FJUT2019暑假第二次周赛 A题

暑假第二次周赛 A题"/>

FJUT2019暑假第二次周赛 A题

# A题

服务器维护
TimeLimit:1500MS MemoryLimit:128MB
64-bit integer IO format:%lld
Problem Description
Dsc最大的梦想就是有一款属于自己的游戏（不可能的），可以把自己的奇思妙想在虚拟的世界中创造出来，假设Dsc成功了，创造出了一款网游（小作坊的那种），现在为了节省成本，Dsc决定自己维护服务器健康，但总会有没时间的时候。

假设从S分钟开始，E分钟结束[S,E]这段时间Dsc是没有时间的，这时候Dsc需要找人来维护服务器，当然是有偿的。在[S,E]这段时间中有n个人有空余时间（不一定是全部[S,E]）第i个人在ai,bi这段时间有空，共需要ci的管理费。

现在请帮Dsc算一算，在他没空的时间内[S,E]，每天都至少有一个人在管理服务器所需要的最少花费是多少，如果没有任何方案使[S,E]时间内每天都至少有一个人在管理服务器则输出-1;

Input
第一行为三个整数n,S,E分别表示有n个人有空，在[S,E]时间内Dsc没空

接下来有n行，每行3个整数ai,bi,ci分别表示第i个人在[ai,bi]时间内有空，雇佣共需要ci管理费

1<=n<=1e5

0<=S<=E<=1e5

S<=ai<=bi<=E

1<=ci<=1e5

Output
Dsc在没空的时间找人管理服务器所需的最少花费，没有方案则输出-1

SampleInput
3 2 4
2 3 2
4 4 1
2 4 4
SampleOutput
3
---

[思路1]：

其实这个题目就是一个DP加上线段树维护即可AC， 根据题意容易得知，需要区间全覆盖，那么我们可以考虑先按L从小到大排序维护从[S, R]所花费的最小价值，可得状态为dp[i]代表从[s, i]区间覆盖的最小价值。
可得状态转移方程为
dp[i] = min(dp[i], quert_min(l - 1, i - 1)) + w)
即可AC

附上代码

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long LL;
const int MAXN = 1e5 + 15;
const LL inf = __LONG_LONG_MAX__;
int n, s, e;
struct Segtree{LL arr[MAXN];struct NODE{int l, r;LL mins;}tree[MAXN << 2];void build(int root, int l, int r){tree[root].l = l, tree[root].r = r;if(l == r){tree[root].mins = arr[l];return ;}int mid = (l + r) >> 1;build(root << 1, l, mid);build(root << 1 | 1, mid + 1, r);tree[root].mins = min(tree[root << 1].mins, tree[root << 1 | 1].mins);return ;}void update(int root, int num, LL val){if(tree[root].l == num && tree[root].r == num){tree[root].mins = val;return ;}int mid = (tree[root].l + tree[root].r) >> 1;if(num <= mid){update(root << 1, num, val);}else{update(root << 1 | 1, num, val);}tree[root].mins = min(tree[root << 1].mins, tree[root << 1 | 1].mins);return ;}LL query(int root, int l, int r){if(tree[root].l >= l && tree[root].r <= r){return tree[root].mins;}int mid = (tree[root].l + tree[root].r) >> 1;if(r <= mid){return query(root << 1, l, r);}else if(l > mid){return query(root << 1 | 1, l, r);}else{return min(query(root << 1, l, mid), query(root << 1 | 1, mid + 1, r));}}
}seg;
struct Person{int l, r;LL value;friend bool operator< (const Person &a, const Person &b){if(a.l == b.l){return a.value < b.value;}else{return a.l < b.l;}}
}arr[MAXN];
LL dp[MAXN];
int main(){ios::sync_with_stdio(false);cin >> n >> s >> e;s += 2, e += 2;for(int i = 0; i < n; i ++){cin >> arr[i].l >> arr[i].r >> arr[i].value;arr[i].l += 2, arr[i].r += 2;}for(int i = s; i <= e; i ++){seg.arr[i] = inf;dp[i] = inf;}for(int i = 0; i < s; i ++){seg.arr[i] = 0;dp[i] = 0;}seg.build(1, 1, e);sort(arr, arr + n);for(int i = 0; i < n; i ++){if(seg.query(1, arr[i].l - 1, arr[i].r - 1) != inf){dp[arr[i].r] = min(dp[arr[i].r], seg.query(1, arr[i].l - 1, arr[i].r - 1) + arr[i].value);seg.update(1, arr[i].r, dp[arr[i].r]);    }}if(dp[e] == inf){cout << "-1\n";}else{cout << dp[e] << endl;}return 0;
}

[思路2]：

对于整个数据进行建图，将区间能互相覆盖的点连接起来，然后跑最短路径

附上代码：

#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <cctype>
#include <cstring>
#include <cmath>
#include <iostream>
#include <sstream>
#include <string>
#include <list>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <algorithm>
#include <functional>
#define lowbit(x) (x&(-x))
#define rep(i,a,n) for (int i=a;i<=n;i++)
#define per(i,a,n) for (int i=a;i>=n;i--)
#define mem(ar,num) memset(ar,num,sizeof(ar))
#define debug(x) cout << #x << ": " << x << endl
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
const int    prime = 999983;
const int    INF = 0x7FFFFFFF;
const LL     INFF =0x7FFFFFFFFFFFFFFF;
const double pi = acos(-1.0);
const double inf = 1e18;
const double eps = 1e-6;
const LL     mod = 1e9 + 7;
const int    maxn = 1e6 + 7;
const int    maxm = 4e6 + 7;int n, m, first[maxn], sign;struct Edge{int to, w, next;
}edge[maxn * 2];void init(){for(int i = 0; i <= n; i++){first[i] = -1;}sign = 0;
}void addEdge(int u,int v,int w){edge[sign].to = v;edge[sign].w = w;edge[sign].next = first[u];first[u] = sign++;
}struct Node{int to;LL cost;Node(){}Node(int tt, LL cc):to(tt),cost(cc){}friend bool operator < (const Node &a, const Node &b){return a.cost > b.cost;}
};LL dist[maxn];
int vis[maxn];LL dijkstra(int s, int t){for(int i = 0; i <= n; i++){dist[i] = INFF;vis[i] = 0;}priority_queue<Node> que;que.push(Node(s,0));while(!que.empty()){Node now = que.top();//printf("%d\n",now.to);que.pop();if(!vis[now.to]){vis[now.to] = 1;dist[now.to] = now.cost;for(int i = first[now.to];~i;i = edge[i].next){int to = edge[i].to;LL w = edge[i].w;if(!vis[to]){que.push(Node(to,now.cost + w));}}}}return dist[t] == INFF ? -1 : dist[t];
}
int main()
{int s, t;scanf("%d%d%d",&m,&s,&t);n = t + 1;init();for(int i = 1; i<= m; i++){int u, v, w;scanf("%d%d%d",&u,&v,&w);addEdge(u,v + 1,w);}for(int i = s + 1; i <= n; i++){addEdge(i ,i - 1, 0);}printf("%lld\n",dijkstra(s, t+1));return 0;
}

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