题解:dp[i]代表着在第i把伞换伞的最小代价,状态转移方程就是dp[i]=min(dp[j]+(a[i].x - a[j].x) * a[j].w) (第i把伞没有在雨中的情况) dp[i]=min(dp[i],dp[j] + (l - a[j].x) * a[j].w )(第i把伞在雨中的转移方程,l的含义详见代码) 最后 ans = min(dp[i] + (mx - a[i].x) * a[i].w, ans); mx为最远的下雨点
#include<bits/stdc++.h>
using namespace std;
const int maxn = 2005;
const long long INF = 1e11 + 5;
int vis[maxn];
struct Node
{
long long x;
long long w;
bool operator < (const Node &p)const
{
return x < p.x;
}
}a[maxn];
long long dp[maxn];
int main()
{
int c, m, n;
cin >> c >> n >> m;
int mx = -1;
for(int i = 1; i <= n; i++)
{
int l, r;
cin >> l >> r;
for(int j = l + 1; j <= r; j++)
{
vis[j] = i;
}
mx = max(r, mx);
}
for(int i = 0; i < m; i++)
{
cin >> a[i].x >> a[i].w;
dp[i] = INF;
}
sort(a, a + m);
long long ans = 1e11 + 5;
for(int i = 0; i < m; i++)
{
if(a[i].x > mx)continue;
int l = -1;
if(!vis[a[i].x])
{
for(int j = a[i].x; j >=0 ;j--)
{
if(vis[j])
{
l = j;
break;
}
}
if(l == -1)
{
dp[i] = 0;
ans = min(dp[i] + (mx - a[i].x) * a[i].w, ans);
//cout << dp[i] << endl;
continue;
}
}
for(int j = 0; j < i; j++)
{
if(dp[j] == INF)continue;
if(vis[a[i].x])
{
dp[i]=min(dp[i],dp[j] + (a[i].x - a[j].x) * a[j].w );
}
else
{
if(l < a[j].x)continue;
dp[i]=min(dp[i],dp[j] + (l - a[j].x) * a[j].w );
}
}
//cout << l << " " << dp[i] << endl;
ans = min(dp[i] + (mx - a[i].x) * a[i].w, ans);
}
if(ans!=INF)cout << ans << endl;
else cout << "-1" << endl;
return 0;
}