UVALive 3530 Martian Mining
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一个n*m的矩形,每个点有两种矿,A矿往W方向,B矿往N方向,每个格子上建两种运送带,一种方向是w的,一种是n的,要求一个格子最多只能有一种带,而且传送带不能断,之后能运到边界才算有效,而且不能曲折,求收集到的A,B总量最大
设dp[ i ][ j ] 为前i行,前j列的最大和,每个格子只可能有一种方向的传送带,要么横,要么竖,那么dp[i][j]=max(dp[i-1][j]+sumr[i][j],dp[i][j-1]+sumc[i][j]);
#include <cstdio>#include <cstring>#include <cstdlib>#include <cmath>#include <iostream>#include <vector>#include <algorithm>#include <queue>using namespace std;#define clr(a, x) memset(a, x, sizeof(a))#define rep(i, n) for (int i = 0; i < (int)(n); i++)#define REP(i,a,b) for(int i=a;i<=(b);i++)typedef long long lld;const int maxn = 550;int dp[maxn][maxn];int sumr[maxn][maxn],sumc[maxn][maxn];int main(){ int n,m; while(~scanf("%d%d",&n,&m)&&(n||m)) { REP(i,1,n) { sumr[i][0]=0; REP(j,1,m) { int a; scanf("%d",&a); sumr[i][j]=sumr[i][j-1]+a; } } REP(i,1,n) { //sum[0][0]=0; REP(j,1,m) { sumc[0][j]=0; int a; scanf("%d",&a); sumc[i][j]=sumc[i-1][j]+a; } } clr(dp,0); REP(i,1,n) REP(j,1,m) dp[i][j]=max(dp[i-1][j]+sumr[i][j],dp[i][j-1]+sumc[i][j]); printf("%d\n",dp[n][m]); } return 0;}
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