题意：找一条从顶部到底部的一条路径，往左下或右下走，使得经过的数字和最大。
分析：递推的经典题目，自底向上递推。当状态保存在a[n][j]时可省去dp数组，空间可优化。
代码1：
/************************************************
* Author :Running_Time
* Created Time :2015-9-1 11:17:04
* File Name :POJ_1163.cpp
************************************************/ #include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std; #define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int N = 1e2 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
int a[N][N];
int dp[N][N]; int main(void) {
int n;
while (scanf ("%d", &n) == 1) {
for (int i=1; i<=n; ++i) {
for (int j=1; j<=i; ++j) {
scanf ("%d", &a[i][j]);
if (i == n) dp[i][j] = a[i][j];
}
}
for (int i=n-1; i>=1; --i) {
for (int j=1; j<=i; ++j) {
dp[i][j] = max (dp[i+1][j], dp[i+1][j+1]) + a[i][j];
}
}
printf ("%d\n", dp[1][1]);
} return 0;
}
代码2(空间优化)：
/************************************************
* Author :Running_Time
* Created Time :2015-9-1 11:18:25
* File Name :POJ_1163_2.cpp
************************************************/ #include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std; #define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int N = 1e2 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
int a[N][N]; int main(void) {
int n;
while (scanf ("%d", &n) == 1) {
for (int i=1; i<=n; ++i) {
for (int j=1; j<=i; ++j) {
scanf ("%d", &a[i][j]);
}
}
for (int i=n-1; i>=1; --i) {
for (int j=1; j<=i; ++j) {
a[n][j] = max (a[n][j], a[n][j+1]) + a[i][j];
}
}
printf ("%d\n", a[n][1]);
} return 0;
}