists of the row and column dimensions in that order on a line followed by m·n integers where m is the row dimension and n is the column dimension. The integers appear in the input in row major order, i.e., the first n integers constitute the first row of the matrix, the second n integers constitute the second row and so on. The integers on a line will be separated from other integers by one or more spaces. Note: integers are not restricted to being positive. There will be one or more matrix specifications in an input file. Input is terminated by end-of-file. Foreachspecificationthenumberofrowswillbebetween1and10inclusive; thenumberofcolumns will be between 1 and 100 inclusive. No path’s weight will exceed integer values representable using 30 bits.
Output
Sample Input
3 4 1 2 8 6
6 1 8 2 7 4
5 9 3 9 9 5
8 4 1 3 2 6
3 7 2 8 6 4
5 6
3 4 1 2 8 6
6 1 8 2 7 4
5 9 3 9 9 5
8 4 1 3 2 6
3 7 2 1 2 3
2 2
9 10
9 10
Sample Output
1 2 3 4 4 5
16
1 2 1 5 4 5
11
1 1
19
题解:和数字三角形一样,水题。
#include <bits/stdc++.h> using namespace std; const int maxn = + , maxm = + ;
const int INF = 0x3f3f3f3f; int n, m;
int val[maxm][maxn], dp[maxm][maxn];
int Next[maxm][maxn]; int read() {
int q = , f = ; char ch = ' ';
while (ch<'' || ch>'') {
if (ch == '-') f = -;
ch = getchar();
}
while ('' <= ch && ch <= '') {
q = q * + ch - '';
ch = getchar();
}
return q * f;
} int main()
{
//freopen("input.txt", "r", stdin);
while (~scanf("%d%d", &m, &n)) {
for (int i = ; i < m; i++) {
for (int j = ; j < n; j++) {
val[i][j] = read();
}
}
//memset(dp, INF, sizeof(dp));
int ans = INF, first = -; for (int j = n - ; j >= ; j--) {
for (int i = ; i < m; i++) {
if (j == n - ) {
dp[i][j] = val[i][j];
}
else {
int row[] = { (i - + m) % m,i,(i + ) % m };
sort(row, row + );
dp[i][j] = INF;
for (int k = ; k < ; k++) {
if (dp[i][j] > dp[row[k]][j + ] + val[i][j]) {
dp[i][j] = dp[row[k]][j + ] + val[i][j];
Next[i][j] = row[k];
}
}
}
if (j == && dp[i][j] < ans) {
ans = dp[i][j];
first = i;
}
}
} printf("%d", first + );
for (int i = Next[first][], j = ; j < n; i = Next[i][j], j++) {
printf(" %d", i + );
}
printf("\n%d\n", ans);
}
return ;
}
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