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ZOJ 2972 Hurdles of 110m(DP)

2016-02-19 17:39 459 查看
题目链接:重点内容http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=2972

跨栏跑分为几段,每段有三种跑步方式,不同方式通过每段路程时间不同,并消耗或补充体力,要求输出最短时间。

二维DP

#include<iostream>
#include<cstdio>
#include<set>
#include<string>
#include<string.h>
#include<cstring>
#include<vector>
#include<map>
#include<queue>
#include<stack>
#include<cctype>
#include<algorithm>
#define mt(a) memset(a,0,sizeof a)
#define fl(a,b,c) fill(a,b,c)

#define inf 1000000000+7

using namespace std;

typedef long long ll;
int ans[200];

int dp[200][500];

int main()
{
int T;
cin >> T;
while (T--)
{
int n, m;
scanf("%d %d", &n, &m);
for (int i = 1; i < 200; i++)
memset(dp[i], inf, sizeof (dp[i]));
memset(dp[0], 0, sizeof(dp[0]));
for (int i = 1; i <= n; i++)
{
int t1, t2, t3, f1, f2;
scanf("%d %d %d %d %d", &t1, &t2, &t3, &f1, &f2);
for (int j = 0; j <= m; j++)
{
if (j >= f1&&dp[i - 1][j] != inf)dp[i][j-f1] = min(dp[i][j-f1], dp[i-1][j] + t1);
dp[i][j] = min(dp[i][j], dp[i - 1][j] + t2);
if (j + f2 >= m)dp[i][m] = min(dp[i][m], dp[i - 1][j] + t3);
else dp[i][j + f2] = min(dp[i][j + f2], dp[i - 1][j] + t3);
}
}
int ans = inf;
for (int i = 0; i <= m; i++)
{
ans = min(ans, dp
[i]);
}
printf("%d\n", ans);
}
return 0;
}
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