貪心+搜尋

貪心法就是遵從某種規律，不斷的貪心將利益最大化的方法。
解決貪心問題的方法是要找出他們之間的規律，並將利益最大化。
舉一個簡單的例子
Iahub got bored, so he invented a game to be played on paper.

He writes n integers a1, a2, …, an. Each of those integers can be either 0 or 1. He’s allowed to do exactly one move: he chooses two indices i and j (1 ≤ i ≤ j ≤ n) and flips all values ak for which their positions are in range [i, j] (that is i ≤ k ≤ j). Flip the value of x means to apply operation x = 1 - x.
The goal of the game is that after exactly one move to obtain the maximum number of ones. Write a program to solve the little game of Iahub.
Input
The first line of the input contains an integer n (1 ≤ n ≤ 100). In the second line of the input there are n integers: a1, a2, …, an. It is guaranteed that each of those n values is either 0 or 1.
Output
Print an integer — the maximal number of 1s that can be obtained after exactly one move.
Examples
Input
5
1 0 0 1 0
Output
4
Input
4
1 0 0 1
Output
4
題目的大意是選擇一個區間【a，b】，將【a，b】之間的所有值都變成1-a[i]，使得最後獲得1的數量最多。

#include<stdio.h>
int main()
{
    int i,j,k,t,n,sum=0,maxx=-99,a[200],b[200];
    scanf("%d",&t);
    for( i=0;i<t;i++)
        scanf("%d",&a[i]);
    for(i=0;i<t;i++)
    {
        for(j=i;j<t;j++)
        {
            for(k=0;k<t;k++)
                b[k]=a[k];
            for(k=i;k<=j;k++)
                b[k]=1-b[k];
            sum=0;
            for(k=0;k<t;k++)
                if(b[k])
                sum++;
            if(sum>maxx)
                maxx=sum;
        }
    }
    printf("%d\n",maxx);
    return 0;
}

搜尋分為深度搜尋和廣度搜尋，深度搜尋是它從某個狀態開始，不斷的轉移狀態知道無法轉移，然後後退到前一步的狀態，繼續轉移到其他狀態，如此不斷的反覆知道找到最優的解。廣度優先搜尋總是先搜尋距離初始狀態近的狀態。廣度的時間複雜度要比深度搜尋的小很多。
舉一個廣搜簡單的例子
You play a computer game. Your character stands on some level of a multilevel ice cave. In order to move on forward, you need to descend one level lower and the only way to do this is to fall through the ice.
The level of the cave where you are is a rectangular square grid of n rows and m columns. Each cell consists either from intact or from cracked ice. From each cell you can move to cells that are side-adjacent with yours (due to some limitations of the game engine you cannot make jumps on the same place, i.e. jump from a cell to itself). If you move to the cell with cracked ice, then your character falls down through it and if you move to the cell with intact ice, then the ice on this cell becomes cracked.
Let’s number the rows with integers from 1 to n from top to bottom and the columns with integers from 1 to m from left to right. Let’s denote a cell on the intersection of the r-th row and the c-th column as (r, c).
You are staying in the cell (r1, c1) and this cell is cracked because you’ve just fallen here from a higher level. You need to fall down through the cell (r2, c2) since the exit to the next level is there. Can you do this?
Input
The first line contains two integers, n and m (1 ≤ n, m ≤ 500) — the number of rows and columns in the cave description.
Each of the next n lines describes the initial state of the level of the cave, each line consists of m characters “.” (that is, intact ice) and “X” (cracked ice).
The next line contains two integers, r1 and c1 (1 ≤ r1 ≤ n, 1 ≤ c1 ≤ m) — your initial coordinates. It is guaranteed that the description of the cave contains character ‘X’ in cell (r1, c1), that is, the ice on the starting cell is initially cracked.
The next line contains two integers r2 and c2 (1 ≤ r2 ≤ n, 1 ≤ c2 ≤ m) — the coordinates of the cell through which you need to fall. The final cell may coincide with the starting one.
Output
If you can reach the destination, print ‘YES’, otherwise print ‘NO’.
Examples
Input
4 6
X…XX
…XX.
.X…X.
…
1 6
2 2
Output
YES
Input
5 4
.X…
…X
X.X.
…
.XX.
5 3
1 1
Output
NO
Input
4 7
…X.XX.
.XX…X.
X…X…
X…
2 2
1 6
Output
YES
題目的大意就是你從一個點出發，每經過一個點後，就不能再通過這個點，如果目標點位“#”’你可以直接通過，如果不是“#”你需要第二次通過。

#include<stdio.h>
#include<string.h>
#include<queue>
using namespace std;
char a[600][600];
int n,m,p,q,ex,ey;
int to[4][2]={0,1,0,-1,1,0,-1,0};
struct ha
{
    int x,y;
}h,s;
int bfs(int x,int y)
{
    queue<ha> q;
    h.x=x;
    h.y=y;
    q.push(h);
    while(!q.empty())
    {
        s=q.front();
        q.pop();
        for(int i=0;i<4;i++)
        {
            h.x=s.x+to[i][0];
            h.y=s.y+to[i][1];
            if(h.x>=1&&h.x<=n&&h.y>=1&&h.y<=m)
            {
                if(h.x==ex&&h.y==ey)
                {
                    if(a[h.x][h.y]=='.')
                        a[h.x][h.y]='X';
                    else
                        return 1;
                    q.push(h);
                }
                else if(a[h.x][h.y]=='.')
                {
                    a[h.x][h.y]='X';
                    q.push(h);
                }
            }
        }
    }
    return 0;
}
int main()
{
    while(~scanf("%d%d",&n,&m))
    {
        if(n==0&&m==0)
            break;
    for(int i=1;i<=n;i++)
        scanf("%s",a[i]+1);
    scanf("%d%d",&p,&q);
    scanf("%d%d",&ex,&ey);
    if(bfs(p,q))
        printf("YES\n");
    else
        printf("NO\n");
    }
    return 0;
}