http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=4028
memset超时
这题竟然是一个差分约束
好吧呢
对于每一个a[i], l <= a[i] <= r
那么设一个源点s
使 l <= a[i] - s <= r 是不是就能建边了
然后对于每一个f[i]
如果前面有一个相等的f[j]
则肯定 a[i] <= a[j] 又能建边了
根据LIS的传递关系
对于每个f[i] 肯定是由上一个等级的传递过来的
即 a[i] > a[j] 是不是又能建边了
不会建边？
请移步： https://www.cnblogs.com/WTSRUVF/p/9153758.html
#include <iostream> #include <cstdio> #include <sstream> #include <cstring> #include <map> #include <cctype> #include <set> #include <vector> #include <stack> #include <queue> #include <algorithm> #include <cmath> #include <bitset> #define rap(i, a, n) for(int i=a; i<=n; i++) #define rep(i, a, n) for(int i=a; i<n; i++) #define lap(i, a, n) for(int i=n; i>=a; i--) #define lep(i, a, n) for(int i=n; i>a; i--) #define rd(a) scanf("%d", &a) #define rlld(a) scanf("%lld", &a) #define rc(a) scanf("%c", &a) #define rs(a) scanf("%s", a) #define rb(a) scanf("%lf", &a) #define rf(a) scanf("%f", &a) #define pd(a) printf("%d\n", a) #define plld(a) printf("%lld\n", a) #define pc(a) printf("%c\n", a) #define ps(a) printf("%s\n", a) #define MOD 2018 #define LL long long #define ULL unsigned long long #define Pair pair<int, int> #define mem(a, b) memset(a, b, sizeof(a)) #define _ ios_base::sync_with_stdio(0),cin.tie(0) //freopen("1.txt", "r", stdin); using namespace std; const int maxn = 110000, INF = 0x7fffffff; int head[maxn], vis[maxn]; LL d[maxn]; int cnt; int n, m, s; int ans[maxn];struct node {int u, v, next;int w; }Node[1000500];void add(int u, int v, int w) {Node[cnt].u = u;Node[cnt].v = v;Node[cnt].w = w;Node[cnt].next = head[u];head[u] = cnt++; }void init() {for(int i = 0; i <= n + 1; i++){head[i] = -1;ans[i] = 0;vis[i] = 0;d[i] = INF;}cnt = 0; }bool spfa() {deque<int> Q;Q.push_front(s);d[s] = 0;vis[s] = 1;while(!Q.empty()){int u = Q.front(); Q.pop_front();vis[u] = 0;for(int i = head[u]; i != -1; i = Node[i].next){int v = Node[i].v;if(d[v] > d[u] + Node[i].w){d[v] = d[u] + Node[i].w;if(!vis[v]){if(Q.empty()) Q.push_front(v);else if(d[v] < d[Q.front()]) Q.push_front(v);else Q.push_back(v);vis[v] = 1;if(++ans[v] > n) return 1;}}}}return 0; }int pre[maxn];int main() {int T;rd(T);while(T--){rd(n);init();s = n + 1;mem(pre, 0);for(int i = 1; i <= n; i++){int f;rd(f);if(pre[f]) add(pre[f], i, 0);if(f > 0) add(i, pre[f - 1], -1);pre[f] = i;}for(int i = 1; i <= n; i++){int l, r;rd(l), rd(r);add(s, i, r);add(i, s, -l);}spfa();for(int i = 1; i <= n; i++){if(i != 1) printf(" ");printf("%lld", d[i]);printf("\n");}}return 0; }
