USACO月赛—JAN12 Silver

题目描述
After several years of record milk production, Farmer John now operates an entire network of N farms (1 <= N <= 100).  Farm i is located at position (x_i, y_i) in the 2D plane, distinct from all other farms, with both x_i and y_i being integers.

FJ needs your help planning his daily delivery route to bring supplies to the N farms.  Starting from farm 1, he plans to visit the farms sequentially (farm 1, then farm 2, then farm 3, etc.), finally returning to farm 1 after visiting farm N.  It tames FJ one minute to make a single step either north, south, east, or west.  Furthermore, FJ wants to visit each farm exactly once during his entire journey (except farm 1, which he visits twice of course).  

Please help FJ determine the minimum amount of time it will take him to complete his entire delivery route.

输入
* Line 1: The number of farms, N.

* Lines 2..1+N: Line i+1 contains two space-separated integers, x_i and y_i  (1 <= x_i, y_i <= 1,000,000).
输出
* Line 1: The minimum number of minutes FJ needs to complete his delivery route, or -1 if it is impossible to find a feasible delivery route that visits every farm exactly once (except farm 1).

样例输入
4
2 2
2 4
2 1
1 3
样例输出
12
提示
OUTPUT DETAILS:



FJ can complete his delivery route in 12 minutes: 2 minutes to go from farm 1 to farm 2, 5 minutes to go from farm 2 to farm 3 (circumventing farm 1), 3 minutes to go from farm 3 to farm 4, and then 2 minutes to return to farm 1.

做法：把每个点的四周的四个点拆出来（显然如果最短路径则不经过该点必经过四周的点），然后做n次最短路径即可。。但中间细节比较多，写的时候要注意。

        这题是个不错的最短路径题，比之前每次都做模板题好多了T_T...不过考试的时候没做出来。。果然还是太水了TTTTTT_______TTTTTT...

题目描述
Farmer John has just received a new shipment of N (1 <= N <= 20) bales of hay, where bale i has size S_i (1 <= S_i <= 100).  He wants to divide the bales between his three barns as fairly as possible.  

After some careful thought, FJ decides that a "fair" division of the hay bales should make the largest share as small as possible.  That is, if B_1, B_2, and B_3 are the total sizes of all the bales placed in barns 1, 2, and 3, respectively (where B_1 >= B_2 >= B_3), then FJ wants to make B_1 as small as possible.

For example, if there are 8 bales in these sizes:

2 4 5 8 9 14 15 20

A fair solution is

Barn 1: 2 9 15   B_1 = 26
Barn 2: 4 8 14   B_2 = 26
Barn 3: 5 20     B_3 = 25

Please help FJ determine the value of B_1 for a fair division of the hay bales.


输入
* Line 1: The number of bales, N.

* Lines 2..1+N: Line i+1 contains S_i, the size of the ith bale.

输出
* Line 1: Please output the value of B_1 in a fair division of the hay  bales.

样例输入
8
14
2
5
15
8
9
20
4
样例输出
26

做法：水dp，和三角形牧场差不多，f(i,j)表示i,j是否可以构成。

题目描述
Farmer John has discovered that his cows produce higher quality milk when they are subject to strenuous exercise.  He therefore decides to send his N cows (1 <= N <= 25,000) to climb up and then back down a nearby mountain!

Cow i takes U(i) time to climb up the mountain and then D(i) time to climb down the mountain.  Being domesticated cows, each cow needs the help of a farmer for each leg of the climb, but due to the poor economy, there are only two farmers available, Farmer John and his cousin Farmer Don.  FJ plans to guide cows for the upward climb, and FD will then guide the cows for the downward climb.  Since every cow needs a guide, and there is only one farmer for each part of the voyage, at most one cow may be climbing upward at any point in time (assisted by FJ), and at most one cow may be climbing down at any point in time (assisted by FD).  A group of cows may temporarily accumulate at the top of the mountain if they climb up and then need to wait for FD's assistance before climbing down.  Cows may climb down in a different order than they climbed up.

Please determine the least possible amount of time for all N cows to make the entire journey.

输入
* Line 1: The number of cows, N.

* Lines 2..1+N: Line i+1 contains two space-separated integers: U(i) and D(i).  (1 <= U(i), D(i) <= 50,000).

输出
* Line 1: A single integer representing the least amount of time for all the cows to cross the mountain.
样例输入
3
6 4
8 1
2 3
样例输出
17
提示
OUTPUT DETAILS:



If cow 3 goes first, then cow 1, and then cow 2 (and this same order is used for both the ascent and descent), this gives a total time of 17.

做法：贪心，每次保证上坡小的尽量在前，下坡小的尽量在后，中间分四种情况讨论下即可。。具体严谨证明偶也不会。。先挖坑QAQ..

Codes：

#include<set>
#include<queue>
#include<vector>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
const int N = 110;
using namespace std;
const int limits = 1000000;
typedef pair<int,int> Points;
const int dx[5] = {0,0,1,-1,0},
          dy[5] = {0,1,0,0,-1};
#define x first
#define y second
#define clr(a,b) memset(a,b,sizeof(a))
#define For(i,n) for(int i=1;i<=n;i++)
#define Rep(i,l,r) for(int i=l;i<=r;i++)
#define Swap(A,B) {Points T = A ; A = T; T = B;}
int n,npcnt,es,stack[5*N],head[N*5],dis[5*N],ans;
set<Points> hash,farm;
bool vis[5*N];
Points p[N],np[5*N];
struct edge{
    int t,next,w;
}E[5*N*5*N];

int Len(Points A,Points B){
    return abs(B.x-A.x)+abs(B.y-A.y);
}

void makelist(int s,int t){
    E[es].t = t; E[es].next = head[s];E[es].w = Len(np[s],np[t]);
    head[s] = es++;
}

bool inseg(Points out,Points A,Points B){
    if(A.x==B.x)
            if(min(A.y,B.y)<out.y&&out.y<max(A.y,B.y)&&out.x==A.x) return true;
    if(A.y==B.y)
            if(min(A.x,B.x)<out.x&&out.x<max(A.x,B.x)&&out.y==A.y) return true;
    return false;
}

bool check(Points A,Points B){
    Points C;//   _| 
    if(A.x>B.x) Swap(A,B);
    C.y = A.y; C.x = B.x;
    bool Flag = A.x==B.x||A.y==B.y||farm.find(C)==farm.end();
    For(i,n)
        if(inseg(p[i],A,C)||inseg(p[i],C,B)){
            Flag = false;
            break;
        }
    if(Flag) return true;
    C.x = A.x; C.y = B.y;
    Flag = A.x==B.x||A.y==B.y||farm.find(C)==farm.end();
    For(i,n)
        if(inseg(p[i],A,C)||inseg(p[i],C,B)){
            Flag = false;
            break;
        }
    return Flag;
}

void init(){
    scanf("%d",&n);
    For(i,n) {
        scanf("%d%d",&p[i].y,&p[i].x);
        hash.insert(p[i]);farm.insert(p[i]);
        np[i] = p[i];
    }npcnt = n;
    clr(head,-1);
    For(i,n)
        For(j,4){
            int newx = p[i].x + dx[j] , newy = p[i].y + dy[j];
            if(newx<1||newx>limits||newy<1||newy>limits) continue;
            if(hash.find(make_pair(newx,newy))==hash.end()) {np[++npcnt] = make_pair(newx,newy);hash.insert(make_pair(newx,newy));}
        }
    For(i,npcnt-1)
      Rep(j,i+1,npcnt){
            if(check(np[i],np[j])) {
                makelist(i,j);
                makelist(j,i);
            }
      } 
}

int SPFA(int s,int t){
    For(i,5*n) dis[i] = 214748364;
    memset(vis,0,sizeof(vis));
    dis[s] = 0;
    stack[stack[0]=1,1] = s;
    while(stack[0]){
        int First = stack[stack[0]--];vis[First] = false;
        for(int p=head[First];p!=-1;p=E[p].next){
            int v = E[p].t;
            if(v<=n&&v!=s&&v!=t) continue;
            if(dis[First]+E[p].w<dis[v]) {
                dis[v] = dis[First] + E[p].w;
                if(!vis[v]) {vis[v] = true; stack[++stack[0]] = v;}
            } 
        }  
    }
    return dis[t];
}

void Work(){
    For(i,n){
        int W = SPFA(i,(i==n)?(1):(i+1)); 
        if(W>21474836) puts("-1");
        if(W>21474836) return;
        ans+=W;
    }
    printf("%d
",ans);
} 

int main(){
   // freopen("date.in","r",stdin);
   // freopen("date.out","w",stdout);
    init();
    Work();
    return 0;
}
--------------------------------------A题--------------------------------------
#include<queue>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
const int N = 30;
const int MAXS = 20*100/3;
#define For(i,n) for(int i=1;i<=n;++i)
#define Rep(i,l,r) for(int i=l;i<=r;++i)

int A[N],f[N][MAXS+200][MAXS+200],sum,ans=2147483647,n;

int main(){
    scanf("%d",&n);
    For(i,n) {scanf("%d",&A[i]),sum+=A[i];};
    memset(f,false,sizeof(f));
    f[0][0][0] = true;
    For(i,n){
        Rep(j,0,MAXS)
          Rep(k,0,MAXS) 
            if(f[i-1][j][k]){
                f[i][j][k]      = true;
                f[i][j+A[i]][k] = true;
                f[i][j][k+A[i]] = true;
            }
    }
    For(i,MAXS){
      For(j,MAXS)
        if(f[n][i][j])
            ans = min(ans,max(i,max(j,sum-i-j)));
    }
    printf("%d
",ans);
    return 0;
}
--------------------------------------B题--------------------------------------
#include<queue>
#include<cstdio>
#include<iostream>
#include<algorithm>
using namespace std;
const int N = 25100;
#define Points pair<int,int>
#define For(i,n) for(int i=1;i<=n;++i)
#define Rep(i,l,r) for(int i=l;i<=r;++i)

Points A[N];
int n;

bool cmp(Points A,Points B){
    if(A.first<A.second){
        if(B.first<B.second) return A.first<B.first;
        else                 return true;
    }else{
        if(B.first>B.second) return A.second>B.second;
        else                 return false;
    }    
}

void solve(){
    sort(A+1,A+n+1,cmp);
    int sum1 = 0 ,sum2 = A[1].first;
    For(i,n){
        sum1+=A[i].first;
        sum2=max(sum1,sum2)+A[i].second;
    }
    printf("%d
",sum2);
}

int main(){
    scanf("%d",&n);
    For(i,n) scanf("%d%d",&A[i].first,&A[i].second);
    solve();
    return 0;
}
----------------------------------C题------------------------------------------