SPOJ - GCJ1C09C 释放囚犯 ——————区间dp

GCJ1C09C - Bribe the Prisoners

Problem

In a kingdom there are prison cells (numbered 1 to P) built to form a straight line segment. Cells number i and i+1 are adjacent, and prisoners in adjacent cells are called “neighbours.” A wall with a window separates adjacent cells, and neighbours can communicate through that window.

All prisoners live in peace until a prisoner is released. When that happens, the released prisoner’s neighbours find out, and each communicates this to his other neighbour. That prisoner passes it on to his other neighbour, and so on until they reach a prisoner with no other neighbour (because he is in cell 1, or in cell P, or the other adjacent cell is empty). A prisoner who discovers that another prisoner has been released will angrily break everything in his cell, unless he is bribed with a gold coin. So, after releasing a prisoner in cell A, all prisoners housed on either side of cell A - until cell 1, cell P or an empty cell - need to be bribed.

Assume that each prison cell is initially occupied by exactly one prisoner, and that only one prisoner can be released per day. Given the list of Q prisoners to be released in Q days, find the minimum total number of gold coins needed as bribes if the prisoners may be released in any order.

Note that each bribe only has an effect for one day. If a prisoner who was bribed yesterday hears about another released prisoner today, then he needs to be bribed again.

Input

The first line of input gives the number of cases, N. N test cases follow. Each case consists of 2 lines. The first line is formatted as

P Q

where P is the number of prison cells and Q is the number of prisoners to be released.
This will be followed by a line with Q distinct cell numbers (of the prisoners to be released), space separated, sorted in ascending order.

Output
For each test case, output one line in the format
Case #X: C
where X is the case number, starting from 1, and C is the minimum number of gold coins needed as bribes.

Limits
1 ≤ N ≤ 100
Q ≤ P
Each cell number is between 1 and P, inclusive.

Large dataset

1 ≤ P ≤ 10000
1 ≤ Q ≤ 100

Input
2
8 1
3
20 3
3 6 14

Output
Case #1: 7
Case #2: 35

Note
In the second sample case, you first release the person in cell 14, then cell 6, then cell 3. The number of gold coins needed is 19 + 12 + 4 = 35. If you instead release the person in cell 6 first, the cost will be 19 + 4 + 13 = 36.

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白书129页的问题，转化为区间dp来解决

• $dp[i][j]$表示将第i个囚犯到第j个囚犯(不含两端)都释放是，所需要的最少金币个数
• 我们把左端点当成0号囚犯，右端点当成q+1号囚犯，这样$dp[0][q+1]$就是答案

#include<bits/stdc++.h>
using namespace std;
const int MAXN=200+10;
int a[MAXN];
int dp[MAXN][MAXN];
int main()
{
        // freopen("../in.txt","r",stdin);
        // freopen("../out.txt","w",stdout);
	int n, p, q;
        scanf("%d",&n);
        int Case = 1;
        while(n--)
        {
                scanf("%d %d",&p,&q);
                for(int i=1;i<=q;++i)
                        scanf("%d",&a[i]);
                a[0] = 0;
                a[q+1] = p+1;
                for(int i=0;i<=q;++i)
                        dp[i][i+1] = 0;
                for(int w=2;w<=q+1;++w){
                        for(int i=0;i+w<=q+1;++i){

                                int j = i+w, t = 0x3f3f3f3f;

                                for(int k = i+1; k < j; ++k)
                                        t = min(t, dp[i][k]+dp[k][j]);

                                dp[i][j] = t + a[j] - a[i] - 2;
                        }
                }
                printf("Case #%d: %d\n",Case++,dp[0][q+1]);
        }
        return 0;
}