AtCoder Regular Contest 105 C - Camels and Bridge

题目链接
题目描述
有n只骆驼过m桥，第i只骆驼负重wiw_iwi，第i座桥的长度为lil_ili,最大负重为viv_ivi，问骆驼队伍长度最短为多少可以通过所有的桥。
思路
n的数据范围很小，可以暴力解决。根据桥的长度从大到小预处理一下其能承受的重量。
dp[i]dp[i]dp[i]表示从第一只骆驼到第i只骆驼的最短距离。
dp[i]=max(dp[j]+len,dp[i])dp[i] = max(dp[j] + len, dp[i])dp[i]=max(dp[j]+len,dp[i])，len表示第j只骆驼到第i只骆驼的最短距离。寻找len的过程中就是在预处理的桥中二分一下第j只到第i只骆驼的重量所返回的最短桥的长度。
代码

#include<bits/stdc++.h>
using namespace std;typedef long long LL;
#define int LL
const int N = 1e5 + 10;int w[10];
bool vis[10];
int b[10], s[10], dp[10];
struct node {int l, v;friend bool operator < (node a, node b) {if(a.l == b.l) {return a.v < b.v;}return a.l < b.l;}
}a[N];
int n, m;
LL res;int find(int x) {int l = 1, r = m;int res = 0;while(l <= r) {int mid = l + r >> 1;if(x > a[mid].v) {res = a[mid].l;l = mid + 1;} else r = mid - 1;}return res;
}void work() {memset(dp, 0, sizeof dp);for(int i = 1; i <= n; i++) {s[i] = s[i - 1] + w[b[i]];}for(int i = 2; i <= n; i++) {for(int j = i - 1; j; j--) {int tmp = s[i] - s[j - 1];int len = find(tmp);dp[i] = max(dp[i], dp[j] + len);}}res = min(res, dp[n]);
}void dfs(int cnt) {if(cnt == n) {work();return;}for(int i = 1; i <= n; i++) {if(!vis[i]) {vis[i] = true;b[cnt + 1] = i;dfs(cnt + 1);vis[i] = false;}}
}void solve() {scanf("%lld%lld", &n, &m);for(int i = 1; i <= n; i++) {scanf("%lld", &w[i]);}for(int i = 1; i <= m; i++) {scanf("%lld%lld", &a[i].l, &a[i].v);}sort(a + 1, a + 1 + m);int mn = 1e9;for(int i = m; i >= 1; i--) {mn = min(mn, a[i].v);a[i].v = mn;}for(int i = 1; i <= n; i++) {if(mn < w[i]) {puts("-1");return;}}res = 2e18;dfs(0);printf("%lld\n", res);
}signed main() {//    freopen("in.txt", "r", stdin);
//    int t; scanf("%d", &t); while(t--)solve();
}

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