Codeforces Round #287 (Div. 2) E. Breaking Good（最短路、dp）

题意:

N≤105个城市,M≤105条边,保证无向图连通且无自环和重边

边分2种,0代表毁坏的,1代表正常的

现要1→n的最短路,且影响值最小,影响值:=最短路上的毁坏边数+不在最短路上的正常边数

输出影响值以及影响的边,状态为新的状态

分析:

设所有的正常边为ALL,最短路为D,最短路上的正常边为X

ans=D−X+ALL−X=D+ALL−2X,由于D和ALL都是常数,我们现在要X最大才能使ans最小

由于边权是1,最短路bfs就可以了,同时dp一下,g[i]:=以i结尾的最短路上最多的正常边数

记录下转移,方便打印路径即可

时间复杂度为O(n+m)

代码:

//
//  Created by TaoSama on 2016-02-13
//  Copyright (c) 2016 TaoSama. All rights reserved.
//
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>

using namespace std;
#define pr(x) cout << #x << " = " << x << "  "
#define prln(x) cout << #x << " = " << x << endl
const int N = 1e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;

int n, m;
struct Edge {
int v, nxt, mark;
} edge[N << 1];
int head
, cnt;

void addEdge(int u, int v, int mark) {
edge[cnt] = (Edge) {v, head[u], mark};
head[u] = cnt++;
}

int f
, g
, pre
;

void bfs() {
queue<int> q; q.push(1);
memset(f, 0x3f, sizeof f);
memset(g, 0, sizeof g);
f[1] = g[1] = 0;
while(q.size()) {
int u = q.front(); q.pop();
for(int i = head[u]; ~i; i = edge[i].nxt) {
int v = edge[i].v, mark = edge[i].mark;
if(f[v] == INF) {
f[v] = f[u] + 1;
g[v] = g[u] + mark;
pre[v] = i;
q.push(v);
} else if(f[v] == f[u] + 1) {
if(g[u] + mark > g[v]) {
g[v] = g[u] + mark;
pre[v] = i;
}
}
}
}
}

int main() {
#ifdef LOCAL
freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);
//  freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);
#endif
ios_base::sync_with_stdio(0);

while(scanf("%d%d", &n, &m) == 2) {
cnt = 0; memset(head, -1, sizeof head);
int all = 0;
for(int i = 1; i <= m; ++i) {
int u, v, mark; scanf("%d%d%d", &u, &v, &mark);
all += mark;
addEdge(u, v, mark);
addEdge(v, u, mark);
}
bfs();
for(int u = n, id; u != 1; u = edge[id ^ 1].v) {
id = pre[u];
if(edge[id].mark == 0)
edge[id].mark = edge[id ^ 1].mark = 2; //to repair
else
edge[id].mark = edge[id ^ 1].mark = 0; //ignore
}
printf("%d\n", f
+ all - 2 * g
);
for(int i = 0; i < cnt; i += 2) {
if(!edge[i].mark) continue;
printf("%d %d %d\n", edge[i ^ 1].v, edge[i].v, edge[i].mark - 1);
}
}
return 0;
}