题意:给定n个区间[ai, bi],每个区间给出一个数ci,求一个整数集合s使得第i个区间里至少有ci个集合s中的数,并且s尽量小。
题解:差分约束
设 f ( i ) f(i) f(i)表示集合s包含了 [ 0 , i − 1 ] [0, i-1] [0,i−1]中的数的数量(因为起点是从0开始的)。
即可得约束:
f ( b + 1 ) − f ( a ) > = c f(b+1)-f(a)>=c f(b+1)−f(a)>=c :(<a, b+1>=c)
f ( a ) − f ( a − 1 ) > = 0 f(a)-f(a-1)>=0 f(a)−f(a−1)>=0
f ( a − 1 ) − f ( a ) > = − 1 f(a-1)-f(a)>=-1 f(a−1)−f(a)>=−1,即单个点要么被包含要么不被包含。
然后跑spfa最长路即可, d [ r + 1 ] d[r+1] d[r+1]即最小的集合,注意端点,还有就是多测。
#define _CRT_SECURE_NO_WARNINGS
#include<iostream>
#include<cstdio>
#include<string>
#include<cstring>
#include<algorithm>
#include<queue>
#include<stack>
#include<cmath>
#include<vector>
#include<fstream>
#include<set>
#include<map>
#include<sstream>
#include<iomanip>
#define ll long long
#define pii pair<int, int>
using namespace std;
const int maxn = 5e4 + 5;
int n, a[maxn], b[maxn], c[maxn], l, r;
struct node {
int v, nxt, w;
}edge[maxn << 2];
int vis[maxn], d[maxn], head[maxn], k, s;
void add(int u, int v, int w) {
edge[++k].nxt = head[u];
edge[k].v = v;
edge[k].w = w;
head[u] = k;
}
void spfa() {
for (int i = l; i <= r + 1; i++)
vis[i] = 0, d[i] = -0x3f3f3f3f;
queue<int>q;
q.push(s);
vis[s] = 1;
d[s] = 0;
while (!q.empty()) {
int u = q.front(); q.pop();
vis[u] = 0;
for (int i = head[u]; i; i = edge[i].nxt) {
int v = edge[i].v, w = edge[i].w;
if (d[v] < d[u] + w) {
d[v] = d[u] + w;
if (vis[v]) continue;
vis[v] = 1;
q.push(v);
}
}
}
}
int main() {
while (~scanf("%d", &n)) {
k = 0;
memset(head, 0, sizeof(head));
l = maxn;
r = 0;
for (int i = 1; i <= n; i++) {
scanf("%d%d%d", &a[i], &b[i], &c[i]);
l = min(l, a[i]);
r = max(r, b[i]);
}
for (int i = 1; i <= n; i++) {
add(a[i], b[i] + 1, c[i]);
}
for (int i = l + 1; i <= r + 1; i++) {
add(i - 1, i, 0);
add(i, i - 1, -1);
}
s = l;
spfa();
printf("%d\n", d[r + 1]);
}
return 0;
}