POJ 1201 Intervals

差分约束的思想，虽然很长时间没写了。。。

用d[i]表示区间[0..i]选的个数，那么对于给出的每个区间有d[b[i]] - d[a[i] - 1] >= c[i],此外还有隐藏约束条件0<=d[i] - d[i - 1]<=1。

考虑图论中的最长路的性质，连边a[i - 1]——>b[i],权为c[i]。对每个节点i连边i——>i+1,权为0；i+1——>i,权为-1，应用spfa跑最长路。

在得出最长路的图中，一定满足上述两个约束条件，那么此时d[n]就代表了满足上述所有约束条件的最小选取个数。

#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<vector>
#include<stack>
#include<queue>
#include<map>
#include<cstdlib>
#include<cmath>
using namespace std;
#define ll long long
#define pb push_back
int d[55555];
bool v[55555];
int g[55555];
int nume;
struct edge
{
    int w,v,nxt;
};
edge e[555555];
int n,m;
int spfa()
{
    queue<int>q;
    while (!q.empty()) q.pop();
    q.push(0);
    memset(d,-1,sizeof(d));
    memset(v,0,sizeof(v));
    d[0] = 0; v[0] = 1;
    while (!q.empty())
    {
        int x = q.front();
        q.pop();
        for (int i = g[x]; i; i=e[i].nxt)
        {
            //cout<<x<<" "<<e[i].v<<" "<<endl;
          if (d[e[i].v] < d[x] + e[i].w)
          {
              //cout<<x<<" "<<e[i].v<<endl;
              d[e[i].v] = d[x] + e[i].w;
              if (!v[e[i].v])
              {
                  q.push(e[i].v);
                  v[e[i].v] = 1;
              }
          }
        }
        v[x] = 0;
    }
    return d[m + 1];
}
void init()
{
    nume = 1;
    memset(g,0,sizeof(g));
}
void addedge(int u,int v,int c)
{
    ++nume;
    e[nume].v = v;
    e[nume].w = c;
    e[nume].nxt = g[u];
    g[u] = nume;
}
int main()
{
    while (scanf("%d",&n) != EOF)
    {
        m = 0;
        init();
        //for (int i = 1; i <= n; i++) scanf("%d%d%d",&t[i].a,&t[i].b,&t[i].c);
        int x,y,z;
        for (int i = 1; i <= n; i++)
        {
            scanf("%d%d%d",&x,&y,&z);
            y++;
            m = max(m,y);
            addedge(x,y,z);
        }
        for (int i = 0; i <= m; i++)
        {
            addedge(i,i + 1,0);
            addedge(i + 1,i,-1);
        }
        printf("%d\n",spfa());
        //for (int i = 0; i <= m + 1; i ++) cout<<i<<" "<<d[i]<<endl;
    }
    return 0;
}

View Code

 

 

 

转载于:https://www.cnblogs.com/yuhao1994/p/3579495.html