poj 1201 Intervals（差分约束）
2014-06-18 10:59:37      个评论    来源：poj 1201 Intervals（差分约束）

Intervals
 Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 20775 Accepted: 7859

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

```5
3 7 3
8 10 3
6 8 1
1 3 1
10 11 1```

Sample Output

`6`

`题意：告诉你n个区间，要在第i区间至少取ci个数，问所有区间总共至少去多少个数？`
`这题的构图很是巧妙，把每个边界看成节点，Xi表示0-i有多少个数被选上，则每个区间[li,ri]可以推出X[ri]-X[li-1]>=Ci，即：X[li-1]-X[ri]<=-Ci。`
`潜在不等式：`
`1、X[i]-X[i+1]<=0;`
`2、X[i+1]-X[i]<=0.`
`本题我多了一个离散化操作。`
```#include
#include
#include
#include
#include
#include
#include
using namespace std;

const int INF = 0x3f3f3f3f;
const int maxn = 2000100;
struct edge{
int u , v , d;
edge(int a = 0 , int b = 0 , int c = 0){
u = a , v = b , d = c;
}
}e[maxn];
int head[maxn] , next[maxn] , cnt , n;
int dis[maxn] , vis[maxn] , vt[maxn] , index;
map mp;
struct interval{
int l , r , value;
interval(int a = 0 , int b = 0 , int c = 0){
l = a , r = b , value = c;
}
};
vector I;

void add(int u , int v , int d){
e[cnt] = edge(u , v , d);
}

void initial(){
I.clear();
for(int i = 0; i < maxn; i++){
dis[i] = INF;
vis[i] = 0;
vt[i] = 0;
}
cnt = 0;
index = 0;
mp.clear();
}

int a , b , c;
for(int i = 0; i < n; i++){
scanf("%d%d%d" , &a , &b , &c);
b++;
I.push_back(interval(a , b , c));
mp[a] = 0;
mp[b] = 0;
}
map::iterator it;
for(it = mp.begin(); it != mp.end(); it++){
it -> second = index++;
}
it = mp.begin();
for(int i = 0; i < index-1; i++){
int tem = it->first;
it++;
add(i , i+1 , it->first - tem);
}
}

bool SPFA(int start){
queue q;
q.push(start);
vt[start]++;
while(!q.empty()){
int u = q.front();
vis[u] = 0;
if(vt[u]>index) return false;
q.pop();
while(Next != -1){
int v = e[Next].v , d = e[Next].d;
if(dis[v] > dis[u]+d){
dis[v] = dis[u]+d;
if(vis[v] == 0){
vis[v] = 1;
vt[v]++;
q.push(v);
}
}
Next = next[Next];
}
}
return true;
}

void computing(){
for(int i = 0; i < I.size(); i++){
}
dis[index-1] = 0;
SPFA(index-1);
printf("%d\n" , -1*dis);
}

int main(){
while(scanf("%d" , &n) != EOF){
initial();
computing();
}
return 0;
}
```