UVA 116 Unidirectional TSP （单向TSP）（dp多段图的最短路）

题意：给一个m行n列(m<=10, n<=100)的整数矩阵，从第一列任何一个位置出发每次往右，右上或右下走一格，最终到达最后一列。要求经过的整数之和最小。第一行的上一行是最后一行，最后一行的下一行是第一行。输出路径上每列的行号。多解时输出字典序最小的。

分析：

1、dp[i][j]---从第i行第j列到最后一列的最小开销。

2、列从右到左，从后一个状态可推知前一个状态的开销。

#pragma comment(linker, "/STACK:102400000, 102400000")
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<deque>
#include<queue>
#include<list>
#define Min(a, b) ((a < b) ? a : b)
#define Max(a, b) ((a < b) ? b : a)
const double eps = 1e-8;
inline int dcmp(double a, double b){
    if(fabs(a - b) < eps) return 0;
    return a > b ? 1 : -1;
}
typedef long long LL;
typedef unsigned long long ULL;
const int INT_INF = 0x3f3f3f3f;
const int INT_M_INF = 0x7f7f7f7f;
const LL LL_INF = 0x3f3f3f3f3f3f3f3f;
const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f;
const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};
const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};
const int MOD = 1e9 + 7;
const double pi = acos(-1.0);
const int MAXN = 100 + 10;
const int MAXT = 10000 + 10;
using namespace std;
int a[20][MAXN];
int dp[20][MAXN];
int path[20][MAXN];//当前位置的下一列所对应行数
int main(){
    int m, n;
    while(scanf("%d%d", &m, &n) == 2){
        memset(dp, INT_INF, sizeof dp);
        memset(path, 0, sizeof path);
        for(int i = 1; i <= m; ++i){
            for(int j = 1; j <= n; ++j){
                scanf("%d", &a[i][j]);
            }
        }
        for(int i = 1; i <= m; ++i) dp[i][n] = a[i][n];
        for(int j = n - 1; j >= 1; --j){
            for(int i = 1; i <= m; ++i){
                int tmp[] = {i - 1, i, i + 1};
                if(i == 1) tmp[0] = m;
                if(i == m) tmp[2] = 1;
                sort(tmp, tmp + 3);
                for(int k = 0; k < 3; ++k){
                    int &cur = tmp[k];
                    if(a[i][j] + dp[cur][j + 1] < dp[i][j]){//保证字典序最小
                        dp[i][j] = a[i][j] + dp[cur][j + 1];
                        path[i][j] = cur;
                    }
                }
            }
        }
        int ans = INT_INF;
        int st = 0;
        for(int i = 1; i <= m; ++i){
            if(dp[i][1] < ans){
                ans = dp[i][1];
                st = i;//第一列行数
            }
        }
        printf("%d", st);
        for(int j = 2; j <= n; ++j){
            printf(" %d", path[st][j - 1]);
            st = path[st][j - 1];
        }
        printf("\n");
        printf("%d\n", ans);
    }
    return 0;
}