Time Limit: 1000MS | | Memory Limit: 65535KB | | 64bit IO Format: %lld & %llu |
Description
Dreams of finding lost treasure almost came true recently. A new machine called 'The Revealer' has been invented and it has been used to detect gold which has been buried in the ground. The machine was used in a cave near the seashore where – it is said – pirates used to hide gold. The pirates would often bury gold in the cave and then fail to collect it. Armed with the new machine, a search party went into the cave hoping to find buried treasure. The leader of the party was examining the soil near the entrance to the cave when the machine showed that there was gold under the ground. Very excited, the party dug a hole two feel deep. They finally found a small gold coin which was almost worthless. The party then searched the whole cave thoroughly but did not find anything except an empty tin trunk. In spite of this, many people are confident that 'The Revealer' may reveal something of value fairly soon.
So,now you are in the point
(1,1) and initially you have 0 gold.In the
n*
m grid there are some traps and you will lose gold.If your gold is not enough you will be die.And there are some treasure and you will get gold.If you are in the point(x,y),you can only walk to point
(x+1,y),(x,y+1),(x+1,y+2)and
(x+2,y+1).Of course you can not walk out of the grid.Tell me how many gold you can get most in the trip.It`s guarantee that
(1,1)is not a trap;
Input
first come
2 integers,
n,m(
1≤n≤1000,
1≤m≤1000)Then follows
n lines with
m numbers
aij
(−100<=aij<=100)
the number in the grid means the gold you will get or lose.
Output
print how many gold you can get most.
Sample Input
3 3
1 1 1
1 -5 1
1 1 1
3 3
1 -100 -100
-100 -100 -100
-100 -100 -100
Sample Output
5
//题意：如果你在位置（x,y）,你下一步只能向(x+1,y),(x,y+1),(x+1,y+2),(x+2,y+1)这四个方向走。
给你一个n*m的矩阵，每个位置都有一个宝藏，正的表示你可以获利那么多的宝藏，
负的表示你要消耗这么多的宝藏，当你的宝藏为负的时候，你就不能再继续向前了。
问在你探寻宝藏的过程中，你获得的宝藏的值最大是多少？
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<algorithm>
#include<iostream>
#define INF 0x3f3f3f3f
#define ull unsingned long long
#define ll long long
#define IN __int64
#define N 1010
#define M 1000000007
using namespace std;
int map[N][N],dp[N][N];
int dx[4]={1,0,1,2};
int dy[4]={0,1,2,1};
int main()
{
int t,n,m;
int i,j,k;
int x,y;
while(scanf("%d%d",&n,&m)!=EOF)
{
memset(dp,-1,sizeof(dp));
memset(map,0,sizeof(map));
for(i=1;i<=n;i++)
{
for(j=1;j<=m;j++)
{
scanf("%d",&map[i][j]);
}
}
int mm=0;
dp[1][1]=map[1][1];
for(i=1;i<=n;i++)
{
for(j=1;j<=m;j++)
{
if(dp[i][j]<0)
continue;
mm=max(dp[i][j],mm);
for(k=0;k<4;k++)
{
x=i+dx[k];
y=j+dy[k];
if(x<1||x>n||y<1||y>m)
continue;
dp[x][y]=max(dp[i][j]+map[x][y],dp[x][y]);
}
}
}
printf("%d\n",mm);
}
return 0;
}
