NOIP2010 题解

机器翻译

　　题解：模拟

#include <cstdio>
#include <cstring>

const int MAXN = 1000;

int n, m, cj, now, MinV, MinId, Pri, a[MAXN+10], in[MAXN+10];

int main(){
    memset(in, -1, sizeof(in));
    scanf("%d %d", &n, &m), now = Pri = 0;
    for (int i=0; i<m; i++){
        scanf("%d", &cj);
        if (~in[cj]) continue;
        Pri ++;
        if (now+1<=n) now ++, in[cj] = i;
        else {
            MinV = 0x3f3f3f3f;
            for (int j=0; j<=MAXN; j++)
                if (in[j]<MinV && in[j]!=-1) MinV = in[j], MinId = j;
            in[MinId] = -1, in[cj] = i; 
        } 
    }
    printf("%d
", Pri);
}

乌龟棋

　　题解：dp

#include <cstdio>
#include <cstring>

const int MAXN = 350+2;
const int MAXE = 40+2;

int n, m, cj, a[MAXN], b[6], dp[MAXE][MAXE][MAXE][MAXE];

int main(){
    memset(dp, 0, sizeof(dp));
    
    scanf("%d %d", &n, &m);
    for (register int i=0; i<n; i++)
        scanf("%d", &a[i]);
    for (register int i=0; i<m; i++)
        scanf("%d", &cj), b[cj] ++;

    dp[0][0][0][0] = a[0];
    for (register int A=0; A<=b[1]; A++)
        for (register int B=0; B<=b[2]; B++)
            for (register int C=0; C<=b[3]; C++)
                for (register int D=0; D<=b[4]; D++){
                    cj = A+B+B+C+C+C+D+D+D+D;
                    if (A && dp[A-1][B][C][D]+a[cj]>dp[A][B][C][D])    dp[A][B][C][D] = dp[A-1][B][C][D]+a[cj];
                    if (B && dp[A][B-1][C][D]+a[cj]>dp[A][B][C][D]) dp[A][B][C][D] = dp[A][B-1][C][D]+a[cj];
                    if (C && dp[A][B][C-1][D]+a[cj]>dp[A][B][C][D]) dp[A][B][C][D] = dp[A][B][C-1][D]+a[cj];
                    if (D && dp[A][B][C][D-1]+a[cj]>dp[A][B][C][D]) dp[A][B][C][D] = dp[A][B][C][D-1]+a[cj];
                }
    printf("%d
", dp[b[1]][b[2]][b[3]][b[4]]);
}

关押罪犯

　　题解：并查集。既然只有两个监狱，那么如果a和b没有在一个监狱，b和c没有在一个监狱，那么a和c一定在一个监狱

　　把怒气值从大到小拍个序，依次处理

　　并查集[1, n]表示和自己一起的，[n+1, 2n]表示不和自己一起的

#include <cstdio>
#include <algorithm>
using std::sort;

const int MAXN = 20000+10;
const int MAXM = 100000+10;

int n, m, f[MAXN*2];

struct Relation{
    int a, b, c;

    friend bool operator < (const Relation& A, const Relation& B){
        return A.c>B.c;
    } 
}r[MAXM];

inline int Find(int x){
    return f[x]==x ? x : f[x] = Find(f[x]);
}

inline int Solve(){
    sort(r, r+m);
    int a, b;
    for (int i=0; i<m; i++){
        a = Find(r[i].a), b = Find(r[i].b);
        if (a==b) return r[i].c;
        else f[a] = Find(r[i].b+n), f[b] = Find(r[i].a+n);
    }
    return 0;
}

int main(){
    scanf("%d %d", &n, &m);
    for (int i=1; i<=2*n; i++)
        f[i] = i;
    for (int i=0; i<m; i++)
        scanf("%d %d %d", &r[i].a, &r[i].b, &r[i].c);

    printf("%d
", Solve());
}

引水入城

　　题解：看懂题是关键...

　　　　　多么痛的领悟：windows下只能bfs，linux下就怎么搞都可以了TAT

#include <cstdio>
#include <cstring>

const int MAXN = 500+10;
const int addX[] = {-1, 1, 0, 0};
const int addY[] = {0, 0, -1, 1};

int n, m, X, Y, Need, Cant, anna[MAXN][MAXN];
bool vis[MAXN][MAXN];

struct Region{
    int l, r;
}rg[MAXN];

inline int findLeft(){
    for (int i=1; i<=m; i++)
        if (vis[n][i]) return i;
    return 0;
}

inline int findRight(){
    for (int i=m; i>=1; i--)
        if (vis[n][i]) return i;
    return 0;
}

inline void DFS(int x, int y){
    if (x<1 || x>n || y<1 || y>m) return;
    vis[x][y] = true;

    for (int i=0; i<4; i++){
        X = x+addX[i], Y = y+addY[i];
        if (1<=X && X<=n && 1<=Y && Y<=m && anna[X][Y]<anna[x][y] && !vis[X][Y])
            DFS(X, Y);
    }
}

inline bool legal(){
    memset(vis, false, sizeof(vis));
    
    for (int i=1; i<=m; i++)
        DFS(1, i);
    
    for (int i=1; i<=m; i++)
            if (not vis[n][i]) ++ Cant;
    
    return !Cant;
}

inline int segmentCover(){
    for (int i=1; i<=m; i++){
        memset(vis, false, sizeof(vis));
        DFS(1, i);
        rg[i].l = findLeft(), rg[i].r = findRight();
    }

    int i = 1, j, k = 0;
    while (i<=m){
        for (j=1; j<=m; j++)
            if (rg[j].l<=i && rg[j].r>k) k = rg[j].r;
        i = k+1, ++ Need;
    }

    return Need;
}

int main(){
    memset(anna, 0x3f, sizeof(anna));
    
    scanf("%d %d", &n, &m), Cant = Need = 0;
    for (int i=1; i<=n; i++)
        for (int j=1; j<=m; j++)
            scanf("%d", &anna[i][j]);

    if (legal()) printf("1
%d", segmentCover());
    else printf("0
%d", Cant);
}