TC训练

SRM 400 DIV 2

1000pt

枚举第一行和第一列，复杂度为O(2^16)

然后贪心和开关问题类似。

#line 2 "LightedPanels.cpp"#include <cstdlib>#include <cctype>#include <cstring>#include <cstdio>#include <cmath>#include <algorithm>#include <vector>#include <string>#include <iostream>#include <sstream>#include <map>#include <set>#include <queue>#include <stack>#include <ctime>using namespace std;typedef long long ll;class LightedPanels {public:    int g[11][11],gg[11][11];    int H , W;    void flip(int x,int y) {        for(int i = -1; i <= 1; ++i)            for(int j = -1; j <= 1; ++j) {                int xx = x + i;                int yy = y + j;                if(xx < 0 || xx >= H || yy < 0 || yy >= W) continue;                gg[xx][yy] = !gg[xx][yy];            }    }    int solve() {        int ans = 0;        for(int i = 1; i < H; ++i)            for(int j = 1; j < W; ++j)                if(gg[i-1][j-1] == 0) {                    ++ans;                    flip(i,j);                }        for(int i = 0; i < H; ++i)            for(int j = 0; j < W; ++j)                if(gg[i][j] != 1) return 65;        return ans;    }    int minTouch(vector <string> board) {        H = board.size();        W = board[0].size();        memset(g,0,sizeof(g));        memset(gg,0,sizeof(gg));        for(int i = 0; i < H; ++i) {            string str = board[i];            for(int j = 0; j < str.size(); ++j)                if(str[j] == '*') g[i][j] = 1;                else g[i][j] = 0;        }        int all = 1<<16,MIN = 65;        for(int s = 0; s < all; ++s) {            int k = 1,step = 0;            for(int x = 0; x < 9; ++x)                for(int y = 0; y < 9; ++y)                    gg[x][y] = g[x][y];            for(int i = 0; i < W; ++i) {                if(s&k) flip(0,i),++step;                k <<= 1;            }            for(int i = 1; i < H; ++i) {                if(s&k) flip(i,0),++step;                k <<= 1;            }            MIN = min(MIN,step+solve());        }        if(MIN >= 65) MIN = -1;        return MIN;    }// END CUT HERE};

SRM 589 DIV 2

500pt卡了好久，想用dp做的,边界处理想了半天。最后看了题解,用贪心过了

1000pt留坑

SRM 590 DIV 2

1000pt留坑

SRM 591 DIV 2

500pt好题

1000pt留坑

SRM 592 DIV 2

1000pt留坑

SRM 593 DIV 2

1000pt留坑

SRM 594 DIV 2

1000pt留坑

SRM 595 DIV 2

1000pt留坑

SRM 596 DIV 2

1000pt留坑

SRM 597 DIV 2

SRM 598 DIV 2

SRM 599 DIV 2

SRM 600 DIV 2

1000pt留坑

SRM 601 DIV 2

1000pt留坑

SRM 602 DIV 2：

1000分的题卡了一天。

既然决定要看题解，就要看的认真。

看一段跳一段，再自己想一段，反而没有真正理解题解的内涵。

好题，以后再做一遍。

SRM 603 DIV 2

SRM 604 DIV 2

SRM 605 DIV 2

1000pt留坑

SRM 607 DIV 2

1000pt留坑

SRM 612 DIV 2

SRM 624 DIV 2

SRM 625 DIV 2

SRM 626 DIV 2

SRM 631 DIV 2

500pt

贪心思想:尽可能的把所有点往前放

class CatsOnTheLineDiv2{        public:        map<int,int>can,now,S;        string getAnswer(vector <int> position, vector <int> count, int time)        {            for(int i = - 2000; i <= 2000; ++i) S[i] = 0,can[i] = 0,now[i] = 0;            int n = position.size();            for(int i = 0; i< n; ++i) S[position[i]] += count[i],can[position[i]] = 1;            for(int i = -1000; i <= 1000; ++i)                if(can[i] == 1){                    for(int j = -time; j < 0; ++j)                        if(S[i] > 0 && now[i+j] == 0) --S[i],++now[i+j];                    if(S[i] != 0 && now[i] != 1) --S[i],now[i] = 1;                    for(int j = 1; j <= time; ++j)                        if(S[i] > 0 && now[i+j] == 0)  --S[i],++now[i+j];            }            for(int i = -2000; i <= 2000; ++i) if(S[i] > 0) return "Impossible";            return "Possible";        }};