101 lines
2.3 KiB
C++
101 lines
2.3 KiB
C++
#include <stdio.h>
|
|
#include <string.h>
|
|
#include <iostream>
|
|
#include <algorithm>
|
|
#include <vector>
|
|
#include <queue>
|
|
#include <set>
|
|
#include <map>
|
|
#include <string>
|
|
#include <math.h>
|
|
#include <stdlib.h>
|
|
#include <time.h>
|
|
using namespace std;
|
|
struct NN1
|
|
{
|
|
int d,e;
|
|
void input()
|
|
{
|
|
scanf("%d%d",&d,&e);
|
|
}
|
|
}node1[3000];
|
|
struct NN2
|
|
{
|
|
int start,finish;
|
|
int r;
|
|
void input()
|
|
{
|
|
scanf("%d%d%d",&start,&finish,&r);
|
|
}
|
|
}node2[3000];
|
|
int a[5010];
|
|
long long f[100010];
|
|
long long f2[5010];
|
|
int dp[5010];
|
|
vector<int>vec[5010];
|
|
vector<int>vec2[5010];
|
|
int main()
|
|
{
|
|
int T;
|
|
int iCase = 0;
|
|
int n,m;
|
|
scanf("%d",&T);
|
|
while(T--)
|
|
{
|
|
iCase++;
|
|
printf("Case #%d:\n",iCase);
|
|
scanf("%d%d",&n,&m);
|
|
int cnt = 0;
|
|
memset(f,0,sizeof(f));
|
|
for(int i = 0;i < n;i++)
|
|
{
|
|
node1[i].input();
|
|
f[node1[i].d] += node1[i].e;
|
|
}
|
|
for(int i = 1;i <= 100000;i++)
|
|
f[i] += f[i-1];
|
|
for(int i = 0;i < m;i++)
|
|
{
|
|
node2[i].input();
|
|
a[cnt++] = node2[i].start;
|
|
a[cnt++] = node2[i].finish;
|
|
}
|
|
sort(a,a+cnt);
|
|
cnt = unique(a,a+cnt) - a;
|
|
map<int,int>mp;
|
|
for(int i = 0;i < cnt;i++)
|
|
mp[a[i]] = i;
|
|
f2[0] = f[a[0]];
|
|
for(int i = 1;i < cnt;i++)
|
|
f2[i] = f[a[i]] - f[a[i-1]];
|
|
for(int i = 0;i < cnt;i++)
|
|
{
|
|
vec[i].clear();
|
|
vec2[i].clear();
|
|
}
|
|
for(int i = 0;i < m;i++)
|
|
{
|
|
node2[i].start = mp[node2[i].start];
|
|
node2[i].finish = mp[node2[i].finish];
|
|
vec[node2[i].start].push_back(node2[i].finish);
|
|
vec2[node2[i].start].push_back(node2[i].r);
|
|
}
|
|
memset(dp,0,sizeof(dp));
|
|
for(int i = cnt-1;i >= 0;i--)
|
|
{
|
|
dp[i] = dp[i+1];
|
|
int sz = vec[i].size();
|
|
for(int j = 0;j < sz;j++)
|
|
dp[i] = max(dp[i],dp[vec[i][j]] + vec2[i][j]);
|
|
}
|
|
long long ans ;
|
|
ans = 0;
|
|
for(int i = 0;i < cnt;i++)
|
|
{
|
|
ans += (long long)dp[i]*f2[i];
|
|
}
|
|
printf("%.2lf\n",(double)ans/100);
|
|
}
|
|
return 0;
|
|
}
|