首页 > 代码库 > HDU1384 Intervals 【差分约束系统】
HDU1384 Intervals 【差分约束系统】
Intervals
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2880 Accepted Submission(s): 1048
Problem Description
You are given n closed, integer intervals [ai, bi] and n integers c1, ..., cn.
Write a program that:
> reads the number of intervals, their endpoints and integers c1, ..., cn from the standard input,
> computes the minimal size of a set Z of integers which has at least ci common elements with interval [ai, bi], for each i = 1, 2, ..., n,
> writes the answer to the standard output
Write a program that:
> reads the number of intervals, their endpoints and integers c1, ..., cn from the standard input,
> computes the minimal size of a set Z of integers which has at least ci common elements with interval [ai, bi], for each i = 1, 2, ..., n,
> writes the answer to the standard output
Input
The first line of the input contains an integer n (1 <= n <= 50 000) - the number of intervals. The following n lines describe the intervals. The i+1-th line of the input contains three integers ai, bi and ci separated by single spaces and such that 0 <= ai <= bi <= 50 000 and 1 <= ci <= bi - ai + 1.
Process to the end of file.
Process to the end of file.
Output
The output contains exactly one integer equal to the minimal size of set Z sharing at least ci elements with interval [ai, bi], for each i = 1, 2, ..., n.
Sample Input
5 3 7 3 8 10 3 6 8 1 1 3 1 10 11 1
Sample Output
6
Author
1384
#include <stdio.h> #include <queue> #include <string.h> #define maxn 50010 #define inf 0x3f3f3f3f using std::queue; int head[maxn], id, dist[maxn]; int left, right; struct Node{ int to, w, next; } E[maxn * 3]; bool vis[maxn]; void addEdge(int u, int v, int c) { E[id].to = v; E[id].w = c; E[id].next = head[u]; head[u] = id++; } int SPFA() { memset(dist, 0x3f, sizeof(dist)); memset(vis, 0, sizeof(vis)); int i, u, v, tmp; u = right; vis[u] = 1; dist[u] = 0; queue<int> Q; Q.push(u); while(!Q.empty()){ u = Q.front(); Q.pop(); vis[u] = 0; for(i = head[u]; i != -1; i = E[i].next){ tmp = dist[u] + E[i].w; v = E[i].to; if(tmp < dist[v]){ dist[v] = tmp; if(!vis[v]){ vis[v] = true; Q.push(v); } } } } return -dist[left]; } int main() { int n, a, b, c, i; while(scanf("%d", &n) == 1){ memset(head, -1, sizeof(head)); left = inf; right = 0; for(i = id = 0; i < n; ++i){ scanf("%d%d%d", &a, &b, &c); addEdge(b + 1, a, -c); if(left > a) left = a; if(right < b) right = b; } ++right; for(i = left; i <= right; ++i){ addEdge(i, i + 1, 1); addEdge(i + 1, i, 0); } printf("%d\n", SPFA()); } return 0; }
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