POJ训练计划1201_Intervals(差分约束)
2014-08-22 21:19
344 查看
解题报告
题目传送门
思路:
解方程组:
(a-1)-b<-c
0<=i-(i-1)<=1
Max-Min>=m
源点为Max,求出dis[Max]-dis[Min]
#include <iostream>
#include <cstring>
#include <cstdio>
#include <deque>
#include <stack>
#define N 100001
#define M 550000
#define inf 0x3f3f3f3f
using namespace std;
int head
,dis
,vis
;
int cnt,n,m;
int maxx=0,minn=inf;
struct node {
int v,w,next;
} edge[M];
void add(int u,int v,int w) {
edge[cnt].v=v,edge[cnt].w=w;
edge[cnt].next=head[u],head[u]=cnt++;
}
void SF() {
stack<int>Q;
memset(dis,inf,sizeof(dis));
memset(vis,0,sizeof(vis));
dis[maxx]=0,vis[maxx]=1;
Q.push(maxx);
while(!Q.empty()) {
int u=Q.top();
Q.pop();
vis[u]=0;
for(int i=head[u]; i!=-1; i=edge[i].next) {
int v=edge[i].v;
if(dis[v]>dis[u]+edge[i].w) {
dis[v]=dis[u]+edge[i].w;
if(!vis[v]) {
vis[v]=1;
Q.push(v);
}
}
}
}
}
int main() {
int i,j,u,v,w;
while(~scanf("%d",&n)) {
cnt=0;
memset(head,-1,sizeof(head));
for(i=1; i<=n; i++) {
scanf("%d%d%d",&u,&v,&w);
if(v>maxx)
maxx=v;
if(u<minn)
minn=u;
add(v,u-1, -w);
}
for(i=1; i<=maxx; i++) {
add(i-1,i,1);
add(i,i-1,0);
}
SF();
printf("%d\n",dis[maxx]-dis[minn-1]);
}
return 0;
}
Intervals
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 end points 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 <= 50000) -- 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 <= 50000 and 1 <= ci <= bi - ai+1.
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
Sample Output
题目传送门
思路:
解方程组:
(a-1)-b<-c
0<=i-(i-1)<=1
Max-Min>=m
源点为Max,求出dis[Max]-dis[Min]
#include <iostream>
#include <cstring>
#include <cstdio>
#include <deque>
#include <stack>
#define N 100001
#define M 550000
#define inf 0x3f3f3f3f
using namespace std;
int head
,dis
,vis
;
int cnt,n,m;
int maxx=0,minn=inf;
struct node {
int v,w,next;
} edge[M];
void add(int u,int v,int w) {
edge[cnt].v=v,edge[cnt].w=w;
edge[cnt].next=head[u],head[u]=cnt++;
}
void SF() {
stack<int>Q;
memset(dis,inf,sizeof(dis));
memset(vis,0,sizeof(vis));
dis[maxx]=0,vis[maxx]=1;
Q.push(maxx);
while(!Q.empty()) {
int u=Q.top();
Q.pop();
vis[u]=0;
for(int i=head[u]; i!=-1; i=edge[i].next) {
int v=edge[i].v;
if(dis[v]>dis[u]+edge[i].w) {
dis[v]=dis[u]+edge[i].w;
if(!vis[v]) {
vis[v]=1;
Q.push(v);
}
}
}
}
}
int main() {
int i,j,u,v,w;
while(~scanf("%d",&n)) {
cnt=0;
memset(head,-1,sizeof(head));
for(i=1; i<=n; i++) {
scanf("%d%d%d",&u,&v,&w);
if(v>maxx)
maxx=v;
if(u<minn)
minn=u;
add(v,u-1, -w);
}
for(i=1; i<=maxx; i++) {
add(i-1,i,1);
add(i,i-1,0);
}
SF();
printf("%d\n",dis[maxx]-dis[minn-1]);
}
return 0;
}
Intervals
Time Limit: 2000MS | Memory Limit: 65536K | |
Total Submissions: 21736 | Accepted: 8180 |
You are given n closed, integer intervals [ai, bi] and n integers c1, ..., cn.
Write a program that:
reads the number of intervals, their end points 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 <= 50000) -- 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 <= 50000 and 1 <= ci <= bi - ai+1.
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
相关文章推荐
- [差分约束]POJ 1201——Intervals
- poj 1201 Intervals( spfa + 差分约束)
- |poj 1201|差分约束|Intervals
- HDU 1384 && POJ 1201--Intervals 【基础差分约束】
- poj 1201 Intervals(第一道差分约束题)
- POJ 1201 Intervals (差分约束)
- POJ 1201 Intervals 差分约束
- poj 1201 Intervals(差分约束,最长路)
- POJ 1201 Intervals (差分约束)
- poj-1201- Intervals-差分约束问题
- POJ 1201 Intervals (差分约束)
- poj 1201 Intervals(差分约束)
- POJ1201 Intervals 【差分约束】
- POJ 1201 Intervals(差分约束)
- POJ 1201 Intervals(图论-差分约束)
- POJ 1201 Intervals(图论-差分约束)
- [POJ 1201]Intervals[差分约束]
- poj 1201 Intervals [差分约束]
- poj 1201 Intervals(差分约束)
- poj 1201 Intervals(差分约束)