NOIP2017 题解(给自己看的) --有坑要填

目录

• D1T1精妙证明:
• D1T3
• D2T2

几道水题就不写了....

D1T1精妙证明:

把ax+by = z 的z按照模a剩余系分类
由于((a,b)=1)所以对于每个(kin[0, a)), (kcdot b)都在不同剩余系内!!(反证法)

那么自然最大的取不到数在(a-1)*b的剩余系内, 也就是((a-1)*b - a = ab-a-b​)

D1T3

orz GXZ:https://www.cnblogs.com/GXZlegend/p/7838900.html
(记忆化搜索?) https://blog.csdn.net/enjoy_pascal/article/details/78592786
70分做法显然

100分做法: 思想是把DP转移抽象为DAG
如果不是DAG: 只要可转移到终点的合法状态不在环内就可以正常dp
否则puts("-1")
p.s. 只要在topsort中环加外向树上的点(illegal points) 入度都大于0 , 所以topsort一遍就ok拉!!!

这道题卡常丧心病狂...
这个程序会re(60分)...不敢开longlong... 求dalao debug..

#include<bits/stdc++.h>
using namespace std;

typedef pair<int, int> pii;
#define FI first
#define SE second
#define PB push_back
#define rep(_i, _st, _ed) for(register int _i = (_st); _i <= (_ed); ++_i)
#define per(_i, _ed, _st) for(register int _i = (_ed); _i >= (_st); --_i)
inline int read(){int ans = 0, f = 1; char c = getchar();while(c < '0' || c > '9') f = (c == '-') ? -1 : f, c = getchar();while('0' <= c && c <= '9') ans = ans*10 + c - '0', c = getchar();return ans;}

const int maxn = 1e5+5, maxm = 3e5+5, mxk = 51;

int n, m, p;
//邻接表
struct graph{
    int v, w, nxt;
}edge[maxm];
int head[maxn]; 
void adde(int u, int v, int w){
    static int cnt = 0;
    edge[++cnt].v = v;
    edge[cnt].w = w;
    edge[cnt].nxt = head[u];
    head[u] = cnt;
}

//精简dijstra
priority_queue<pii> pq;
int dis[maxn]; bool vis[maxn];
void dijstra(){
    memset(dis, 0x3f, sizeof dis);
    memset(vis, 0, sizeof vis);
    dis[1] = 0; 
    pq.push(make_pair(0, 1));
    while(!pq.empty()){
        int u = pq.top().second; pq.pop();
        if(vis[u]) continue;
        vis[u] = 1;
        for(int e = head[u]; e; e = edge[e].nxt){
            int v = edge[e].v, w = edge[e].w;
            if(dis[u] + w < dis[v]) {
                dis[v] = dis[u] + w;
                pq.push(make_pair(-dis[v], v));
            }
        }
    }
}

int T, k; 

const int mxhsh = maxn*mxk;
int f[mxhsh], ind[mxhsh], le[mxhsh], ri[mxhsh], cnt, pt[mxhsh], q[mxhsh];
#define hsh(_k, _u) ((_k)*n + _u)
#define mod(_cur) if(_cur > p) _cur -= p;

signed main(){
    T = read();
    while(T--){
        n = read(), m = read(), k = read(), p = read();
        memset(head, 0, sizeof head), memset(edge, 0, sizeof edge);
        rep(i, 1, m){
            int u = read(), v = read(), w = read();
            adde(u, v, w);
        }
		dijstra();
        //构建状态转移图 (该题中为一种扩展最短路图)
        cnt = 1; memset(ind, 0, sizeof ind);
        rep(u, 1, n) rep(x, 0, k){
            le[hsh(x, u)] = cnt;
            for(int e = head[u]; e; e = edge[e].nxt){
                int v = edge[e].v, w = edge[e].w, x2;
                x2 = dis[u] + w - dis[v] + x;
                if(x2 > k || x2 < 0) continue;
                
                ++ind[hsh(x2, v)];     
				//printf("u=%d v=%d w=%d, fr = %d to = %d, ind = %d
",u, v, w,  hsh(x, u), hsh(x2, v), ind[hsh(x2, v)]);
                            
                pt[cnt++] = hsh(x2, v);
            }
            ri[hsh(x, u)] = cnt - 1;
        }

        //拓扑排序 + 状态转移
        memset(f, 0, sizeof f);
        f[hsh(0, 1)] = 1;

        int fr = 1, bk = 0;
        rep(i, 1, hsh(k, n)) if(!ind[i]) q[++bk] = i;//这句必须有！！常数再大也要加！！！
        
        while(bk >= fr){
            int u = q[fr++];
            rep(i, le[u], ri[u]){
                int v = pt[i];
                f[v] += f[u]; mod(f[v]);

                if( (--ind[v]) == 0) q[++bk] = v;
            }
        }

        //统计答案
        int infini = 0, ans = 0;
        rep(x, 0, k){
            if(ind[hsh(x, n)]) infini = 1;
            ans += f[hsh(x, n)]; mod(ans);
        }
        if(infini) puts("-1");
        else printf("%d
", ans);
    }
    return 0;
}

D2T2

正解(O(3^nn^2))....
注释很详细了

//这题卡常!!!
#include<bits/stdc++.h>
using namespace std;

typedef pair<int, int> pii;
#define FI first
#define SE second
#define PB push_back
#define rep(_i, _st, _ed) for(register int _i = (_st); _i <= (_ed); ++_i)
#define per(_i, _ed, _st) for(register int _i = (_ed); _i >= (_st); --_i)
inline int read(){int ans = 0, f = 1; char c = getchar();while(c < '0' || c > '9') f = (c == '-') ? -1 : f, c = getchar();while('0' <= c && c <= '9') ans = ans*10 + c - '0', c = getchar();return ans;}

#define min(a, b) ((a) < (b)) ? (a) : (b) //注意在这个宏定义中a, b都会算两次
int f[13][10005];
int n, m, mat[15][15], log_2[10005], mincost[15];
signed main(){
    n = read(), m = read();
    rep(i, 0, n) rep(j, 0, n) mat[i][j] = 1e7;
    rep(i, 1, m){
        int u = read(), v = read(), w = read();
        u--, v--;
        mat[v][u] = mat[u][v] = min(mat[u][v], w);
    }
    
    log_2[0] = 1;
    rep(i, 1, n) log_2[(1 << i)] = i;

    int mxsta = (1 << n) - 1, ans = 1e8;
    memset(f, 0x2f, sizeof f);
    rep(u, 0, n-1) f[0][(1<<u)] = 0;//这一步非常机智, 使得程序无需枚举起始点

    rep(l, 1, n) rep(sta, 0, mxsta){
        int rev = mxsta ^ sta;

        //求补集中每个节点向原图连边的mincost
        for(int cur = rev; cur; cur -= cur&(-cur)){
            int d = log_2[cur&(-cur)];
            mincost[d] = 1e8;
            rep(j, 0, n - 1)
                //当前集合中存在j
                if(sta & (1 << j))  
                    mincost[d] = min(mincost[d], mat[d][j] * l);
        }
        //不重复枚举补集的子集
        for(int sub = rev; sub; sub = (sub-1) & rev){
            int cost = 0;//将该子集连入树中的花费(dep == l)
            
            //枚举补集的子集中的每一个元素
            for(int cur = sub; cur; cur -= cur&(-cur)){
                int d = log_2[cur&(-cur)];
                //当前元素
                cost += mincost[d];
            }
            //cout<<cost<<endl;
            f[l] [sta | sub] = min( f[l] [sta | sub], f[l-1][sta] + cost);
        }
    }


    rep(i, 0, n){
        ans = min(ans, f[i][mxsta]);
        //printf("u=%lld i=%lld ans = %lld
", u, i, f[i][mxsta]);            
    }
    cout<<ans<<endl;
    return 0;
}