[Codeforces Round #369 (Div. 2) C. Coloring Trees] DP

[Codeforces Round #369 (Div. 2) C. Coloring Trees] DP

题目链接：[Codeforces Round #369 (Div. 2) C. Coloring Trees]
题意描述：给定N棵树，有M种颜料，每个树最初的颜色分别是c1,c2,…,cn(0≤ci≤M)，ci为0表示的是第i棵树没有还染色，现在要对所有没有染过色的树进行染色(即ci=0)，并且给定第i棵树染第j种颜料的费用是pij，在刚好将数段划分为K段的情况下求最小的花费， (1≤K≤N≤100,1≤M≤100)。
解题思路：
用dp[i][j][k]表示前i棵树染色并且最后一个的颜色是j，划分的数段数为k的最小花费。
那么，写出如下递推式：

dp[i][j][k]={min{dp[i−1][t][t==j?k:k−1]}min{dp[i−1][t][t==j?k:k−1]+pij},if ci!=0,otherwise

t表示上一棵树的颜色，负责度：O(N∗M2∗K)

#include <set>#include <stack>#include <queue>#include <cmath>#include <cstdio>#include <string>#include <cstring>#include <iostream>#include <algorithm>using namespace std;//#pragma comment(linker, "/STACK:1024000000,1024000000")#define FIN             freopen("input.txt","r",stdin)#define FOUT            freopen("output.txt","w",stdout)#define fst             first#define snd             second//typedef __int64 LL;typedef long long LL;typedef pair<int, int> PII;const LL INF = 0x3f3f3f3f3f3f3f3f;const int MAXN = 100 + 5;int N, M, K;LL C[MAXN], P[MAXN][MAXN];LL dp[MAXN][MAXN][MAXN];int main() {#ifndef ONLINE_JUDGE    FIN;#endif // ONLINE_JUDGE    while (~scanf ("%d %d %d", &N, &M, &K) ) {        for (int i = 1; i <= N; i++) {            scanf ("%I64d", &C[i]);        }        for (int i = 1; i <= N; i++) {            for (int j = 1; j <= M; j++) {                scanf ("%I64d", &P[i][j]);            }        }        memset (dp,  0x3f, sizeof (dp) );        if (C[1] == 0) {            for (int j = 1; j <= M; j++) {                dp[1][j][1] = P[1][j];            }        } else {            dp[1][C[1]][1] = 0;        }        for (int i = 2; i <= N; i++) {            if (C[i]) {                int j = C[i];                for (int k = 1; k <= K; k++) {                    for (int t = 1; t <= M; t++) {                        if (j == t)  dp[i][j][k] = min (dp[i][j][k], dp[i - 1][t][k]);                        else dp[i][j][k] = min (dp[i][j][k], dp[i - 1][t][k - 1]);                    }                }            } else {                for (int j = 1; j <= M; j++) {                    for (int k = 1; k <= K; k++) {                        if (C[i -  1]) {                            int t = C[i -  1];                            if (t == j) dp[i][j][k] = min (dp[i][j][k], dp[i - 1][t][k] + P[i][j]);                            else dp[i][j][k] = min (dp[i][j][k], dp[i - 1][t][k - 1] + P[i][j]);                        } else {                            for (int t = 1; t <= M; t++) {                                if (t == j) dp[i][j][k] = min (dp[i][j][k], dp[i - 1][t][k] + P[i][j]);                                else dp[i][j][k] = min (dp[i][j][k], dp[i - 1][t][k - 1] + P[i][j]);                            }                        }                    }                }            }        }        LL ans = INF;        for (int j = 1; j <= M; j++) {            ans = min (ans,  dp[N][j][K]);        }        if (ans == INF) ans = -1;        printf ("%I64d\n", ans);    }    return 0;}