SRM 562 DIV2

1、第一个，水题，直接求最大值即可，略。

2、根据题目，clipboard最大的时候是50，则可以得到，当粘贴超过50次以后，每次增长的个数是肯定是固定个数。

　　写的时候为了保险，把50换到了100多。

#include <iostream>
#include <string>
#include <vector>
#include <cstdlib>
#include <map>
#include <algorithm>
#include <stack>
#include <queue>
#include <cmath>
using namespace std;
typedef long long ll;
const int sz_pix = 160;
int num[sz_pix];
bool judge[sz_pix][sz_pix];
class PastingPaintingDivTwo {
public:
    long long countColors(vector<string> clipboard, int T) {
        int r = clipboard.size();
        int c = clipboard[0].size();
        for (int i = 0; i < sz_pix; i++)
            for (int j = 0; j < sz_pix; j++) {
                num[i] = 0;
                judge[i][j] = false;
            }
        for (int i = 0; i < 105; i++) {
            for (int a = 0; a < r; a++) {
                for (int b = 0; b < c; b++) {
                    if ((!judge[i + a][i + b]) && (clipboard[a][b] == 'B')) {
                        judge[i + a][i + b] = true;
                        num[i]++;
                    }
                }
            }
        }
        if (T > 103) {
            ll inc = num[102] ;
            ll remain = T - 103;
            ll tmp = remain * inc;
            ll res = tmp;
            for (int i = 0; i < 103; i++) {
                res += num[i];
            }
            return res;
        }
        ll res = 0;
        for (int i = 0; i < T; i++) {
            res += num[i];
        }
        return res;

    }
};

3、动态规划+状态压缩的问题。

　　

#include <iostream>
#include <string>
#include <vector>
#include <cstdlib>
#include <map>
#include <algorithm>
#include <stack>
#include <queue>
#include <cmath>
using namespace std;
typedef long long ll;
ll num[15][15][1 << 14];

bool path[15][15];
class RandomOption {
public:
    double getProbability(int keyCount, vector<int> badLane1,
            vector<int> badLane2) {

        for (int i = 0; i < keyCount; i++)
            num[0][i][1 << i] = 1;

        int sz = badLane1.size();
        for (int i = 0; i < sz; i++) {
            int s = badLane1[i];
            int e = badLane2[i];
            path[s][e] = path[e][s] = true;
        }
        for (int i = 1; i < keyCount; i++) {
            for (int j = 0; j < keyCount; j++) {
                for (int pre = 0; pre < (1 << keyCount); pre++) {
                    for (int k = 0; k < keyCount; k++) {
                        if (path[j][k] || (pre & (1 << k)))
                            continue;
                        int now = pre | (1 << k);
                        num[i][k][now] += num[i - 1][j][pre];
                    };
                }
            }

        }
        double all = 1;
        int keyCount1 = keyCount;
        while (keyCount1) {
            all *= keyCount1;
            keyCount1--;
        }
        double pos = 0;
        for (int i = 0; i < keyCount; i++) {
            pos += num[keyCount - 1][i][(1 << keyCount) - 1];
        }

        double res = pos / all;
        return res;
    }
};