UVA - 1632 Alibaba 区间dp

题意：给定n个点，其中第i个点的坐标是xi，且它会在di秒后消失。Alibaba可以从任意位置出发，求访问完所有点的最短时间。无解输出No solution。

思路：dp(i,j,0)表示访问完区间[i,j]后停留在i点的最短时间，dp(i,j,1)表示访问完区间[i,j]后停留在j点的最短时间。转移方程如下:

dp[i][j][0] = min(dp[i+1][j][0]+dis[i+1]-dis[i], dp[i+1][j][1]+dis[j]-dis[i]);
dp[i][j][1] = min(dp[i][j-1][0]+dis[j]-dis[i], dp[i][j-1][1]+dis[j]-dis[j-1]);

AC代码

#include <cstdio>
#include <cmath>
#include <cctype>
#include <bitset>
#include <algorithm>
#include <cstring>
#include <utility>
#include <string>
#include <iostream>
#include <map>
#include <set>
#include <vector>
#include <queue>
#include <stack>
using namespace std;
#pragma comment(linker, "/STACK:1024000000,1024000000")
#define eps 1e-10
#define inf 0x3f3f3f3f
#define PI pair<int, int>
typedef long long LL;
const int maxn = 1e4 + 5;
int dp[maxn][maxn][2]; //0-stop at i 1-stop at j
int dis[maxn], dead[maxn];
int main() {
int n;
while(scanf("%d", &n) == 1) {
for(int i = 0; i < n; ++i) {
scanf("%d%d", &dis[i], &dead[i]);
}
memset(dp, 0, sizeof(dp));
for(int i = n-1; i >= 0; --i)
for(int j = i+1; j < n; ++j) {
dp[i][j][0] = min(dp[i+1][j][0]+dis[i+1]-dis[i], dp[i+1][j][1]+dis[j]-dis[i]);
if(dp[i][j][0] >= dead[i]) dp[i][j][0] = inf;
dp[i][j][1] = min(dp[i][j-1][0]+dis[j]-dis[i], dp[i][j-1][1]+dis[j]-dis[j-1]);
if(dp[i][j][1] >= dead[j]) dp[i][j][1] = inf;
}
int res = min(dp[0][n-1][0], dp[0][n-1][1]);
if(res == inf) printf("No solution\n");
else printf("%d\n", res);
}
return 0;
}

如有不当之处欢迎指出！