题意:给定若干区间,并为每个区间指定一个数值。如区间为[Ai,Bi],数值为Ci。求一个最小集合S,使得每个区间[Ai,Bi]都至少包含其中的Ci个数。
思路:差分约束思想。s[bi+1]-s[ai]>=ci; bi+1的原因是bi也要选进来。对于这种闭区间,要注意边界的选择。此外还必须有:s[i]-s[i-1]<=1和s[i]-s[i-1]>=0。
一个技巧:i->i-1和i-1->i这些边不要加进来,可以直接判断!再有需要注意的是,dis数组要初始化为负无穷,而不是0,因为求的是最长路。
#include <stdio.h>
#include <string.h>
#define N 50010
#define INF 0x3ffffff
#define min(a,b) a<b?a:b
#define max(a,b) a>b?a:b
struct edge{
int y,w,next;
}e[N];
int n,begin,end,top;
int q[1500000],dis[N],first[N],visited[N];
void init(){
top = 0;
begin=INF,end=-INF;
memset(first,-1,sizeof(first));
memset(visited,0,sizeof(visited));
}
void add(int x,int y,int w){
e[top].y = y;
e[top].w = w;
e[top].next = first[x];
first[x] = top++;
}
int relax(int a,int b,int w){
if(dis[a]+w > dis[b]){
dis[b] = dis[a] + w;
return 1;
}
return 0;
}
void spfa(){
int i,front,rear;
front = rear = -1;
for(i = begin;i<=end;i++)
dis[i] = -INF;
dis[begin] = 0;
q[++rear] = begin;
visited[begin] = 1;
while(front < rear){
int now = q[++front];
visited[now] = 0;
for(i = first[now];i!=-1;i=e[i].next)
if(relax(now,e[i].y,e[i].w) && !visited[e[i].y]){
q[++rear] = e[i].y;
visited[e[i].y] = 1;
}
if(now>begin && dis[now]-1 > dis[now-1]){//s[i]-s[i-1]<=1
dis[now-1] = dis[now]-1;
if(!visited[now-1]){
q[++rear] = now-1;
visited[now-1] = 1;
}
}
if(now<end && dis[now] > dis[now+1]){//s[i]-s[i-1]>=0
dis[now+1] = dis[now];
if(!visited[now+1]){
visited[now+1] = 1;
q[++rear] = now+1;
}
}
}
}
int main(){
freopen("a.txt","r",stdin);
while(scanf("%d",&n)!=EOF){
int i,a,b,c;
init();
for(i = 0;i<n;i++){
scanf("%d %d %d",&a,&b,&c);
add(a,b+1,c);
begin = min(begin,a);
end = max(end,b+1);
}
spfa();
printf("%d\n",dis[end]-dis[begin]);
}
return 0;
}
i->i+1这些点加入图中版本:
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <cstdlib>
using namespace std;
#define clc(s,t) memset(s,t,sizeof(s))
#define INF 0x3fffffff
#define N 50005
int n,be,en;
struct edge{
int y,w,next;
}e[N*3];
int first[N],top;
int dis[N],used[N];
void add(int x,int y,int w){
e[top].y = y;
e[top].w = w;
e[top].next = first[x];
first[x] = top++;
}
int relax(int x,int y,int w){
if(dis[x] + w > dis[y]){
dis[y] = dis[x]+w;
return 1;
}
return 0;
}
int spfa(int s,int t){
int i,now;
queue<int> q;
for(i = s;i<=t;i++)
dis[i] = -INF;
dis[s] = 0;
clc(used, 0);
used[s] = 1;
q.push(s);
while(!q.empty()){
now = q.front();
q.pop();
used[now] = 0;
for(i = first[now];i!=-1;i=e[i].next){
if(e[i].y<s || e[i].y>t)
continue;
if(relax(now,e[i].y,e[i].w) && !used[e[i].y]){
used[e[i].y] = 1;
q.push(e[i].y);
}
}
}
return dis[t];
}
int main(){
int i,a,b,c;
top = 0;
clc(first,-1);
be = N;
en = 0;
scanf("%d",&n);
for(i = 0;i<n;i++){
scanf("%d %d %d",&a,&b,&c);
add(a,b+1,c);
be = min(be,a);
en = max(en,b+1);
}
for(i = be;i<=en;i++){
add(i,i+1,0);
add(i+1,i,-1);
}
printf("%d\n",spfa(be,en));
return 0;
}