ural 1029 Ministry

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类型：DP

题目：http://acm.timus.ru/problem.aspx?space=1&num=1029

思路：很容易想到f(i, j) = min(f(i - 1, j), f(i, j - 1), f(i, j + 1)) + val[i][j]

实现时，可以分两步dp，对于每一行先从左往右dp，然后再从右往左dp，这样可以得到正确解

利用上一层的最优解，假如当前j是由j - 1得到，那么从左往右dp可以得到解

假如当前j是由j + 1得到，那么从右往左dp可以得到最优解

用pre数组记录路径

// ural 1029 Ministry// wa wa ac 0.046s#include <iostream>#include <string>#include <cstdio>#include <cstring>using namespace std;#define FOR(i,a,b) for(i = (a); i < (b); ++i)#define FORE(i,a,b) for(i = (a); i <= (b); ++i)#define FORDE(i,a,b) for(i = (a); i >= (b); --i)int n, m;int endn[100 * 510], da[110][510], pre[110][510], dp[110][510];void input() {    int i, j;    scanf("%d %d", &m, &n);    FORE(i, 1, m) FORE(j, 1, n) {        scanf("%d", &da[i][j]);        dp[i][j] = 0x7f7f7f7f;    }}void solve() {    int i, j;    FORE(i, 1, n)        dp[1][i] = da[1][i], pre[1][i] = -2;    FORE(i, 2, m) {        FORE(j, 1, n) {            dp[i][j] = dp[i - 1][j] + da[i][j];            if(j == 1)                continue;            //zuo            if(dp[i][j] > dp[i][j - 1] + da[i][j])                dp[i][j] = dp[i][j - 1] + da[i][j], pre[i][j] = -1;        }        FORDE(j, n - 1, 1) {            //you            if(dp[i][j] > dp[i][j + 1] + da[i][j])                dp[i][j] = dp[i][j + 1] + da[i][j], pre[i][j] = 1;        }    }}void output() {    int i, j;    int minend = 0x7f7f7f7f, mini = -1;    FORE(i, 1, n)        if(minend > dp[m][i])            minend = dp[m][i], mini = i;    j = mini;    i = m;    int cnt = 0;    while(pre[i][j] != -2) {        endn[cnt++] = j;        switch(pre[i][j]) {            case -1:--j;break;            case 1: ++j;break;            case 0: --i;break;        }    }    endn[cnt++] = j;    printf("%d", endn[--cnt]);    FORDE(i, cnt - 1, 0)        printf(" %d", endn[i]);    printf("\n");}int main() {    input();    solve();    output();    return 0;}/*3 41   100 1000 01   1   1000 0100 0   1000 100*/