HDU 1384 Intervals （差分约束）

Sample Input

53 7 38 10 36 8 11 3 110 11 1

 

Sample Output

6

题意：给你n个数u，v，w；要求在[u,v]区间至少取w个数（整数），求最少要取多少个数。

S[v+1] - S[u] >= w, S[i+1] - S[i] >=0&&<=1,S[i] - S[i+1] <=-1.

在u,v+1之间建一条边，跑一遍SPFA即可。

#include <iostream>#include <stdio.h>#include <string.h>#include<string>#include<math.h>#include<queue>#include<stack>#include <algorithm>#include<vector>#include<map>using namespace std;#define M 100#define inf 0x3fffffff#define maxn 50005#define ll long longstruct edge{int v,num,next;}e[maxn*4];int edge_num,a,b,vis[maxn],d[maxn];int head[maxn];void add(int a,int b,int c){e[edge_num].num=b;e[edge_num].v=c;e[edge_num].next=head[a];head[a]=edge_num++;}int n;void SPFA(){queue<int>q;memset(vis,0,sizeof(vis));memset(d,-50005,sizeof(d));/*for(int i=a;i<=b;i++)d[i]=-50005;*/while(!q.empty()) q.pop();q.push(a);d[a]=0;vis[a]=1;while(!q.empty()){int u=q.front();q.pop();for(int i=head[u];i!=-1;i=e[i].next){int v=e[i].num;if(d[v]<d[u]+e[i].v){d[v]=d[u]+e[i].v;if(!vis[v]){q.push(v);vis[v]=1;}}}vis[u]=0;}}int main(){while(~scanf("%d",&n)){edge_num=0;memset(head,-1,sizeof(head));int u,v,w;a=inf,b=0;while(n--){scanf("%d%d%d",&u,&v,&w);if(u<a)a=u;if(v+1>b)b=v+1;add(u,v+1,w);}for(int i=a;i<b;i++){add(i+1,i,-1);add(i,i+1,0);}SPFA();printf("%d\n",d[b]);}return 0;}