洛谷P2680 运输计划

大概就是二分+树上差分...

题意：给你树上m条路径，你要把一条边权变为0，使最长的路径最短。

最大的最小，看出二分(事实上我并没有看出来...)

然后二分k，对于所有大于k的边，树上差分求出最长公共边，然后看是否可以。

(yy的解法②：边按照长度排序，然后二分。删除最长公共边。据logeadd juru说是三分)

代码量3.6k，180行，还是有点长的。

#include <cstdio>
#include <algorithm>
#include <cstring>
const int N = 300010;

inline void read(int &x) {
    char c = getchar();
    x = 0;
    while(c > '9' || c < '0') {
        c = getchar();
    }
    while(c <= '9' && c >= '0') {
        x = (x << 3) + (x << 1) + c - 48;
        c = getchar();
    }
    return;
}

struct Edge {
    int v, nex, len;
}edge[N << 1]; int top;

int e[N], n, m, lm, fa[N][20], d[N], lenth[N]; /// 点
int l[N], r[N], mid[N], len[N]; /// 路径
bool use[N]; /// 树上差分
int num, large, R, f[N];

inline void add(int x, int y, int z) {
    edge[++top].v = y;
    edge[top].len = z;
    edge[top].nex = e[x];
    e[x] = top;
    return;
}

inline void DFS1(int x, int f) {
    fa[x][0] = f;
    for(int i = e[x]; i; i = edge[i].nex) {
        int y = edge[i].v;
        if(y != f) {
            lenth[y] = lenth[x] + edge[i].len;
            d[y] = d[x] + 1;
            DFS1(y, x);
        }
    }
    return;
}

inline void getlca() {
    while((1 << lm) < n) {
        lm++;
    }
    DFS1(1, 0);
    for(int i = 1; i <= lm; i++) {
        for(int x = 1; x <= n; x++) {
            fa[x][i] = fa[fa[x][i - 1]][i - 1];
        }
    }
    return;
}

inline int lca(int x, int y) {
    if(d[x] > d[y]) {
        std::swap(x, y);
    }
    int t = lm;
    while(t > -1 && d[y] > d[x]) {
        if(d[fa[y][t]] >= d[x]) {
            y = fa[y][t];
        }
        t--;
    }
    if(x == y) {
        return x;
    }
    t = lm;
    while(t > -1 && fa[x][0] != fa[y][0]) {
        if(fa[x][t] != fa[y][t]) {
            x = fa[x][t];
            y = fa[y][t];
        }
        t--;
    }
    return fa[x][0];
}

inline int DFS(int x) {
    int cnt = 0;
    for(int i = e[x]; i; i = edge[i].nex) {
        int y = edge[i].v;
        if(y == fa[x][0]) {
            continue;
        }
        int temp = DFS(y);
        cnt += temp;
        if(temp == num) {
            large = std::max(large, edge[i].len);
        }
    }
    cnt += f[x];
    return cnt;
}

inline bool check(int k) {
    num = 0;
    memset(f, 0, sizeof(f));
    for(int i = 1; i <= m; i++) {
        bool t = len[i] > k;
        use[i] = t;
        num += t;
        if(t) {
            f[l[i]]++;
            f[r[i]]++;
            f[mid[i]] -= 2;
        }
    }
    large = 0;
    DFS(1);
    return R - large <= k;
}

inline int getlong(int i) {
    int x = l[i];
    int ans = 0;
    while(x != mid[i]) {
        ans = std::max(ans, lenth[x] - lenth[fa[x][0]]);
        x = fa[x][0];
    }
    x = r[i];
    while(x != mid[i]) {
        ans = std::max(ans, lenth[x] - lenth[fa[x][0]]);
        x = fa[x][0];
    }
    return ans;
}

int main() {
    scanf("%d%d", &n, &m);
    int x, y, z;
    for(int i = 1; i < n; i++) {
        //scanf("%d%d%d", &x, &y, &z);
        read(x);
        read(y);
        read(z);
        add(x, y, z);
        add(y, x, z);
    }
    getlca();
    int dr = 0, dl = 0, dm, A = 1;
    for(int i = 1; i <= m; i++) {
        //scanf("%d%d", &l[i], &r[i]);
        read(l[i]);
        read(r[i]);
        mid[i] = lca(l[i], r[i]);
        len[i] = lenth[l[i]] + lenth[r[i]] - 2 * lenth[mid[i]];
        if(len[i] > dr) {
            dr = len[i];
            A = i;
        }
    }
    R = dr;
    dl = dr - getlong(A);
    if(dl < 0) {
        printf("ERROR ");
    }
    while(dl < dr) {
        dm = (dr + dl) / 2;
        if(check(dm)) {
            //printf("check %d 1 \n", dm);
            dr = dm;
        }
        else {
            //printf("check %d 0 \n", dm);
            dl = dm + 1;
        }
    }
    printf("%d", dr);
    return 0;
}