Watermelon Full of Water [ZOJ 4778]

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=4778

优先队列优化下的DP,dp[i]记录的是吃到i天的最小花费,最后dp[N]即为答案。

const int MM = 111111;
#define debug puts("wrong")
typedef long long int64;
//typedef __int64 int64;
int64 N, cnt;
struct Info{int64 x,y;}p[MM];
struct Tpoint{
    int64 val,dead;
    //Tpoint(int v,int d):val(v),dead(d) {}
    bool friend operator<(Tpoint x,Tpoint y) {
        if(x.val!=y.val) return x.val>y.val;
        else return x.dead>y.dead;
    }
};
priority_queue<Tpoint>que;

void get_data() {
    int64 i,j,k;
    for(i=1;i<=N;i++) scanf("%lld",&p[i].x);
    for(i=1;i<=N;i++) scanf("%lld",&p[i].y);
}
int64 dp[MM];
void solve() {
    int64 i,j,k,tmp, mx;
    Tpoint tt, t1, t2;
    while(!que.empty()) que.pop();
    t1.val=p[1].x, t1.dead=p[1].y;
    dp[1]=p[1].x;
    que.push(t1);
    for(i=2;i<=N;i++) {                
            t1.val=dp[i-1]+p[i].x;
            t1.dead=i+p[i].y-1;    
            que.push(t1);
        while(!que.empty() && que.top().dead<i) que.pop();
        dp[i]=que.top().val;
    }
    printf("%lld\n", dp[N]);
}
int main() {
    while(scanf("%lld",&N)!=EOF) get_data(),solve();
    return 0;
}