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Description
Thanks a lot for helping Harry Potter in finding the Sorcerer‘s Stone of Immortality in October. Did we not tell you that it was just an online game ? uhhh! now here is the real onsite task for Harry. You are given a magrid S ( a magic grid ) having R rows and C columns. Each cell in this magrid has either a Hungarian horntail dragon that our intrepid hero has to defeat, or a flask of magic potion that his teacher Snape has left for him. A dragon at a cell (i,j) takes away |S[i][j]| strength points from him, and a potion at a cell (i,j) increases Harry‘s strength by S[i][j]. If his strength drops to 0 or less at any point during his journey, Harry dies, and no magical stone can revive him.
Harry starts from the top-left corner cell (1,1) and the Sorcerer‘s Stone is in the bottom-right corner cell (R,C). From a cell (i,j),
Harry can only move either one cell down or right i.e., to cell (i+1,j) or cell (i,j+1) and he can not move outside the magrid. Harry has used magic before starting his journey to determine which cell contains what, but lacks the basic simple mathematical skill to
determine what minimum strength he needs to start with to collect the Sorcerer‘s Stone. Please help him once again.
The first line contains the number of test cases T. T cases follow. Each test case consists of R C in the first line followed by the description of the grid
in R lines, each containing C integers. Rows are numbered 1 to R from top to bottom and columns are numbered 1 to C from left to right. Cells with
S[i][j] < 0 contain dragons, others contain magic potions.
Output T lines, one for each case containing the minimum strength Harry should start with from the cell (1,1) to have a positive strength through
out his journey to the cell (R,C).
1 ≤ T ≤ 5
2 ≤ R, C ≤ 500
-10^3 ≤ S[i][j] ≤ 10^3
S[1][1] = S[R][C] = 0
3
2 3
0 1 -3
1 -2 0
2 2
0 1
2 0
3 4
0 -2 -3 1
-1 4 0 -2
1 -2 -3 0
2
1
2
Case 1 : If Harry starts with strength = 1 at cell (1,1), he cannot maintain a positive strength in any possible path. He needs at least strength = 2 initially.
Case 2 : Note that to start from (1,1) he needs at least strength = 1.
Hint
| Added by: | Varun Jalan |
| Date: | 2011-12-15 |
| Time limit: | 0.336s |
| Source limit: | 50000B |
| Memory limit: | 1536MB |
| Cluster: | Cube (Intel G860) |
| Languages: | All |
| Resource: | Anil Kishore - ICPC Asia regionals, Amritapuri 2011 |
Source
http://acm.hust.edu.cn/vjudge/contest/view.action?cid=121703#problem/A
My Solution
训练的时候刚开始想到的是记忆化搜索， 但无论怎么优化还是TLE 3，没办法，想想递推怎么写
但是转化方程还是有点小问题， WA5
然后后来才想明白
只要 dp[i][j] = max(dp[i+1][j], dp[i][j+1]) + s[i][j];
if(dp[i][j] > 0) dp[i][j] = 0;
这里不要讨论s[i][j]的正负，都是直接加上s[i][j]就好了
然后处理好边界就好了
dp检查的时候应当着重与转移方程啊⊙﹏⊙‖∣
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int maxn = 500 + 8;
int s[maxn][maxn], r, c, dp[maxn][maxn];
//int pq; //!!!!!!在这些最小值里，找出最大的一个
/*
void dfs(int a, int b, int sum, int maxmin)
{
if(maxmin < pq) return;
if(a == r-1 && b == c - 1) {
pq = max(maxmin, pq);return;
}
if(a + 1 < r) dfs(a + 1, b, sum + s[a+1][b], maxmin = min(sum + s[a+1][b], maxmin));
if(b + 1 < c) dfs(a, b+1, sum + s[a][b+1], maxmin = min(sum + s[a][b+1], maxmin));
}
*/
int main()
{
#ifdef LOCAL
freopen("a.txt", "r", stdin);
//freopen("b.txt", "w", stdout);
#endif // LOCAL
int T;
scanf("%d", &T);
while(T--){
//memset(dp, 0x3f3f, sizeof dp);
scanf("%d%d", &r, &c);
for(int i = 0; i < r; i++){
for(int j = 0; j < c; j++){
scanf("%d", &s[i][j]);
}
}
/*TLE 5
pq = 10000000;int sumt = 0;
for(int i = 0; i < c; i++){
sumt += s[0][i];
pq = min(pq, sumt);
}
for(int j = 1; j < r; j++){
sumt += s[j][c-1];
pq = min(pq, sumt);
}
dfs(0, 0, s[0][0], 10000000);
*/
for(int i = r - 1; i >= 0; i--){
if(i == r - 1) dp[r-1][c-1] = 0;
else {
dp[i][c-1] = s[i][c-1] + dp[i+1][c-1];
if(dp[i][c-1] > 0) dp[i][c-1] = 0;
}
for(int j = c - 2; j >= 0; j--){
if(i != r - 1) {
dp[i][j] = max(dp[i+1][j], dp[i][j+1]) + s[i][j];
if(dp[i][j] > 0) dp[i][j] = 0;
}
else {
dp[i][j] = s[i][j] + dp[i][j+1];
if(dp[i][j] > 0) dp[i][j] = 0;
}
}
}
/*
for(int i = 0; i < r; i++){
for(int j = 0; j < c; j++){
printf("%d ", dp[i][j]);
}
printf("\n");
}
*/
if(dp[0][0] > 0)printf("1\n");
else {printf("%d\n", -dp[0][0]+1);}
}
return 0;
}
Thank you!
------from ProLights
UESTC 2016 Summer Training #2 Div.2 A dp、递推、多阶段问题
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原文地址：http://blog.csdn.net/prolightsfxjh/article/details/51893917
