Codeforces Round #369 (Div. 2) C. Coloring Trees(dp)
2016-08-30 16:49
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题目链接:Codeforces Round #369 (Div. 2) C. Coloring Trees
题意:
有n个树,每个树有一个颜色,如果颜色值为0,表示没有颜色,一共有m个颜色,第j种颜色涂第i棵树需要花费pij,颜色一样且相邻的分为一组
现在要将所有颜色为0的树涂上颜色,使得这些树恰好可以分为k组,问最小的花费
题解:
考虑dp[i][j][k],表示考虑第i棵树涂第j种颜色,当前分为k组的最小花费,然后状态转移看代码,注意的是dp的初始状态
#include<bits/stdc++.h> #define F(i,a,b) for(int i=a;i<=b;++i) using namespace std; typedef long long ll; int n,m,d; ll dp[111][111][111],v[101][101],inf=1e18; int dt[101]; inline void up(ll &a,ll b){if(a>b)a=b;} int main(){ scanf("%d%d%d",&n,&m,&d); F(i,1,n)scanf("%d",dt+i); F(i,1,n)F(j,1,m)scanf("%lld",&v[i][j]); F(i,0,n)F(j,0,m)F(k,0,d)dp[i][j][k]=inf; if(dt[1])dp[1][dt[1]][1]=0; else F(i,1,m)dp[1][i][1]=v[1][i]; F(i,2,n)F(j,1,m)F(k,1,d)if(dp[i-1][j][k]<inf) { if(dt[i])up(dp[i][dt[i]][k+(dt[i]!=j)],dp[i-1][j][k]); else F(ii,1,m)up(dp[i][ii][k+(ii!=j)],dp[i-1][j][k]+v[i][ii]); } ll ans=inf; F(i,1,m)up(ans,dp [i][d]); if(ans==inf)puts("-1"); else printf("%lld\n",ans); return 0; }View Code
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