2013
12-21

# Destroy the Well of Life

In the game of DotA (Defense of the Ancient), there are two opposite legions called The Sentinel and The Scourage.
Now The Scourage has made a big success and The Sentinel is at stake!
So The Sentinel must destroy the Well of Life of The Scourage.
The chief of The Sentinel, Prophet, asks EarthShaker to do this arduous task.There are N Wells of Life of The Scourage (The Wells of Life are numbered from 1 to N), and EarthShaker’s task is to destroy the Nth Well of Life.
The following information is known about each Well of Life:
Wi � the weight of water on i-th Well of Life before it is destroyed.
Li � if the weight of water on i-th Well of Life is more than Li, the i-th Well of Life will be destroyed and the water of it will pours to the (i + 1)-th Well of Life.
Pi � EarthShaker has a skill called Echo-Slam, the i-th Well of Life will be immediately destroyed when he uses Echo-Slam to it and the water of it will pours to the (i + 1)-th Well of Life. For the i-th Well of Life, the energy that EarthShaker need to use Echo-Slam to destroy it is Pi.

Can you tell EarthShaker the minimum amount of energy needed to destroy the Nth Well of Life?

The input consists of several test cases. There is a single number on the first line, the number of cases. There are no more than 10 cases.
Each case contains a natural number N on the first line, 1<=N<=100,000.
Following N lines contains three numbers Wi, Li, Pi (Wi<=Li, 0<=Wi, Li, Pi <=20,000), representing the information of the i-th Well of Life.

For each case, print a number in a line representing the least amount of energy EarthShaker needed to use, so as to destroy the Nth Well of Life. Use the format in the sample.

1
3
1000 1000 1
0 1000 2
2 10 100

Case 1: Need to use 3 mana points.

题目分析：

#include<stdio.h>
int l[100001];
int r[100001];
int sum[100001];
int main()
{
//freopen("e:1.txt","r",stdin);
int n,T,t=0;
scanf("%d",&T);

while(T--)
{
int i,max;
scanf("%d",&n);
for(i = 0;i < n;i++)
scanf("%d%d%d",&l[i],&r[i],&sum[i]);
int min=sum[n-1],msum;
for(i=0;i<n-1;i++)
{
max=0;
msum=0;
for(int j = i;j < n;j++)
{
max+=l[j];
if(max <= r[j]) msum+=sum[j];

if(msum>min) break;
}
if(msum < min)
min=msum;
}
printf("Case %d: Need to use %d mana points.\n",++t,min);
}
return 0;
}


1. 这道题这里的解法最坏情况似乎应该是指数的。回溯的时候
O(n) = O(n-1) + O(n-2) + ….
O(n-1) = O(n-2) + O(n-3)+ …
O(n) – O(n-1) = O(n-1)
O(n) = 2O(n-1)

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